Bhagya

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.4

Question 1.
Find the slope, x-intercept, y-intercept of each of the following lines, i. 2x + 3y-6 = 0 ii. 3x-y-9 = 0 iii. x + 2y = 0
Solution:
i. Given equation of the line is 2x + 3y – 6 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 2, b = 3, c = -6

ii. Given equation of the line is 3x – y – 9 = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 3, b = – 1, c = – 9

iii. Given equation of the line is x + 2y = 0.
Comparing this equation with ax + by + c = 0,
we get
a = 1, b = 2, c = 0

Question 2.
Write each of the following equations in ax + by + c = 0 form.
i. y = 2x – 4
ii. y = 4
iii. \(\frac{x}{2}+\frac{y}{4}=1\)
iv. \(\frac{x}{3}-\frac{y}{2}=0\)
i. y = 2x – 4
∴ 2x – y – 4 = 0 is the equation in ax + by + c = 0 form.

ii. y = 4
∴ 0x + 1y – 4 = 0 is the equation in ax + by + c = 0 form.

iii. \(\frac{x}{2}+\frac{y}{4}=1\)
∴ \(\frac{2 x+y}{4}\)
∴ 2x + y – 4 = 0 is the equation in ax + by + c = 0 form.

iv. \(\frac{x}{3}-\frac{y}{2}=0\)
∴ 2x – 3y = 0
∴ 2x – 3y + 0 = 0 is the equation in ax + by + c = 0 form.
[Note: Answer given in the textbook is ‘2x – 3y – 6 = 0’. However, as per our calculation it is ‘2x-3y + 0 = 0’.]

Question 3.
Show that the lines x – 2y – 7 = 0 and 2x – 4y + 15 = 0 are parallel to each other.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x – 4y + 15 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{-4}=\frac{1}{2}\)
Since m₁ = m₂
the given lines are parallel to each other.

Question 4.
Show that the lines x – 2y – 7 = 0 and 2x + y + 1 = 0 are perpendicular to each other. Find their point of intersection.
Solution:
Let m₁ be the slope of the line x – 2y – 7 = 0.
∴ m₁ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}\)
Let m₂ be the slope of the line 2x + y + 1 = 0.
∴ m₂ = \(\frac{-\text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2\)
Since m₁ x m₁ = \(\frac{1}{2}\) x (- 2) = -1,
the given lines are perpendicular to each other. Consider,
x – 2y – 7 = 0 …(i)
2x + y + 1 =0 …(ii)
Multiplying equation (ii) by 2, we get
4x + 2y + 2 = 0 …(iii)
Adding equations (i) and (iii), we get
5x – 5 = 0
∴ x = 1
Substituting x = 1 in equation (ii), we get
2 + y + 1 = 0
∴ y = – 3
∴ The point of intersection of the given lines is (1,-3).

Question 5.
If the line 3x + 4y = p makes a triangle of area 24 square units with the co-ordinate axes, then find the value of p.
Solution:
Let the line 3x + 4y = p cuts the X and Y axes at points A and B respectively.
3x + 4y = p

This equation is of the form \(\frac{x}{a}+\frac{y}{b}=1\),
where a = \(\frac{p}{3}\) and b = \(\frac{p}{4}\)
∴ A (a, 0) ≡ (\(\frac{p}{3}\), 0) and B ≡ (0, b) = (0, \(\frac{p}{4}\))
∴ OA = \(\frac{p}{3}\) and OB = \(\frac{p}{4}\)
Given, A (∆OAB) = 24 sq. units
∴ \(\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24\)
∴ \(\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24\)
∴ p² = 576
∴ p = ± 24

Question 6.
Find the co-ordinates of the foot of the perpendicular drawn from the point A(- 2,3) to the line 3x-y -1 = 0.
Solution:
Let M be the foot of perpendicular drawn from
point A(- 2,3) to the line
3x-y- 1 = 0 …(i)
Slope of the line 3x-y – 1 = 0 is \(\frac{-3}{-1}\) =3.
Since AM ⊥ to line (i),
slope of AM = \(\frac{-1}{3}\)
∴ Equation of AM is
y – 3 = \(\frac{-1}{3}\)(x + 2)

∴ 3(y – 3) = – 1(x + 2)
∴ 3y – 9 = -x – 2
∴ x + 3y – 7 = 0 …………(ii)
The foot of perpendicular i.e., point M, is the point of intersection of equations (i) and (ii).
By (i) x 3 + (ii), we get 10x -10 = 0
∴ x = 1
Substituting x = 1 in (ii), we get
1 + 3y – 7 = 0
∴ 3y = 6
∴ y = 2
∴ The co-ordinates of the foot of the perpendicular Mare (1,2).

Question 7.
Find the co-ordinates of the circumcentre of the triangle whose vertices are A(- 2, 3), B(6, -1), C(4,3),e
Solution:
Here, A(-2, 3), B(6, -1), C(4, 3) are the vertices of ∆ABC.
Let F be the circumcentre of AABC.
Let FD and FE be the perpendicular bisectors of the sides BC and AC respectively.
∴ D and E are the midpoints of side BC and AC respectively.

Since FD passes through (5, 1) and has slope 1/2 equation of FD is
y – 1 = \(\frac{1}{2}\)(x-5)
∴ 2 (y – 1) = x – 5
∴ 2y – 2 = x – 5
∴ x – 2y – 3 = 0 …(i)
Since both the points A and C have same y co-ordinates i.e. 3,
the given points lie on the line y = 3.
Since the equation FE passes through E(1, 3),
the equation of FE is x = 1. .. .(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value of x in (i), we get
1 – 2y -3 = 0
∴ y = -1
∴ Co-ordinates of circumcentre F ≡ (1, – 1).

Question 8.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(3, – 2), B(7,6), C (-1,2).
Solution:
Let O be the orthocentre of ∆ABC.
Let AD and BE be the altitudes on the sides BC and AC respectively.

Slope of side BC = \(\frac{2-6}{-1-7}=\frac{-4}{-8}=\frac{1}{2}\)
∴ Slope of AD = – 2 [∵ AD ⊥ BC]
∴ Equation of line AD is
y – (-2) = (- 2) (x – 3)
∴ y + 2 = -2x + 6
∴ 2x + y -4 = 0 …(i)
Slope of side AC = \(\frac{-2-2}{3-(-1)}=\frac{-4}{4}\) = -1
∴ Slope of BE = 1 …[ ∵ BE ⊥ AC]
∴ Equation of line BE is
y – 6 = 1(x – 7)
∴ y – 6 = x – 1
∴ x = y + 1 …(ii)
Substituting x = y + 1 in (i), we get
2(y + 1) + y – 4 = 0
∴ 2y + 2 + y – 4 = 0
∴ 3y – 2 = 0
∴ y = \(\frac{2}{3} in (ii), we get
Substituting y = [latex]\frac{2}{3}\) in (ii), we get
x = \(\frac{2}{3}+1=\frac{5}{3}\)
∴ Co-ordinates of orthocentre, O = \(\left(\frac{5}{3}, \frac{2}{3}\right)\)

Question 9.
Show that the lines 3 – 4y + 5 = 0, lx – 8y + 5 = 0 and 4JC + 5y – 45 = 0 are concurrent. Find their point of concurrence.
Solution:
The number of lines intersecting at a point are called concurrent lines and their point of intersection is called the point of concurrence. Equations of the given lines are
3x – 4y + 5 = 0 …(i)
7x-8y + 5 = 0 …(ii)
4x + 5y – 45 = 0 …(iii)
By (i) x 2 – (ii), we get
– x + 5 = 0
∴ x = 5
Substituting x = 5 in (i), we get
3(5) – 4y + 5 = 0
∴ -4y = – 20
∴ y = 5
∴ The point of intersection of lines (i) and (ii) is given by (5, 5).
Substituting x = 5 and y = 5 in L.H.S. of (iii), we get
L.H.S. = 4(5) + 5(5) – 45
= 20 + 25 – 45
= 0
= R.H.S.
∴ Line (iii) also passes through (5, 5).
Hence, the given three lines are concurrent and the point of concurrence is (5, 5).

Question 10.
Find the equation of the line whose x-intercept is 3 and which ¡s perpendicular to the line 3x – y + 23 = 0.
Solution:
Slope of the line 3x – y + 23 = 0 is 3.
∴ Slope of the required line perpendicular to
3x – y + 23 = 0 is \(\frac{-1}{3}\)
Since the x-intercept of the required line is 3, it passes through (3, 0).
∴ The equation of the required line is ‘
y – 0 = \(\frac{-1}{3}\)(x – 3)
∴ 3y = x + 3
∴ x + 3y = 3

Question 11.
Find the distance of the origin from the line 7x + 24y – 50 = 0.
Solution:
Let p be the perpendicular distance of origin
fromtheline7x + 24y – 50 = 0
Here, a = 7, b = 24, c = -50

Question 12.
Find the distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0.
Solution:
Let p be the perpendicular distance of the point A(- 2, 3) from the line 12x – 5y – 13 = 0
Here, a = 12, b = – 5, c = – 13, x₁ = -2, y₁ = 3

Question 13.
Find the distance between parallel lines 4x – 3y + 5 = 0 and 4xr – 3y + 7 = 0.
Solution:
Equations of the given parallel lines are 4x – 3y + 5 = 0 and 4x – 3y + 1 = 0
Here, a = 4, b = – 3, c₁ = 5 and c₂ = 7
∴ Distance between the parallel lines

Question 14.
Find the distance between the parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.
Solution:
Equations of the given parallel lines are 3x + 2y + 6 = 0 and
9x + 6y – 1 = 0 i.e., 3x + 2y – \(\frac{7}{3}\) =0
Here, a = 3, b = 2, c₁ = 6 and c₂ = \(\frac{-7}{3}\)
∴ Distance between the parallel lines

Question 15.
Find the points on the line x + y – 4 = 0 which are at a unit distance from the line 4JC + 3y = 10.
Solution:
Let P(x₁, y₁) be a point on the line x + y – 4 = o.
∴ x₁ + y₁ – 4 = 0
∴ y₁ = 4 – x₁ …(i)
Also, distance of P from the line 4x + 3y- 10 = 0 is 1

∴ 5 = | x₁ + 2 |
∴ x₁ + 2 = ± 5
∴ x₁ + 2 = 5 or x₁ + 2 = – 5
∴ x₁ = 3 or x₁ = – 7
From (i), when x₁ = 3, y₁ = 1
and when x₁ = -7, y₁ = 11
∴ The required points are (3, 1) and (-7, 11).
[Note: The question has been modified]

Question 16.
Find the equation of the line parallel to the X-axis and passing through the point of intersection of lines x + y – 2 = 0 and 4x + 3y = 10.
Solution:
Let u = x + y – 2 = 0 and v = 4x + 3y – 10 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)
∴ x + y – 2 + 4kx + 3ky – 10k = 0
∴ x + 4kx + y + 3ky – 2 – 10k = 0
∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0
But, this line is parallel to X-axis.
∴ Its slope = 0
∴ \(\frac{-(1+4 k)}{1+3 k}\) = 0
∴ 1 + 4k = 0
∴ k = \(\frac{-1}{4}\)
Substituting the value of k in (i), we get
(x + y – 2) + (4x + 3y – 10) = 0
∴ 4(x +y – 2) – (4x + 3y -10 ) = 0
∴ 4x + 4y – 8 – 4x – 3y + 10 = 0
∴ y + 2 = 0, which is the equation of the required line.
[Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

Question 17.
Find the equation of the line passing through the point of intersection of lines x + y – 2 = 0 and 2xr – 3y + 4 = 0 and making intercept 3 on the X-axis.
Solution:
Let u ≡ x + y – 2 = 0 and v ≡ 2x – 3y + 4 = 0
Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.
∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)
But, x-intercept of line is 3.
∴ It passes through (3, 0).
Substituting x = 3 and y = 0 in (i), we get
(3 + 0 – 2) + k(6 – 0 + 4) = 0
∴ 1 + 10k = 0
k = \(\frac{-1}{10}\)
Substituting the value of k in (i), we get (x + y – 2) + \(\left(\frac{-1}{10}\right)\) (2x – 3y + 4) = 0
∴ 10(x + y – 2) – (2x – 3y + 4) = 0
∴ 10x + 10y -20 — 2x + 3y-4 = 0
∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Question 18.
If A(4, 3), B(0, 0) and C(2, 3) are the vertices of ΔABC, then find the equation of bisector of angle BAC.
Solution:
Let the bisector of ∠ BAC meets BC at point D.
∴ Point D divides seg BC in the ratio l(AB) : l(AC)

∴ 18 (y – 3) = 6 (x – 4)
∴ 3(y – 3) = x – 4
∴ 3y – 9 = x – 4
∴ x – 3y + 5 = 0

Question 19.
D(- 1, 8), E(4, – 2), F(- 5, – 3) are midpoints of sides BC, CA and AB of AABC. Find
i. equations of sides of ΔABC.
ii. co-ordinates of the circumcentre of ΔABC.
Solution:
Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of ΔABC.
Given, points D, E and F are midpoints of sides BC, CA and AB respectively of ΔABC.

∴ x₁ + x₂ = -10 …………. (v)
and y₁ + y₂ = – 6 …………(vi)
For x-coordinates:
Adding (i), (iii) and (v), we get
2x₁ + 2x₂ + 2x₃ = – 4
∴ x₁ + x₂ + x₃ = -2 …………..(vii)
Solving (i) and (vii), we get x₁ = 0
Solving (iii) and (vii), we get x₂ = – 10
Solving (v) and (vii), we get x₃ = 8

For y-coordinates:
Adding (ii), (iv) and (vi), we get 2y₁ + 2y₂ + 2y₃ = 6
y₁ + y₂ + y₃ = 3 …….(viii)
Solving (ii) and (viii), we get y₁ = -13
Solving (iv) and (viii), we get y₂ = 7
Solving (vi) and (viii), we get y₃ = 9
∴ Vertices of AABC are A(0, – 13), B(- 10, 7), C(8, 9)

∴ 8(y + 13) = 22x
∴ 4(y + 13) = 11x
∴ 11x – 4y – 52 = 0

ii. Here, A(0, – 13), B(- 10, 7), C(8, 9) are the vertices of ΔABC.
Let F be the circumcentre of AABC.
Let FD and FE be perpendicular bisectors of the sides BC and AC respectively.
D and E are the midpoints of side BC and AC.

∴ Slope of FD = -9 … [ ∵ FD ⊥ BC]
Since FD passes through (-1, 8) and has slope -9, equation of FD is
y – 8 = -9 (x +1)
∴ y – 8 = -9x – 9
∴ y = -9x – 1
Also, slope of AC = \(\frac{-13-9}{0-8}=\frac{11}{4}\)
∴ Slope of FE = \(\frac{-4}{11}\) [ ∵ FE ⊥ AC]
Since FE passes through (4, -2) and has slope -4
\(\frac{-4}{11}\), equation of FE is
(y + 2) = \(\frac{-4}{11}\) (x – 4)
∴ 11(y + 2) = -4(x – 4)
∴ 11y + 22 = – 4x + 16
∴ 4x + 11y = -6 …………(ii)
To find co-ordinates of circumcentre, we have to solve equations (i) and (ii).
Substituting the value ofy in (ii), we get
4x + 11(-9x- 1) = – 6
∴ 4x – 99x -11 = – 6
∴ -95x = 5
∴ x = \(\frac{-1}{19}\)
Substituting the value of x in (i), we get
y = -9(\(\frac{-1}{19}\)) – 1 = \(\frac{-10}{19}\)
∴ Co-ordinates of circumcentre F ≡ \(\left(\frac{-1}{19}, \frac{-10}{19}\right)\)

Question 20.
0(0, 0), A(6, 0) and B(0, 8) are vertices of a triangle. Find the co-ordinates of the incentre of ∆OAB.
Solution:
Let bisector of ∠O meet AB at point D and bisector of ∠A meet BO at point E
∴ Point D divides seg AB in the ratio l(OA): l(OB)
and point E divides seg BO in the ratio l(AB): l(AO)
Let I be the incentre of ∠OAB.

∴ Point D divides AB internally in 6 : 8
i.e. 3 :4

∴ y = x …(i)
Now, by distance formula,
l(AB) = \(\begin{aligned}
&=\sqrt{(6-0)^{2}+(0-8)^{2}} \\
&=\sqrt{36+64}=10
\end{aligned}\)
l(AO) = \(\sqrt{(6-0)^{2}+(0-0)^{2}}\) = 6
∴ Point E divides BO internally in 10 : 6 i.e. 5:3

∴ -2y = x – 6
∴ x + 2y = 6 …(ii)
To find co-ordinates of incentre, we have to solve equations (i) and (ii).
Substituting y = x in (ii), we get
x + 2x = 6
∴ x = 2
Substituting the value of x in (i), we get
y = 2
∴ Co-ordinates of incentre I ≡ (2, 2).

Alternate Method:
Let I be the incentre.
I lies in the 1st quadrant.
OPIR is a square having side length r.
Since OA = 6, OP = r,
PA = 6 – r
Since PA = AQ,
AQ = 6 – r …(i)
Since OB = 8, OR = r,
BR = 8 – r
∴ BR = BQ
∴ BQ = 8 – r …(ii)

AB = BQ + AQ
Also, AB = \(\begin{aligned}
&=\sqrt{\mathrm{OA}^{2}+\mathrm{OB}^{2}} \\
&=\sqrt{6^{2}+8^{2}} \\
&=\sqrt{100}=10
\end{aligned}\)
∴ BQ + AQ= 10
∴ (8 – r) + (6 – r) = 10
∴ 2r = 14- 10 = 4
∴ r = 2
∴ I = (2,2)

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.3

Question 1.
Write the equation of the line:
i. parallel to the X-axis and at a distance of 5 units from it and above it.
ii. parallel to the Y-axis and at a distance of 5 units from it and to the left of it.
iii. parallel to the X-axis and at a distance of 4 units from the point (- 2,3).
Solution:
i. Equation of a line parallel to X-axis is y = k. Since the line is at a distance of 5 units above the X-axis, k = 5
∴ The equation of the required line is y = 5.

ii. Equation of a line parallel to the Y-axis is x = h. Since the line is at a distance of 5 units to the left of the Y-axis, h = -5
∴ The equation of the required line is x = -5.
[Note: Answer given in the textbook is ‘y = -5
However, we found that ‘x = – 5’.]

iii. Equation of a line parallel to the X-axis is of the form y = k (k > 0 or k < 0).
Since the line is at a distance of 4 units from the point (- 2, 3),
k = 4 + 3 = 7 or k = 3- 4 = -1
∴ The equation of the required line is y = 1 or y = – 1.

Question 2.
Obtain the equation of the line:
i. parallel to the X-axis and making an intercept of 3 units on the Y-axis.
ii. parallel to the Y-axis and making an intercept of 4 units on the X-axis.
Solution:
i. Equation of a line parallel to X-axis with y-intercept ‘k’ isy = k.
Here, y-intercept = 3
∴ The equation of the required line is y = 3.

ii. Equation of a line parallel to Y-axis with x-intercept ‘h’ is x = h.
Here, x-intercept = 4
∴ The equation of the required line is x = 4.

Question 3.
Obtain the equation of the line containing the point:
i. A(2, – 3) and parallel to the Y-axis.
ii. B(4, – 3) and parallel to the X-axis.
Solution:
i. Equation of a line parallel to Y-axis is of the form x = h.
Since the line passes through A(2, – 3), h = 2
∴ The equation of the required line is x = 2.

ii. Equation of a line parallel to X-axis is of the formy = k.
Since the line passes through B(4, – 3), k = -3
∴ The equation of the required line is y = – 3.

Question 4.
Find the equation of the line:
i. passing through the points A(2, 0) and B(3,4)
ii. passing through the points P(2, 1) and Q(2,-1)
Solution:
i. The required line passes through the points A(2, 0) and B(3,4).
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
Here, (x₁y₁) = (2,0) and (x₁,y₂) = (3,4)
∴ The equation of the required line is
∴ \(\frac{y-0}{4-0}=\frac{x-2}{3-2}\)
∴ \(\frac{y}{4}=\frac{x-2}{1}\)
∴ y = 4(x – 2)
∴ y = 4x – 8
∴ 4x – y – 8 = 0

ii. The required line passes through the points P(2, 1) and Q(2,-1).
Since both the given points have same
x co-ordinates i.e. 2,
the given points lie on the line x = 2.
∴ The equation of the required line is x = 2.

Question 5.
Find the equation of the line:
i. containing the origin and having inclination 60°.
ii. passing through the origin and parallel to AB, where A is (2,4) and B is (1,7).
iii. having slope 1/2 and containing the point (3, -2)
iv. containing the point A(3, 5) and having slope 2/3
v. containing the point A(4, 3) and having inclination 120°.
vi. passing through the origin and which bisects the portion of the line 3JC + y = 6 intercepted between the co-ordinate axes.
Solution:
i. Given, Inclination of line = θ = 60°
Slope of the line (m) = tan θ = tan 60°
= \(\sqrt{3}\)
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
.‘. The equation of the required line is y = \(\sqrt{3}\) x

ii. Given, A (2, 4) and B (1, 7)
Slope of AB = \(\frac{7-4}{1-2}\) = -3 1-2
Since the required line is parallel to line AB, slope of required line (m) = slope of AB
∴ m = – 3 and the required line passes through the origin.
Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is y = – 3x

iii. Given, slope(m) = \(=\frac{1}{2}\) and the line passes through (3, – 2).
Equation of the line in slope point form is
y-y ₁= m(x-x₁)
∴ The equation of the required line is
[y-(- 2)]=\(\frac{1}{2}\)(x-3)
∴ 2(y + 2)=x – 3
∴ 2y + 4 = x – 3
∴ x – 2y – 7 = 0

iv. Given, slope(m) = \(\frac{2}{3}\) and the line passes through (3, 5).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is y – 5 = \(\frac{2}{3}\)(x-3)
∴ 3 (y – 5) = 2 (x – 3)
∴ 3y – 15 = 2x – 6
∴ 2x – 3y + 9 = 0

v. Given, Inclination of line = θ = 120°
Slope of the line (m) = tan θ = tan 120°
= tan (90° + 30°)
= – cot 30°
= – \(\sqrt{3}\)
and the line passes through A(4, 3).
Equation of the line in slope point form is y-y₁ = m(x -x₁)
∴ The equation of the required line is
y- 3 = –\(\sqrt{3}\)(x-4)
∴ y – 3 = –\(\sqrt{3}\) x + 4\(\sqrt{3}\)
∴ \(\sqrt{3}\)x + y – 3 -4\(\sqrt{3}\) = 0

vi.

Given equation of the line is 3x +y = 6.
∴ \(\frac{x}{2}+\frac{y}{6}=1\)
This equation is of the form \(\frac{x}{\mathrm{a}}+\frac{y}{\mathrm{~b}}\) = 1,
where a = 2, b = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2, 0) and B(0, 6) respectively. Required line is passing through the midpoint of AB.
∴ Midpoint of AB = ( \(\frac{2+0}{2}, \frac{0+6}{2}\) ) = (1,3)
∴ Required line passes through (0, 0) and (1,3).
Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-0}{3-0}=\frac{x-0}{1-0}\)
\(\frac{y}{3}=\frac{x}{1}\)
∴ y = 3x
∴ 3x – y = 0

Alternate Method:
Given equation of the line is 3x + y = 6 …(i)
Substitute y = 0 in (i) to get a point on X-axis.
∴ 3x + 0 = 6
∴ x = 2
Substitute x = 0 in (i) to get a point on Y-axis.
∴ 3(0) + 7 = 6
∴ y = 6
∴ The line 3x + y = 6 intersects the X-axis and Y-axis at A(2,0) and B(0,6) respectively.
Let M be the midpoint of AB.
M = \(\left(\frac{2+0}{2}, \frac{0+6}{2}\right)\) = (1,3)
Slope of OM (m) = \(\frac{3-0}{1-0}\) = 3
Equation of OM is of the formy = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 6.
Line y = mx + c passes through the points A(2,1) and B(3,2). Determine m and c.
Solution:
Given, A(2, 1) and B(3,2)
Equation of the line in two point form is \(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the required line is
\(\frac{y-1}{2-1}=\frac{x-2}{3-2}\)
∴ \(\frac{y-1}{1}=\frac{x-2}{1}\)
∴ y – 1 = x – 2
∴ y = x – 1
Comparing this equation with y = mx + c, we get
m = 1 and c = – 1

Alternate Method:
Points A(2, 1) and B(3, 2) lie on the line y = mx + c.
∴ They must satisfy the equation.
∴ 2m + c = 1 …(i)
and 3m + c = 2 …(ii)
equation (ii) – equation (i) gives m = 1
Substituting m = 1 in (i), we get 2(1) + c = 1
∴ c = 1 – 2 = – 1

Question 7.
Find the equation of the line having inclination 135° and making x-intercept 7.
Solution:
Given, Inclination of line = 0 = 135°
∴ Slope of the line (m) = tan 0 = tan 135°
= tan (90° + 45°)
= – cot 45° = – 1 x-intercept of the required line is 7.
∴ The line passes through (7, 0).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
∴ The equation of the required line is y — 0 = – 1 (x – 7)
∴ y = -x + 7
∴ x + y – 7 = 0

Question 8.
The vertices of a triangle are A(3, 4), B(2, 0) and C(- 1, 6). Find the equations of the lines containing
i. side BC
ii. the median AD
iii. the midpoints of sides AB and BC.
Solution:
Vertices of AABC are A(3, 4), B(2, 0) and C(- 1, 6).
i. Equation of the line in two point form is
\(\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{x-x_{1}}{x_{2}-x_{1}}\)
∴ The equation of the side BC is
\(\frac{y-0}{6-0}=\frac{x-2}{-1-2}\)
\(\frac{y}{6}=\frac{x-2}{-3}\)
∴ – 3y = 6x – 12
∴ 6x + 3y – 12 = 0
∴ 2x + y – 4 = 0

ii. Let D be the midpoint of side BC.
Then, AD is the median through A.
∴ D = \(\left(\frac{2-1}{2}, \frac{0+6}{2}\right)=\left(\frac{1}{2}, 3\right)\)
The median AD passes through the points
A(3,4) and D( \(\frac{1}{2}\) , 3)

∴ The equation of the median AD is
\(\frac{y-4}{3-4}=\frac{x-3}{\frac{1}{2}-3}\)
\(\frac{y-4}{-1}=\frac{x-3}{-\frac{5}{2}}\)
\(\frac{5}{2}\)(y-4) = x – 3
∴ 5y – 20 = 2x – 6
∴ 2x – 5y + 14 = 0

iii. Let D and E be the midpoints of side AB and side BC respectively.
The equation of the line DE is

∴ -4(y-2) = 2x-5
∴ 2x + 4y – 13 = 0

Question 9.
Find the x and y-intercepts of the following lines:
i. \(\frac{x}{3}+\frac{y}{2}=1\)
ii. \(\frac{3 x}{2}+\frac{2 y}{3}=1\)
iii. 2x – 3y + 12 = 0
Solution:
i. Given equation of the line is latex]\frac{x}{3}+\frac{y}{2}=1[/latex]
This is of the form \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = 3, y-intercept = 2

ii. Given equation of the line is \(\frac{3 x}{2}+\frac{2 y}{3}\) = 1
∴ \(\frac{x}{\left(\frac{2}{3}\right)}+\frac{y}{\left(\frac{3}{2}\right)}\) = 1
This is of the form = \(\frac{x}{a}+\frac{y}{b}\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = \(\frac{2}{3}\) and y-intercept = \(\frac{3}{2}\)

iii. Given equation of the line is 2x – 3y + 12 = 0
∴ 2x – 3y = – 12
∴ \(\frac{2 x}{(-12)}-\frac{3 y}{(-12)}=1\)
∴ \(\frac{x}{-6}+\frac{y}{4}=1\)
This is of the form \(\) = 1,
where x-intercept = a, y-intercept = b
∴ x-intercept = – 6 and y-intercept = 4

Question 10.
Find equations of the line which contains the point A(l, 3) and the sum of whose intercepts on the co-ordinate axes is zero.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be
\(\frac{x}{a}+\frac{y}{b}=1\) ………..(i)
Since, the sum of the intercepts of the line is zero.
∴ a + b = 0
∴ b = – a
Substituting b = – a in (i), we get
\(\frac{x}{a}+\frac{y}{(-a)}=1\)
x – y = a .. .(ii)
Since, the line passes through A(1, 3).
∴ 1 – 3 = a
∴ a = – 2
Substituting the value of a in (ii), equation of the required line is
∴ x – y = – 2,
∴ x – y + 2 = 0

Case II: Line passing through origin.
Slope of line passing through origin and
A(1, 3) is m = \(\frac{3-0}{1-0}\) = 3
∴ Equation of the line having slope m and passing through origin (0, 0) is / = mx.
∴ The equation of the required line is y = 3x
∴ 3x – y = 0

Question 11.
Find equations of the line containing the point A(3, 4) and making equal intercepts on the co-ordinate axes.
Solution:
Case I: Line not passing through origin.
Let the equation of the line be \(\frac{x}{a}+\frac{y}{b}=1\) …………(i)
This line passes through A(3, 4).
∴ \(\frac{3}{a}+\frac{4}{b}=1\)……………..(ii)
Since, the required line make equal intercepts on the co-ordinate axes.
∴ a = b …(iii)
Substituting the value of b in (ii), we get
\(\frac{3}{a}+\frac{4}{a}=1\)
∴ \(\frac{7}{a}=1\)
∴ a = 7
∴ b = 7 …[From (iii)]
Substituting the values of a and b in (i), equation of the required line is
\(\frac{x}{7}+\frac{y}{7}=1\) = 1
∴ x + y = 7

Case II: Line passing through origin.
Slope of line passing through origin and A(3,4) is m = \(=\frac{4-0}{3-0}=\frac{4}{3}\)
∴ Equation of the line having slope m and passing through origin (0, 0) is y = mx.
∴ The equation of the required line is 4
y = \(\frac{4}{3}\)x
∴ 4x – 3y = 0

Question 12.
Find the equations of the altitudes of the triangle whose vertices are A(2, 5), B(6, – 1 ) and C(- 4, – 3).
Solution:

A(2, 5), B(6, – 1), C(- 4, – 3) are the vertices of ∆ABC.
Let AD, BE and CF be the altitudes through the vertices A, B and C respectively of ∆ABC.
∴ Slope of AD = -5 …[∵AD ⊥ BC]
Since altitude AD passes through (2, 5) and has slope – 5,
equation of the altitude AD is y – 5 = -5 (x – 2)
∴ y – 5 = – 5x + 10
∴ 5x +y -15 = 0
Now, slope of AC = \(\frac{-3-5}{-4-2}=\frac{-8}{-6}=\frac{4}{3}\)
Slope of BE = \(\frac{-3}{4}\)
…[∵ BE ⊥ AC]
Since altitude BE passes through (6,-1) and has slope \(\frac{-3}{4}\),
equation of the altitude BE is
y-(-1) = \(\frac{-3}{4}\) (x – 6)
∴ 4 (y + 1) = – 3 (x – 6)
∴ 4y + 4 =-3x+ 18
∴ 3x + 4y – 14 = 0
Also, slope of AB = \(\frac{-1-5}{6-2}=\frac{-6}{4}=\frac{-3}{2}\)
∴ Slope of CF = \({2}{3}\) ….[∵ CF ⊥ AB]
Since altitude CF passes through (- 4, – 3) and has slope , \(\frac{2}{3}\)
equation of the altitude CF is
y-(-3) = \(\frac{2}{3}\)[x-(-4)]
∴ 3 (y + 3) = 2 (x + 4)
∴ 3y + 9 = 2x + 8
∴ 2x – 3y – 1 = 0

Question 13.
Find the equations of perpendicular bisectors of sides of the triangle whose vertices are P(-1, 8), Q(4, – 2) and R(- 5, – 3).
Solution:

Let A, B and C be the midpoints of sides PQ, QR and PR respectively of APQR.
A is the midpoint of side PQ.

Slope of perpendicular bisector of PQ is \(\frac{1}{2}\) and it passes through (\(\frac{3}{2}\)), 3).
Equation of the perpendicular bisector of side PQ is
y – 3 = \(\frac{1}{2}\)(x – \(\frac{3}{2}\))
y – 3 = (\(\frac{1}{2}\left(\frac{2 x-3}{2}\right)\))
∴ 4(y – 3) = 2x – 3
∴ 4y – 12 = 2x – 3
∴ 2x – 4y + 9 = 0
B is the midpoint of side QR
∴ B = \(\left(\frac{4-5}{2}, \frac{-2-3}{2}\right)=\left(\frac{-1}{2}, \frac{-5}{2}\right)\)
Slope of side QR = \(\frac{-3-(-2)}{-5-4}=\frac{-1}{-9}=\frac{1}{9}\)
∴ Slope of perpendicular bisector of QR is -9 and it passes through \(\left(-\frac{1}{2},-\frac{5}{2}\right)\)
∴ Equation of the perpendicular bisector of side QR is

∴ 2y + 5 = -18x – 9
∴ 18x + 2y + 14 = 0
∴ 9x + y + 7 = 0
C is the midpoint of side PR.

Equation of the perpendicular bisector of PR is \(y-\frac{5}{2}=-\frac{4}{11}(x+3)\)
∴ \(11\left(\frac{2 y-5}{2}\right)\) =-4(x + 3)
∴ 11(2y – 5) = – 8 (x + 3)
∴ 22y – 55 = – 8x – 24
∴ 8x + 22y -31 = 0

Question 14.
Find the co-ordinates of the orthocentre of the triangle whose vertices are A(2, – 2), B(l, 1) and C(-1,0).
Solution:
Let O be the orthocentre of AABC.
Let AM and BN be the altitudes of sides BC and AC respectively.
Now, slope of BC = \(\frac{0-1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}\)
Slope of AM = -2 ,..[∵ AM ⊥ BC]
Since AM passes through (2, – 2) and has slope -2,
equation of the altitude AM is y – (- 2) = – 2 (x – 2)
∴ y + 2 = -2x + 4
∴ 2x + y – 2 = 0 …(i)

Also, slope of AC = \(\frac{0-(-2)}{-1-2}=\frac{2}{-3}\)
∴ Slope of BN = \(\frac{3}{2}\) …[∵ BN ⊥ AC]
Since BN passes through (1,1) and has slope \(\frac{3}{2}\), equation of the altitude BN is
y – 1 = \(\frac{3}{2}\)(x-1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0 …(ii)
To find co-ordinates of orthocentre, we have to solve equations (i) and (ii).
By (i) x 2 + (ii), we get
7x – 5 = 0
∴ x = \(\frac{5}{7}\)
substituting x = \(\frac{5}{7}\) in eq (i), we get
2(\(\frac{5}{7}\)) + y – 2 = 0
∴ y = -2(\(\frac{5}{7}\)) + 2
∴ y = \(\frac{-10+14}{7}=\frac{4}{7}\)
∴ Coordinates of orthocentre O = \(\left(\frac{5}{7}, \frac{4}{7}\right)\)

Question 15.
N(3, – 4) is the foot of the perpendicular drawn from the origin to line L. Find the equation of line L.
Solution:

Slope of ON = \(\frac{-4-0}{3-0}=\frac{-4}{3}\)
Since line L ⊥ ON,
slope of the line L is \(\frac{3}{4}\) and it passes through point N(3, -4).
Equation of the line in slope point form is y – y₁ = m(x – x₁)
Equation of line L is
y-(-4) = \(\frac{3}{4}\)(x-3)
∴ 4(y + 4) = 3(x – 3)
∴ 4y + 16 = 3x – 9
∴ 3x – 4y – 25 = 0

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.2 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.2

Question 1.
Find the slope of each of the lines which passes through the following points:
i. A(2, -1), B(4,3)
ii. C(- 2,3), D(5, 7)
iii. E(2,3), F(2, – 1)
iv. G(7,1), H(- 3,1)
Solution:
i. Here, A = (2, -1) andB = (4, 3)
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-(-1)}{4-2}=\frac{4}{2}\) = 2

ii. Here, C = (-2, 3) and D = (5, 7)
Slope of line CD = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{7-3}{5-(-2)}=\frac{4}{7}\)

iii. Here, E s (2, 3) and F = (2, -1)
Slope of line EF = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-1-3}{2-2}=\frac{-4}{0}\), which ix not defined.

Alternate Method:
Points E and F have same x co-ordinates i.e. 2.
Points E and F lie on a line parallel to Y-axis.
∴ The slope of EF is not defined.

iv. Here, G = (7, 1) and H = (-3, 1)
Slope of line GH = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{1-1}{-3-7}\) = o

Alternate Method:
Points G and H have same y co-ordinate i.e. 1.
∴ Points G and H lie on a line parallel to the
X-axis.
∴ The slope of GH is 0.

Question 2.
If the x and x-intercepts of line L are 2 and 3 respectively, then find the slope of line L.
Solution:
Given, x-intercept of line L is 2 and y-intercept of line L is 3
∴ The line L intersects X-axis at (2, 0) and Y-axis at (0,3).
∴ The line L passes through (2, 0) and (0, 3).
Slope of line L = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-2}=\frac{-3}{2}\)

Question 3.
Find the slope of the line whose inclination is 30°.
Solution:
Given, inclination (θ) = 30°
∴ Slope of the line = tanθ = tan30° = \(\frac{1}{\sqrt{3}}\)

Question 4.
Find the slope of the line whose inclination is \(\frac{\pi}{4}\)
Solution:
Given, inclination (0) = \(\frac{\pi}{4}\)
∴ Slope of the line = tan θ = tan\(\frac{\pi}{4}\) = 1

Question 5.
A line makes intercepts 3 and 3 on the co-ordinate axes. Find the inclination of the line.
Solution:
Given, x-intercept of line is 3 and y-intercept of line is 3
∴ The line intersects X-axis at (3, 0) and Y-axis at (0, 3).
∴ The line passes through (3, 0) and (0,3).
∴ Slope of line = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{0-3}\) = -1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan \(\frac{\pi}{4}\)
= tan(π-\(\frac{\pi}{4}\) ) …[v tan(π – θ) = -tan θ]
tan θ = tan \(\frac{3\pi}{4}\)
θ = \(\frac{3\pi}{4}\)
The inclination of the line is \(\frac{3\pi}{4}\).
[Note: Answer given in the textbook is ‘-1 However, as per our calculation it is \(\frac{3\pi}{4}\)]

Question 6.
Without using Pythagoras theorem, show that points A (4, 4), B (3, 5) and C (- 1, – 1) are the vertices of a right-angled triangle.
Solution:
Given, A(4,4), B(3, 5), C (-1, -1).
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-4}{3-4}\) = – 1
Slope of BC = \(\frac{-1-5}{-1-3}=\frac{-6}{-4}=\frac{3}{2}\)
Slope of AC = \(\frac{-1-4}{-1-4}\) = 1
Slope of AB x slope of AC = – 1 x 1 = – 1
∴ side AB ⊥ side AC
∴ ∆ABC is a right angled triangle right angled at A.
∴ The given points are the vertices of a right angled triangle.

Question 7.
Find the slope of the line which makes angle of 45° with the positive direction of the Y-axis measured anticlockwise.
Solution:

Since the line makes an angle of 45° with positive direction of Y-axis in anticlockwise direction,
Inclination of the line (0) = (90° + 45°)
∴ Slope of the line = tan(90° + 45°)
= – cot 45°
= -1

Question 8.
Find the value of k for which the points P(k, -1), Q(2,1) and R(4,5) are collinear.
Solution:
Given, points P(k, – 1), Q (2, 1) and R(4, 5) are collinear.
∴ Slope of PQ = Slope of QR .
∴ \(\frac{1-(-1)}{2-k}=\frac{5-1}{4-2}\)
∴ \(\frac{2}{2-k}=\frac{4}{2}\)
∴ 4 = 4 (2 – k)
∴ 1 = 2 – k
∴ k = 2 – 1 = 1

Question 9.
Find the acute angle between the X-axis and the line joining the points A(3, -1) and B(4, – 2).
Solution:
Given, A (3, – 1) and B (4, – 2)
Slope of AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-(-1)}{4-3}\) = – 1
But, slope of a line = tan θ
∴ tan θ = – 1
= – tan 45°
= tan (180° -45°)
… [∵ tan (180° – θ) = -tan θ]
= tan 135°
∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis.
Then, α + 0 = 180°
α = 180°- 135° = 45°
∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

Question 10.
A line passes through points A(xi, y0 and B(h, k). If the slope of the line is m, then show that k – y₁ = m (h – x₁).
Solution:
Given, A(x₁, y₁), B(h, k) and
slope of line AB = m
Slope of line AB = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
∴ m = \(\frac{\mathrm{k}-y_{1}}{\mathrm{~h}-x_{1}}\)
∴ k – y₁ = m (h – x₁)

Question 11.
If the points A(h, 0), B(0, k) and C(a, b) lie on a line, then show that \(\frac{a}{h}+\frac{b}{k}\) = 1. ‘
Solution:
Given, A(h, 0), B(0, k) and C(a, b)
Since the points A, B and C lie on a line, they are collinear.
∴ Slope of AB = slope of BC

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 5 Straight Line Ex 5.1 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 5 Straight Line Ex 5.1

Question 1.
If A(1, 3) and B(2, 1) are points, find the equation of the locus of point P such that PA = PB.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(1, 3), B(2, 1) and
PA = PB
∴ PA² = PB²
∴ (x – 1)² + ( y – 3)² = (x – 2)² + (y – 1)²
∴ x² – 2x + 1 + y² – 6y + 9 = x² – 4x + 4 + y² – 2y + 1
-2x – 6y + 10 = -4x – 2y + 5
∴ 2x – 4y + 5 = 0
∴ The required equation of locus is 2x – 4y + 5 = 0.

Question 2.
A(- 5,2) and B(4,1). Find the equation of the locus of point P, which is equidistant from A and B.
Solution:
Let P(x, y) be any point on the required locus.
P is equidistant from A(- 5, 2) and B(4, 1).
∴ PA = PB
∴ PA² = PB²
∴ (x + 5)² + (y – 2)² = (x – 4)² + (y – 1)²
∴ x² + 10x + 25 + y² — 4y + 4
= x² – 8x + 16 + y² – 2y + 1
∴ 10x – 4y + 29 = -8x – 2y + 17
∴ 18x – 2y + 12 = 0
∴ 9x – y + 6 = 0
The required equation of locus is 9x -y + 6 = 0.

Question 3.
If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.
Solution:
Let P(x, y) be any point on the required locus.
Given, A(2, 0), B(0, 3) and AP = 2BP
∴ AP² = 4BP²
∴ (x – 2)² + (y – 0)² = 4[(x – 0)² + (y – 3)²]
∴ x² – 4x + 4 + y² = 4(x² + y² – 6y + 9)
x² – 4x + 4 + y² = 4x² + 4y² – 24y + 36
∴ 3x² + 3 y² + 4x – 24y + 32 = 0
∴ The required equation of locus is
3x² + 3y² + 4x – 24y + 32 = 0.
[Note: Answer given in the textbook , is
‘3x² + 3y² + 4x + 24y + 32 = O’.
However, as per our calculation it is ‘3x² + 3y² + 4x – 24y + 32 = 0’.]

Question 4.
If A(4,1) and B(5,4), find the equation of the locus of point P such that PA² = 3PB².
Solution:
Let P(x, y) be any point on the required locus. Given, A(4,1), B(5,4) and PA2 = 3PB2
∴ (x – 4)2 + (y – 1)2 = 3[(x – 5)² + (y – 4)²]
∴ x² – 8x + 16 + y² – 2y + 1 = 3(x² – 10x + 25 + y² – 8y + 16)
∴ x² – 8x + y² – 2y + 17 = 3x² -30x + 75 + 3y² – 24y + 48
∴ 2x² + 2y² – 22x – 22y + 106 = 0
∴ x² + y² – 11x – 11y + 53 = 0
∴ The required equation of locus is
x² + y² – 11x – 11y + 53 = 0.

Question 5.
A(2, 4) and B(5, 8), find the equation of the
locus of point P such that PA² – PB² = 13.
Solution:
Let P(x, y) be any point on the required locus. Given, A(2,4), B(5, 8) and PA²-PB² = 13
∴ [(x -2)² + (y – 4)²] – [(x -5)² + (y- 8)²] = 13
∴ (x² – 4x + 4 + y² – 8y + 16) – (x² – 10x + 25 + y² – 16y + 64) =13
∴ x² – 4x+ y² – 8y + 20 – x² + 10x – y² + 16y – 89 = 13
∴ 6x + 8y- 69 = 13
∴ 6x + 8y – 82 = 0
∴ 3x + 4y – 41 = 0
∴ The required equation of locus is 3x + 4y- 41 = 0.

Question 6.
A(1, 6) and B(3, 5), find the equation of the locus of point P such that segment AB subtends right angle at P. (∠APB = 90°)
Solution:
Let P(x, y) be any point on the required locus. Given,
A(l, 6) and B(3, 5),
∠APB = 90°
∴ ΔAPB is a right angled triangle,
By Pythagoras theorem,

AP² + PB² = AB² P (x,y)
∴ [(x – 1)² + (y – 6)²] + [(x – 3)² + (y – 5)²] = (1 – 3)² + (6 -5)²
∴ x² — 2x + 1 + y² — 12y + 36 + x² – 6x + 9 + y² – 10y + 25 = 4 + 1
∴ 2x² + 2y² – 8x – 22y + 66 = 0
∴ x² + y² – 4x – 11y + 33 = 0
∴ The required equation of locus is x² + y² – 4x – 11y + 33 = 0.
[Note: Answer given in the textbook is
‘3x² + 4y² – 4x – 11y + 33 = 0’.
However, as per our calculation it is ‘x² + y² – 4x – 11y + 33 = O

Question 7.
If the origin is shifted to the point 0′(2, 3), the axes remaining parallel to the original axes, find the new co-ordinates of the points
i. A(1, 3) ii. B(2,5)
Solution:
Origin is shifted to (2, 3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + handy = Y + k
x = X + 2 andy = Y + 3 …(i)

i. Given, A(x, y) = A( 1, 3)
x = X + 2 andy = Y + 3 …[From(i)]
∴ 1 = X + 2 and 3 = Y + 3 X = – 1 and Y = 0
∴ The new co-ordinates of point A are (- 1,0).

ii. Given, B(x, y) = B(2, 5)
x = X + 2 and y = Y + 3 …[From(i)]
∴ 2 = X + 2 and 5 = Y + 3
∴ X = 0 and Y = 2
∴ The new co-ordinates of point B are (0, 2).

Question 8.
If the origin is shifted to the point O'(1, 3), the axes remaining parallel to the original axes, find the old co-ordinates of the points
i. C(5,4) ii. D(3,3)
Solution:
Origin is shifted to (1,3) = (h, k)
Let the new co-ordinates be (X, Y).
x = X + h andy = Y + k
∴ x = X+1 andy = Y + 3 …(i)

i. Given, C(X, Y) = C(5, 4)
x = X +1 andy = Y + 3 …[From(i)]
∴ x = 5 + 1 = 6 andy = 4 + 3 = 7
∴ The old co-ordinates of point C are (6, 7).

ii. Given, D(X, Y) = D(3, 3)
x = X + 1 andy = Y + 3 …[From(i)]
∴ x = 3 + 1 = 4 and y = 3 + 3 = 6
∴ The old co-ordinates of point D are (4, 6).

Question 9.
If the co-ordinates A(5, 14) change to B(8, 3) by shift of origin, find the co-ordinates of the point, where the origin is shifted.
Solution:
Let the origin be shifted to (h, k).
Given, A(x, y) = A(5,14), B(X, Y) = B(8, 3)
Since x = X + h andy = Y + k,
5 = 8 + hand 14 = 3 + k ,
∴ h = – 3 and k = 11
The co-ordinates of the point, where the origin is shifted are (- 3, 11).

Question 10.
Obtain the new equations of the following loci if the origin is shifted to the point 0′(2,2), the direction of axes remaining the same:
i. 3x-y + 2 = 0
11. x²+y²-3x = 7
iii. xy – 2x – 2y + 4 = 0
iv. y² – 4x – 4y + 12 = 0
Solution:
Given, (h,k) = (2,2)
Let (X, Y) be the new co-ordinates of the point (x,y).
∴ x = X + handy = Y + k
∴ x = X + 2 andy = Y + 2
i. Substituting the values of x and y in the equation 3x -y + 2 = 0, we get
3(X + 2) – (Y + 2) + 2 = 0
∴ 3X + 6-Y-2 + 2 = 0
∴ 3 X – Y + 6 = 0, which is the new equation of locus.

ii. Substituting the values of x and y in the equation
x² + y² – 3x = 7, we get
(X + 2)² + (Y + 2)² – 3(X + 2) = 7
∴ X² + 4X + 4 + Y² + 4Y + 4 – 3X – 6 = 7
∴ X² + Y² + X + 4Y – 5 = 0, which is the new
equation of locus.

iii; Substituting the values of x and y in the equation xy – 2x – 2y + 4 = 0, we get
(X + 2) (Y + 2) – 2(X + 2) – 2(Y + 2) + 4 = 0
∴ XY + 2X + 2Y + 4 – 2X – 4-2Y- 4 + 4 = 0
∴ XY = 0, which is the new equation of locus.

iv. Substituting the values of x and y in the equation y² – 4x – 4y + 12 = 0, we get
(Y + 2)² – 4(X + 2) – 4(Y + 2) + 12 = 0
∴ Y² + 4Y + 4 – 4X – 8 – 4Y -8 + 12 = 0
∴ Y² – 4X = 0, which is the new equation of locus.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(B)

(I) Select the correct option from the given alternatives.

Question 1.
Given A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 2
\end{array}\right]\), I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) if A – λI is a singular matrix, then ___________
(a) λ = 0
(b) λ² – 3λ – 4 = 0
(c) λ² + 3λ – 4 = 0
(d) λ² – 3λ – 6 = 0
Answer:
(b) λ² – 3λ – 4 = 0
Hint:

Question 2.
Consider the matrices A = \(\left[\begin{array}{ccc}
4 & 6 & -1 \\
3 & 0 & 2 \\
1 & -2 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
0 & 1 \\
-1 & 2
\end{array}\right]\), C = \(\left[\begin{array}{l}
3 \\
1 \\
2
\end{array}\right]\). Out of the given matrix products, ___________
(i) (AB)^TC
(ii) C^TC(AB)^T
(iii) C^TAB
(iv) A^TABB^TC
(a) Exactly one is defined
(b) Exactly two are defined
(c) Exactly three are defined
(d) all four are defined
Answer:
(c) Exactly three are defined
Hint:
A is of order 3 × 3, B is of order 3 × 2 and C is of order 3 × 1.
(AB)^TC is of order 2 × 1.
C^TC and (AB)^T are of different orders.
C^TC (AB)^T is not defined.
C^TAB is of order 1 × 2.
A^TABB^TC is of order 3 × 1.

Question 3.
If A and B are square matrices of equal order, then which one is correct among the following?
(a) A + B = B + A
(b) A + B = A – B
(c) A – B = B – A
(d) AB = BA
Answer:
(a) A + B = B + A
Hint:
Matrix addition is commutative.
∴ A + B = B + A

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & 2 \\
2 & 1 & -2 \\
a & 2 & b
\end{array}\right]\) is a matrix satisfying the equation AA^T = 9I, where I is the identity matrix of order 3, then the ordered pair (a, b) is equal to ___________
(a) (2, -1)
(b) (-2, 1)
(c) (2, 1)
(d) (-2, -1)
Answer:
(d) (-2, -1)

Question 5.
If A = \(\left[\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right]\) and |A³| = 125, then α = ___________
(a) ±3
(b) ±2
(c) ±5
(d) 0
Answer:
(a) ±3
Hint:
|A³| = 125
|A|³ = 53 …….[∵ |Aⁿ| = |A|ⁿ, n ∈ N]
∴ |A| = 5
\(\left|\begin{array}{ll}
\alpha & 2 \\
2 & \alpha
\end{array}\right|=5\)
α² – 4 = 5
α² = 9
∴ α = ± 3

Question 6.
If \(\left[\begin{array}{ll}
5 & 7 \\
x & 1 \\
2 & 6
\end{array}\right]-\left[\begin{array}{cc}
1 & 2 \\
-3 & 5 \\
2 & y
\end{array}\right]=\left[\begin{array}{cc}
4 & 5 \\
4 & -4 \\
0 & 4
\end{array}\right]\), then ___________
(a) x = 1, y = -2
(b) x = -1, y = 2
(c) x = 1, y = 2
(d) x = -1, y = -2
Answer:
(c) x = 1, y = 2

Question 7.
If A + B = \(\left[\begin{array}{ll}
7 & 4 \\
8 & 9
\end{array}\right]\) and A – B = \(\left[\begin{array}{ll}
1 & 2 \\
0 & 3
\end{array}\right]\), then the value of A is ___________
(a) \(\left[\begin{array}{ll}
3 & 1 \\
4 & 3
\end{array}\right]\)
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
6 & 2 \\
8 & 6
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
7 & 6 \\
8 & 12
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{ll}
4 & 3 \\
4 & 6
\end{array}\right]\)

Question 8.
If \(\left[\begin{array}{cc}
x & 3 x-y \\
z x+z & 3 y-w
\end{array}\right]=\left[\begin{array}{ll}
3 & 2 \\
4 & 7
\end{array}\right]\), then ___________
(a) x = 3, y = 7, z = 1, w = 14
(b) x = 3, y = -5, z = -1, w = -4
(c) x = 3, y = 6, z = 2, w = 7
(d) x = -3, y = -7, z = -1, w = -14
Answer:
(a) x = 3, y = 7, z = 1, w = 14

Question 9.
For suitable matrices A, B, the false statement is ___________
(a) (AB)^T = A^TB^T
(B) (A^T)^T = A
(C) (A – B)^T = A^T – B^T
(D) (A + B)^T = A^T + B^T
Answer:
(a) (AB)^T = A^TB^T
Hint:
(AB)^T = B^TA^T

Question 10.
If A = \(\left[\begin{array}{cc}
-2 & 1 \\
0 & 3
\end{array}\right]\) and f(x) = 2x² – 3x, then f(A) = ___________
(a) \(\left[\begin{array}{cc}
14 & 1 \\
0 & -9
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
-14 & 1 \\
0 & 9
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-14 & -1 \\
0 & -9
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{cc}
14 & -1 \\
0 & 9
\end{array}\right]\)
Hint:

(II) Answer the following questions.

Question 1.
If A = diag[2, -3, -5], B = diag[4, -6, -3] and C = diag [-3, 4, 1], then find
i. B + C – A
ii. 2A + B – 5C.
Solution:

Question 2.
If f(α) = A = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\), find
i. f(-α)
ii. f(-α) + f(α)
Solution:

Question 3.
Find matrices A and B, where
(i) 2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) and A + 3B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\)
(ii) 3A – B = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
1 & 0 & 5
\end{array}\right]\) and A + 5B = \(\left[\begin{array}{ccc}
0 & 0 & 1 \\
-1 & 0 & 0
\end{array}\right]\)
Solution:
i. Given equations are
2A – B = \(\left[\begin{array}{cc}
1 & -1 \\
0 & 1
\end{array}\right]\) …….(i)

Question 4.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & -2 \\
-1 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
-3 & 4 & 1 \\
2 & -1 & -3
\end{array}\right]\), verify
i. (A + 2B^T)^T = A^T + 2B
ii. (3A – 5B^T)^T = 3A^T – 5B.
Solution:

Question 5.
If A = \(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) and A + A^T = I, where I is a unit matrix of order 2 × 2, then find the value of α.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
3 & 2 \\
-1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
1 & 3 & 2 \\
4 & -1 & -3
\end{array}\right]\), show that AB is singular.
Solution:

Question 7.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\), show that AB and BA are both singular matrices.
Solution:

Question 8.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 0 \\
2 & 3 & 4 \\
0 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
2 & 2 & -4 \\
-4 & 2 & -4 \\
2 & -1 & 5
\end{array}\right]\), show that BA = 6I.
Solution:

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
0 & 3
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
3 & -2
\end{array}\right]\), verify that |AB| = |A|.|B|.
Solution:

Question 10.
If Aα = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that Aα . Aβ = Aα+β
Solution:

Question 11.
If A = \(\left[\begin{array}{cc}
1 & \omega \\
\omega^{2} & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
\omega^{2} & 1 \\
1 & \omega
\end{array}\right]\), where ω is a complex cube root of unity, then show that AB + BA + A – 2B is a null matrix.
Solution:
ω is the complex cube root of unity.
ω³ = 1
ω³ – 1 = 0
(ω – 1) (ω² + ω + 1) = 0
ω = 1 or ω² + ω + 1 = 0
But, ω is a complex number.
1 + ω + ω² = 0 …….(i)
AB + BA + A – 2B

which is a null matrix.

Question 12.
If A = \(\left[\begin{array}{lrr}
2 & -2 & -4 \\
-1 & 3 & 4 \\
1 & -2 & -3
\end{array}\right]\), show that A² = A.
Solution:

Question 13.
If A = \(\left[\begin{array}{ccc}
4 & -1 & -4 \\
3 & 0 & -4 \\
3 & -1 & -3
\end{array}\right]\), show that A² = I.
Solution:

Question 14.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
-4 & 2
\end{array}\right]\), show that A² – 5A – 14I = 0.
Solution:

Question 15.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), show that A – 4A + 3I = 0.
Solution:

Question 16.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & -4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & x \\
y & 0
\end{array}\right]\) and (A + B)(A – B) = A² – B², find x and y.
Solution:
(A + B)(A – B) = A² – B²
A² – AB + BA – B² = A² – B²
-AB + BA = 0
AB = BA

By equality of matrices, we get
2 – 4x = -3x
∴ x = 2 and 2y = 2x
y = x
∴ y = 2
∴ x = 2, y = 2

Question 17.
If A = \(\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
0 & -1 \\
1 & 0
\end{array}\right]\), show that (A + B)(A – B) ≠ A² – B².
Solution:
We have to prove that
(A + B) . (A – B) ≠ A² – B²
i.e., to prove that A(A – B) + B(A – B) ≠ A² – B²
i.e., to prove that A² – AB + BA – B² ≠ A² – B²
i.e., to prove that AB ≠ BA.

∴ AB ≠ BA

Question 18.
If A = \(\left[\begin{array}{ll}
2 & -1 \\
3 & -2
\end{array}\right]\), find A³.
Solution:

∴ A² = I
Multiplying throughout by A, we get
A³ = A . I
∴ A³ = A

Question 19.
Find x, y if,

Solution:

Question 20.
Find x, y, z if

Solution:

Question 21.
If A = \(\left[\begin{array}{ccc}
2 & 1 & -3 \\
0 & 2 & 6
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & 0 & -2 \\
3 & -1 & 4
\end{array}\right]\), find AB^T and A^TB.
Solution:

Question 22.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\), show that (AB)^T = B^TA^T.
Solution:

Question 23.
If A = \(\left[\begin{array}{ll}
3 & -4 \\
1 & -1
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
1+2 n & -4 n \\
n & 1-2 n
\end{array}\right]\), for all n ∈ N.
Solution:

Question 24.
If A = \(\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\), prove that Aⁿ = \(\left[\begin{array}{cc}
\cos n \theta & \sin n \theta \\
-\sin n \theta & \cos n \theta
\end{array}\right]\), for all n ∈ N.
Solution:

Question 25.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in Rupees) of these crops by both the farmers for the month of April and may 2008 is given below,

Find
(i) the total sale in rupees for two months of each farmer for each crop.
(ii) the increase in sales from April to May for every crop of each farmer.
Solution:
(i) Total sale for Shantaram:
For rice = 15000 + 18000 = ₹ 33000.
For wheat = 13000 + 15000 = ₹ 28000.
For groundnut = 12000 + 12000 = ₹ 24000.
Total sale for Kantaram:
For rice = 18000 + 21000 = ₹ 39000
For wheat = 15000 + 16500 = ₹ 31500
For groundnut = 8000 + 16000 = ₹ 24000

Alternate method:
Matrix form

∴ The total sale of April and May of Shantaram in ₹ is ₹ 33000 (rice), ₹ 28000 (wheat), ₹ 24000 (groundnut), and that of Kantaram in ₹ is ₹ 39000(rice), ₹ 31500(wheat), and ₹ 24000 (groundnut).

(ii) Increase in sale from April to May for Shantaram:
For rice = 18000 – 15000 = ₹ 3000
For wheat = 15000 – 13000 = ₹ 2000
For groundnut = 12000 – 12000 = ₹ 0
Increase in sale from April to May for Kantaram:
For rice = 21000 – 18000 = ₹ 3000
For wheat = 16500 – 15000 = ₹ 1500
For groundnut = 16000 – 8000 = ₹ 8000

Alternate method:
Matrix form

∴ The increase in sales for Shantaram from April to May in each crop is ₹ 3000 (rice), ₹ 2000(wheat), 0 (groundnut), and that for Kantaram is ₹ 3000 (rice), ₹ 1500 (wheat), and ₹ 8000 (groundnut).

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.7 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.7

Question 1.
Find A^T, if
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
2 & -6 & 1 \\
-4 & 0 & 5
\end{array}\right]\)
Solution:
i. A = \(\left[\begin{array}{cc}
1 & 3 \\
-4 & 5
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{rr}
1 & -4 \\
3 & 5
\end{array}\right]\)

ii. A = \(\left[\begin{array}{ccc}
-4 & 0 & 5
2 & -6 & 1 \\
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\)

[Note: Answer given in the textbook is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
6 & 0 \\
1 & 5
\end{array}\right]\). However, as per our calculation it is A^T = \(\left[\begin{array}{cc}
2 & -4 \\
-6 & 0 \\
1 & 5
\end{array}\right]\). ]

Question 2.
If [a_ij]_3×3 where a_ij = 2(i – j), find A and
A^T. State whether A and A^T are symmetric or skew-symmetric matrices?
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given a_ij = 2 (i — j)
∴ a₁₁ = 2(1-1) = 0,
a₁₂ = 2(1-2) = -2,
a₁₃ = 2(1-3) = -4,
a₂₁ = 2(2-1) = 2,
a₂₂ = 2(2-2) = 0,
a₂₃=2(2-3) = -2,
a₃₁ = 2(3-1) = 4,
a₃₂ = 2(3-2) = 2,
a₃₃=2(3-3) = 0

∴ A^T = -A and A = -A^T
∴ A and A^T both are skew-symmetric matrices.

Questionn 3.
If A = \(\left[\begin{array}{cc}
5 & -3 \\
4 & -3 \\
-2 & 1
\end{array}\right]\), prove that (2A)^T = 2A^T.
Solution:

From (i) and (ii), we get
(2A)^T = 2A^T

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\), prove that (3A)^T = 3A^T.
Solution:

From (i) and (ii), we get
(3A)^T = 3A^T

Question 5.
If A = \(\left[\begin{array}{ccc}
0 & 1+2 i & 1-2 \\
-1-2 i & 0 & -7 \\
2-i & 7 & 0
\end{array}\right]\),
where i = \(\sqrt{-1}\), prove that A^T = – A.
Solution:

Question 6.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\) , B = \(\left[\begin{array}{cc}
2 & 1 \\
4 & -1 \\
-3 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 4 \\
-2 & 3
\end{array}\right]\) then show that
i. (A + B)^T = AT + B^T
ii. (A – C)^T = A^T – C^T
Solution:

From (i) and (ii), we get
(A + B)^T = AT + B^T
[Note: The question has been modified.]

From (i) and (ii), we get
(A – C)^T = A^T – C^T</sup

Question 7.
If A = \(\left[\begin{array}{cc}
5 & 4 \\
-2 & 3
\end{array}\right]\) and \(\left[\begin{array}{cc}
-1 & 3 \\
4 & -1
\end{array}\right]\) then find C^T, such that 3A – 2B + C = I, where I is the unit matrix of order 2.
Solution:
3A – 2B + C = I
∴ C = I + 2B – 3A

Question 8.
If A = \(\left[\begin{array}{ccc}
7 & 3 & 0 \\
0 & 4 & -2
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
0 & -2 & 3 \\
2 & 1 & -4
\end{array}\right]\), then find
i. A^T + 4B^T
ii. 5A^T – 5B^T
Solution:

ii. ii. 5A^T – 5B^T = 5(A^T – B^T)

Question 9.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
3 & 1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{rrr}
2 & 1 & -4 \\
3 & 5 & -2
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
0 & 2 & 3 \\
-1 & -1 & 0
\end{array}\right]\), verify that (A + 2B + 3C)^T = A^T + 2B^T + 3C^T.
Solution:
A + 2B + 3C

∴ A^T + 2B^T + 3C^T = \(\left[\begin{array}{cc}
5 & 6 \\
8 & 8 \\
2 & -2
\end{array}\right]\)
From (i) and (ii), we get
(A + 2B + 3C)^T = A^T + 2B^T + 3C^T

Question 10.
If A = \(\left[\begin{array}{ccc}
-1 & 2 & 1 \\
-3 & 2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 & 1 \\
-3 & 2 \\
-1 & 3
\end{array}\right]\), prove that (A + B^T)^T = A^T + B.
prove that (A + B^T)^T = A^T + B
Solution:


From (i) and (ii), we get
(A + B^T)^T = A^T + B

Question 11.
Prove that A + A^T is a symmetric and A – A^T is a skew symmetric matrix, where
i. A = \(\left[\begin{array}{ccc}
1 & 2 & 4 \\
3 & 2 & 1 \\
-2 & -3 & 2
\end{array}\right]\)
ii. A = \(\left[\begin{array}{ccc}
5 & 2 & -4 \\
3 & -7 & 2 \\
4 & -5 & -3
\end{array}\right]\)
Solution:

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.


∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

∴ (A + A^T)^T = A + A^T, i.e., A + A^T = (A + A^T)^T
∴ A + A^T is a symmetric matrix.

∴ (A – A^T)^T = – (A – A^T),
i.e., A – A^T = -(A – A^T)^T
∴ A – A^T is skew symmetric matrix.

Question 12.
Express the following matrices as the sum of a symmetric and a skew symmetric matrix.
i. \(\left[\begin{array}{cc}
4 & -2 \\
3 & -5
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
3 & 3 & -1 \\
-2 & -2 & 1 \\
-4 & -5 & 2
\end{array}\right]\)
Solution:
A square matrix A can be expressed as the sum of a symmetric and a skew symmetric matrix as
A = \(\frac{1}{2}\) (A + A^T) + \(\frac{1}{2}\) (A – A^T)

P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
A = P + Q
A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

∴ P is symmetric matrix …[∵ a_ij = a_ji]
and Q is a skew symmetric matrix [∵ -a_ij = -a_ji]
∴ A = P + Q
∴ A = \(\left[\begin{array}{cc}
4 & \frac{1}{2} \\
\frac{1}{2} & -5
\end{array}\right]+\left[\begin{array}{ll}
0 & \frac{-5}{2} \\
\frac{5}{2} & 0
\end{array}\right]\)

Question 13.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
3 & -2 \\
4 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
0 & 3 & -4 \\
2 & -1 & 1
\end{array}\right]\), verify that
i. (AB)^T = B^TA^T
ii. (BA)^T = A^TB^T
Solution:


From (i) and (ii), we get
(AB)^T = B^TA^T

From (i) and (ii) we get
(BA)^T = A^TB^T

Question 14.
If A = \(\left[\begin{array}{cc}
\cos \alpha & \sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\), show that A^TA = I, where I is the unit matrix of order 2.
Solution:


∴ A^TA = I, where I is the unit matrix of order 2.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.5 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.5

Question 1.
If A = \(\left[\begin{array}{cc}
2 & -3 \\
5 & -4 \\
-6 & 1
\end{array}\right]\), B = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & 2 \\
0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
4 & 3 \\
-1 & 4 \\
-2 & 1
\end{array}\right]\)
Show that
i. A+B=B+A
ii. (A + B) + C = A + (B + C)
Solution:

From (i) and (ii), we get
A + B = B + A

Question 2.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
5 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & -3 \\
4 & -7
\end{array}\right]\) then find the matrix A – 2B + 6I, where I is the unit matrix of order 2.
Solution:

Question 3.
If A = \(\), B = \(\) then find the matrix C such that A + B + C is a zero matrix.
Solution:
A+ B + C is a zero matrix.
∴ A + B + C = O
C = -(A + B)

Question 4.
If A = \(\left[\begin{array}{cc}
1 & -2 \\
3 & -5 \\
-6 & 0
\end{array}\right]\) B = \(\left[\begin{array}{cc}
-1 & -2 \\
4 & 2 \\
1 & 5
\end{array}\right]\) and C = \(\left[\begin{array}{cc}
2 & 4 \\
-1 & -4 \\
-3 & 6
\end{array}\right]\) , find the matrix X such that 3A – 4B + 5X = C.
Solution:
3A-4B + 5X = C
∴ 5X = C + 4B – 3A

Question 5.
Solve the following equations for X and Y, if 3X – Y = \(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\) and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\)
Solution:
Given equations are
\(=\left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]\)……………….. (i)
and X – 3Y = \(\left[\begin{array}{ll}
0 & -1 \\
0 & -1
\end{array}\right]\) ………………(ii)
By (i) x 3 – (ii) we get

Question 6.
Find the matrices A and B, if 2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\)
Solution:
Given equations are
2A – B = \(=\left[\begin{array}{ccc}
6 & -6 & 0 \\
-4 & 2 & 1
\end{array}\right]\) ……………….. (i)
and A – 2B = \(\left[\begin{array}{ccc}
3 & 2 & 8 \\
-2 & 1 & -7
\end{array}\right]\) ……………….(ii)
By (i) – (ii) x 2, we get

Question 7.
Simplify \(\cos \theta\left[\begin{array}{cc}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Quesiton 8.
If A = \(\left[\begin{array}{cc}
1 & 2 i \\
-3 & 2
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
2 i & 1 \\
2 & -3
\end{array}\right]\) where i =\(\sqrt{-1}\), find A + B and A – B. Show that A + B is singular. Is A – B singular? Justify your answer.
Solution:

Question 9.
Find x and y, if \(\left[\begin{array}{ccc}
2 x+y & -1 & 1 \\
3 & 4 y & 4
\end{array}\right]+\left[\begin{array}{ccc}
-1 & 6 & 4 \\
3 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
3 & 5 & 5 \\
6 & 18 & 7
\end{array}\right]\)
Solution:

∴ By equality of matrices, we get
2x + y – 1 = 3 and 4y = 18
∴ 2x + y = 4 and y = \(\frac{18}{4}=\frac{9}{2}\)
∴ 2x + \(\frac{9}{2}\) = 4
∴ 2x = 4 – \(\frac{9}{2}\)
∴ 2x = \(\frac{1}{2}=\)
∴ x = –\(\frac{1}{4}=\) and y = \(\frac{9}{2}=\)

Question 10.
If \(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\), find a, b, c and d.
Solution:
\(\left[\begin{array}{ll}
2 a+b & 3 a-b \\
c+2 d & 2 c-d
\end{array}\right]=\left[\begin{array}{cc}
2 & 3 \\
4 & -1
\end{array}\right]\)

∴ By equality of matrices, we get
2a + b = 2 ….(i)
3a – b = 3 ….(ii)
c + 2d = 4 ….(iii)
2c – d = -1 ….(iv)
Adding (i) and (ii), we get
5a = 5
∴ a = 1
Substituting a = 1 in (i), we get
2(1) + b = 2
∴ b = 0
By (iii) + (iv) x 2, we get
5c = 2
∴ c = \(\frac{2}{5}\)
Substituting c = \(\frac{2}{5}\) in (iii), we get
\(\frac{2}{5}\) + 2d = 4
∴ 2d = 4 – \(\frac{2}{5}\)
∴ 2d = \(\frac{18}{5}\)
∴ d = \(\frac{9}{5}\)
[Note: Answer given in the textbook is d = \(\frac{3}{5}\).
However, as per our calculation it is d = \(\frac{9}{5}\).]

Question 11.
There are two book shops owned by Suresh and Ganesh. Their sales (in Rupees) for books in three subjects – Physics, Chemistry and Mathematics for two months, July and August 2017 are given by two matrices A and B.

i. Find the increase in sales in Rupees from July to August 2017.
ii. If both book shops got 10% profit in the month of August 2017, find the profit for each bookseller in each subject in that month.
Solution:
i. Increase in sales in rupees from July to August 2017
For Suresh:
Increase in sales for Physics books
= 6650 – 5600= ₹ 1050
Increase in sales for Chemistry books
= 7055 – 6750 = ₹ 305
Increase in sales for Mathematics books
= 8905 – 8500 = ₹ 405
For Ganesh:
Increase in sales for Physics books
= 7000 – 6650 = ₹ 350
Increase in sales for Chemistry books
= 7500 – 7055 = ₹ 445
Increase in sales for Mathematics books
= 10200 – 8905 = ₹ 1295
[Note: Answers given in the textbook are 1760, 2090. However, as per our calculation they are 1050, 305, 405, 350, 445, 1295.]

ii. Both book shops got 10% profit in the month of August 2017.
For Suresh:
Profit for Physics books = \(\frac{6650 \times 10}{100}\) = ₹ 665
Profit for Chemistry books = \(\frac{7055 \times 10}{100}\) = ₹ 705.50
Profit for Mathematics books = \(\frac{8905 \times 10}{100}\) = ₹ 890.50

For Ganesh:
Profit for Physics books = \(\frac{7000 \times 10}{100}\) = ₹ 700
Profit for Chemistry books = \(\frac{7500 \times 10}{100}\) = ₹ 750
Profit for Mathematics books = \(\frac{10200 \times 10}{100}\) = ₹ 1020
[Note: Answers given in the textbook for Suresh’s profit in Chemistry and Mathematics books are ? 675 and ?850 respectively. However, as per our calculation profit amounts are ₹ 705.50 and ₹ 890.50 respectively.]

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.4 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.4

Question 1.
Construct a matrix A = [a_ij]_{3 x 2} whose elements ay are given by
i. a_ij = \(\frac{(\mathbf{i}-\mathbf{j})^{2}}{5-\mathbf{i}}\)
ii. a_ij = i – 3j
iii. a_ij \(\frac{(i+j)^{3}}{5}\)
Solution:

[Note: Answer given in the textbook is A = \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{2} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\)
However, as per our calculation it is \(\left[\begin{array}{ll}
0 & \frac{1}{4} \\
\frac{1}{3} & 0 \\
2 & \frac{1}{2}
\end{array}\right]\) ].

ii. a_ij = i – 3j
∴ a₁₁ = 1 – 3(1) = 1 – 3 = -2,
a₁₂= 1 – 3(2) = 1 – 6 = -5,
a₂₁ = 2 – 3(1) = 2 – 3 =-1,
a₂₂ = 2 – 3(2) = 2 – 6 = – 4
a₃₁ = 3 – 3(1) = 3-3 = 0,
a₃₂ = 3 – 3(2) = 3 – 6 = -3
∴ A = \(\left[\begin{array}{cc}
-2 & -5 \\
-1 & -4 \\
0 & -3
\end{array}\right]\)

iii. a_ij = \(\frac{(i+j)^{3}}{5}\)

Question 2.
Classify the following matrices as a row, a column, a square, a diagonal, a scalar, a unit, an upper triangular, a lower triangular, a symmetric or a skew- symmetric matrix.
i. \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & -2 & 4 \\
0 & 0 & -5 \\
0 & 0 & 0
\end{array}\right]\)
As every element below the diagonal is zero in matrix A.
∴ A is an upper triangular matrix.

ii. \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = \(\left[\begin{array}{ccc}
0 & -4 & -7 \\
4 & 0 & 3 \\
7 & -3 & 0
\end{array}\right]\)
∴ A^T = \(-\left[\begin{array}{ccc}
0 & 4 & 7 \\
-4 & 0 & -3 \\
-7 & 3 & 0
\end{array}\right]\)
∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

iii. \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{c}
5 \\
4 \\
-3
\end{array}\right]\)
∴ As matrix A has only one column.
∴ A is a column matrix.

iv. \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
9 & \sqrt{2} & -3
\end{array}\right]\)
As matrix A has only one row.
∴ A is a row matrix.

v. \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
6 & 0 \\
0 & 6
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero and diagonal elements same.
∴ A is a scalar matrix.

vi. \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & 0 \\
3 & -1 & 0 \\
-7 & 3 & 1
\end{array}\right]\)
As every element above the diagonal is zero in matrix A.
∴ A is a lower triangular matrix.

vii. \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & \frac{1}{3}
\end{array}\right]\)
As matrix A has all its non-diagonal elements zero.
∴ A is a diagonal matrix.

viii. \(\left[\begin{array}{ccc}
10 & -15 & 27 \\
-15 & 0 & \sqrt{34} \\
27 & \sqrt{34} & \frac{5}{3}
\end{array}\right]\)
Solution:

∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.

ix. \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Solution:
A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
In matrix A, all the non-diagonal elements are zero and diagonal elements are one.
∴ A is a unit (identity) matrix.

x. \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\)
∴ A^T = A, i/e., A = A^T
∴ A is a symmetric matrix.
∴ A is a symmetric matrix.

Question 3.
Which of the following matrices are singular or non-singular?
i. \(\left[\begin{array}{ccc}
\mathbf{a} & \mathbf{b} & \mathbf{c} \\
\mathbf{p} & \mathbf{q} & \mathbf{r} \\
\mathbf{2 a}-\mathbf{p} & \mathbf{2 b}-\mathbf{q} & \mathbf{2 c}-\mathbf{r}
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
Solution:

ii. Let A = \(\left[\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
5 & 0 & 5 \\
1 & 99 & 100 \\
6 & 99 & 105
\end{array}\right|\)
Applying C₂ → C₂ + C₁
|A| = \(\left|\begin{array}{ccc}
5 & 5 & 5 \\
1 & 100 & 100 \\
6 & 105 & 105
\end{array}\right|\)
= 0 … [∵ C₂ and C₃ are identical]
∴ A is a singular matrix.

iii. Let A = \(\left[\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ccc}
3 & 5 & 7 \\
-2 & 1 & 4 \\
3 & 2 & 5
\end{array}\right| \)
= 3(5 – 8) – 5(-10 – 12) + 7(-4 – 3)
= -9 + 110 – 49 = 52 ≠ 0
∴ A is a non-singular matrix.

iv. Let A = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{cc}
7 & 5 \\
-4 & 7
\end{array}\right]\) = 49 + 20 = 69 ≠ 0

Question 4.
Find k, if the following matrices are singular.
i. \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
7 & 3 \\
-2 & k
\end{array}\right]\)
Since A is a singular matrix,
|A|=0
∴ \(\left|\begin{array}{cc}
7 & 3 \\
-2 & \mathrm{k}
\end{array}\right|\) = o
∴ 7k + 6 = 0
∴ 7k = -6
k = -6/7

ii. Let A = \(\left[\begin{array}{ccc}
4 & 3 & 1 \\
7 & k & 1 \\
10 & 9 & 1
\end{array}\right]\)
Since A is a singular matrix,
|A|= 0
∴ \(\left|\begin{array}{ccc}
4 & 3 & 1 \\
7 & \mathrm{k} & 1 \\
10 & 9 & 1
\end{array}\right|\) = 0
∴ 4(k – 9) – 3(7 – 10) + 1(63 – 10k) = 0
∴ 4k – 36 + 9 + 63 – 10k = 0
∴ -6k + 36 = 0
∴ 6k = 36
∴ k = 6

iii. Let A = \(\left[\begin{array}{ccc}
\mathbf{k}-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right]\)
Since A is a singular matrix
|A| = 0
∴ \(\left|\begin{array}{ccc}
k-1 & 2 & 3 \\
3 & 1 & 2 \\
1 & -2 & 4
\end{array}\right|\)
∴ (k – 1)(4 + 4) – 2(12 – 2) + 3 (-6 – 1) = 0
∴ 8k-8-20-21 =0
∴ 8k = 49
∴ k = 49/8

Question 5.
If A = \(\left[\begin{array}{lll}
5 & 1 & -1 \\
3 & 2 & 0
\end{array}\right]\), find (A^T)^T.
Solution:

Question 6.
If A = \(\), find (A^T)^T.
Solution:

Question 7.
Find a, b, c, if \(\left[\begin{array}{ccc}
1 & \frac{3}{5} & a \\
b & -5 & -7 \\
-4 & c & 0
\end{array}\right]\) is a symmetric matrix.
Solution:

Question 8.
Find x, y, z, if \(\) is a symmetric matrix.
Solution:

Question 9.
For each of the following matrices, using its transpose, state whether it is symmetric, skew-symmetric or neither.
i. \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
ii. \(\left[\begin{array}{ccc}
2 & 5 & 1 \\
-5 & 4 & 6 \\
-1 & -6 & 3
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
0 & 1+2 \mathbf{i} & \mathbf{i}-2 \\
-1-2 \mathbf{i} & 0 & -7 \\
2-\mathbf{i} & 7 & 0
\end{array}\right]\)
Solution:
i. Let A = \(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T =\(\left[\begin{array}{ccc}
1 & 2 & -5 \\
2 & -3 & 4 \\
-5 & 4 & 9
\end{array}\right]\)
∴ A^T = A, i.e., A = A^T
∴ A is a symmetric matrix.

ii.

∴ A ≠ A^T, i.e., A ≠ -A^T
∴ A is neither a symmetric nor skew-symmetric matrix.

iii.

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Question 10.
Construct the matrix A = [a_ij]_{3 x 3}, where a_ij = i – j. State whether A is symmetric or skew-symmetric.
Solution:
A = [a_ij]_{3 x 3}
∴ A = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
Given, a_ij = i – j
a₁₁ = 1-1 = 0, a₁₂ = 1-2 = – 1, a₁₃ = 1 – 3 = – 2,
a₂₁ – 2 – 1 = 1, a₂₂ = 2 – 2 = 0, a₂₃ =2 – 3 = – 1,
a₃₁ = 3 – 1 = 2, a₃₂ = 3 – 2 = 1, a₃₃ = 3 – 3 = 0

∴ A^T = -A, i.e., A = -A^T
∴ A is a skew-symmetric matrix.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A) Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Miscellaneous Exercise 4(A)

I. Select the correct option from the given alternatives.

Question 1.
The determinant D = \(\left|\begin{array}{ccc}
a & b & a+b \\
b & c & b+c \\
a+b & b+c & 0
\end{array}\right|\) = 0, if
(a) a, b, c are in A.P.
(b) a, b, c are in G.P.
(c) a, b, c are in H.P.
(d) α is a root of ax² + 2bx + c = 0
Answer:
(b) a, b, c are in G.P.
Hint:
Applying R₃ → R₃ – (R₁ + R₂), we get
\(\left|\begin{array}{llc}
a & b & a+b \\
b & c & b+c \\
0 & 0 & -(a+2 b+c)
\end{array}\right|=0\)
∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0
∴ (-ac + b²) (a + 2b + c) = 0
∴ -ac + b² = 0 or a + 2b + c = 0
∴ b² = ac
∴ a, b, c are in G.P.

Question 2.
If \(\left|\begin{array}{lll}
x^{k} & x^{k+2} & x^{k+3} \\
y^{k} & y^{k+2} & y^{k+3} \\
z^{k} & z^{k+2} & z^{k+3}
\end{array}\right|\) = (x – y) (y – z) (z – x) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\) then
(a) k = -3
(b) k = -1
(c) k = 1
(d) k = 3
Answer:
(b) k = -1
Hint:

Question 3.
Let D = \(\left|\begin{array}{ccc}
\sin \theta \cdot \cos \phi & \sin \theta \cdot \sin \phi & \cos \theta \\
\cos \theta \cdot \cos \phi & \cos \theta \cdot \sin \phi & -\sin \theta \\
-\sin \theta \cdot \sin \phi & \sin \theta \cdot \cos \phi & 0
\end{array}\right|\) then
(a) D is independent of θ
(b) D is independent of φ
(c) D is a constant
(d) \(\frac{d D}{d}\) at θ = \(\frac{\pi}{2}\) is equal to 0
Answer:
(b) D is independent of φ

Question 4.
The value of a for which the system of equations a³x + (a + 1)y + (a + 2)³ z = 0, ax + (a + 1)y + (a + 2)z = 0 and x + y + z = 0 has a non zero solution is
(a) 0
(b) -1
(c) 1
(d) 2
Answer:
(b) -1
Hint:
The given system of equations will have a non-zero solution, if

Question 5.
\(\left|\begin{array}{lll}
b+c & c+a & a+b \\
q+r & r+p & p+q \\
y+z & z+x & x+y
\end{array}\right|=\)
(a) 2 \(\left|\begin{array}{lll}
c & b & a \\
r & q & p \\
z & y & x
\end{array}\right|\)
(b) 2 \(\left|\begin{array}{lll}
b & a & c \\
q & p & r \\
y & x & z
\end{array}\right|\)
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
(d) 2 \(\left|\begin{array}{lll}
a & c & b \\
p & r & q \\
x & z & y
\end{array}\right|\)
Answer:
(c) 2 \(\left|\begin{array}{lll}
a & b & c \\
p & q & r \\
x & y & z
\end{array}\right|\)
Hint:

Question 6.
The system 3x – y + 4z = 3, x + 2y – 3z = -2 and 6x + 5y + λz = -3 has atleast one solution when
(a) λ = -5
(b) λ = 5
(c) λ = 3
(d) λ = -13
Answer:
(a) λ = -5
Hint:
The given system of equations will have more than one solution if

Question 7.
If x = -9 is a root of \(\left|\begin{array}{lll}
x & 3 & 7 \\
2 & x & 2 \\
7 & 6 & x
\end{array}\right|=0\), has other two roots are
(a) 2, -7
(b) -2, 7
(c) 2, 7
(d) -2, -7
Answer:
(c) 2, 7
Hint:

Question 8.
If \(\left|\begin{array}{ccc}
6 i & -3 i & 1 \\
4 & 3 i & -1 \\
20 & 3 & i
\end{array}\right|\) = x + iy, then
(a) x = 3, y = 1
(b) x = 1, y = 3
(c) x = 0, y = 3
(d) x = 0, y = 0
Answer:
(d) x = 0, y = 0

Question 9.
If A(0, 0), B(1, 3) and C(k, 0) are vertices of triangle ABC whose area is 3 sq.units, then the value of k is
(a) 2
(b) -3
(c) 3 or -3
(d) -2 or 2
Answer:
(d) -2 or 2

Question 10.
Which of the following is correct?
(a) Determinant is a square matrix
(b) Determinant is number associated to matrix
(c) Determinant is a number associated with a square matrix
(d) None of these
Answer:
(c) Determinant is a number associated with a square matrix

II. Answer the following questions.

Question 1.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)
Solution:
(i) \(\left|\begin{array}{ccc}
2 & -5 & 7 \\
5 & 2 & 1 \\
9 & 0 & 2
\end{array}\right|\)

= 2(4 – 0) + 5(10 – 9) + 7(0 – 18)
= 2(4) + 5(1) + 7(-18)
= 8 + 5 – 126
= -113

(ii) \(\left|\begin{array}{ccc}
1 & -3 & 12 \\
0 & 2 & -4 \\
9 & 7 & 2
\end{array}\right|\)

= 1(4 + 28) + 3(0 + 36) + 12(0 – 18)
= 1(32) + 3(36) + 12(-18)
= 32 + 108 – 216
= -76

Question 2.
Evaluate determinant along second column \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 2 & -2 \\
0 & 1 & -2
\end{array}\right|\)
Solution:

Question 3.
Evaluate:
(i) \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
400 & 600 & 1000 \\
48 & 47 & 18
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
101 & 102 & 103 \\
106 & 107 & 108 \\
1 & 2 & 3
\end{array}\right|\)
by using properties.
Solution:

Question 4.
Find the minors and cofactors of elements of the determinants.
(i) \(\left|\begin{array}{ccc}
-1 & 0 & 4 \\
-2 & 1 & 3 \\
0 & -4 & 2
\end{array}\right|\)
(ii) \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right|\)
Solution:

Question 5.
Find the values of x, if
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:
(i) \(\left|\begin{array}{ccc}
1 & 4 & 20 \\
1 & -2 & -5 \\
1 & 2 x & 5 x^{2}
\end{array}\right|=0\)
⇒ 1(-10x² + 10x) – 4(5x² + 5) + 20(2x + 2) = 0
⇒ -10x² + 10x – 20x² – 20 + 40x + 40 = 0
⇒ -30x² + 50x + 20 = 0
⇒ 3x² – 5x – 2 = 0 …..[Dividing throughout by (-10)]
⇒ 3x² – 6x + x – 2 = 0
⇒ 3x(x – 2) + 1(x – 2) = 0
⇒ (x – 2) (3x + 1) = 0
⇒ x – 2 = 0 or 3x + 1 = 0
⇒ x = 2 or x = \(-\frac{1}{3}\)

(ii) \(\left|\begin{array}{ccc}
1 & 2 x & 4 x \\
1 & 4 & 16 \\
1 & 1 & 1
\end{array}\right|=0\)
⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0
⇒ 1(-12) – 2x(-15) + 4x(-3) = 0
⇒ -12 + 30x – 12x = 0
⇒ 18x = 12
⇒ x = \(\frac{12}{18}=\frac{2}{3}\)

Question 6.
By using properties of determinant, prove that \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|=0\)
Solution:

Question 7.
Without expanding the determinants, show that

Solution:

Question 8.
If \(\left|\begin{array}{lll}
a & 1 & 1 \\
1 & b & 1 \\
1 & 1 & c
\end{array}\right|=0\) then show that \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=1\)
Solution:

Question 9.
Solve the following linear equations by Cramer’s Rule.
(i) 2x – y + z = 1, x + 2y + 3z = 8, 3x + y – 4z = 1
(ii) \(\frac{1}{x}+\frac{1}{y}=\frac{3}{2}, \quad \frac{1}{y}+\frac{1}{z}=\frac{5}{6}, \quad \frac{1}{z}+\frac{1}{x}=\frac{4}{3}\)
(iii) 2x + 3y + 3z = 5, x – 2y + z = -4, 3x – y – 2z = 3
(iv) x + y + 2z = 7, 3x + 4y – 5z = 5, 2x – y + 3z = 12
Solution:
(i) Given equations are
2x – y + z = 1
x + 2y + 3z = 8
3x + y – 4z = 1
D = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-8 – 3) – (-1)(-4 – 9) + 1(1 – 6)
= 2(-11) + 1(-13) + 1(-5)
= -22 – 13 – 5
= -40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
1 & -1 & 1 \\
8 & 2 & 3 \\
1 & 1 & -4
\end{array}\right|\)
= 1(-8 – 3) – (-1)(-32 – 3) + 1(8 – 2)
= 1(-11) + 1(-35) + 1(6)
= -11 – 35 + 6
= -40

Dy = \(\left|\begin{array}{ccc}
2 & 1 & 1 \\
1 & 8 & 3 \\
3 & 1 & -4
\end{array}\right|\)
= 2(-32 – 3) -1(-4 – 9) + 1(1 – 24)
= 2(-35) – 1(-13) + 1(-23)
= -70 + 13 – 23
= -80

Dz = \(\left|\begin{array}{ccc}
2 & -1 & 1 \\
1 & 2 & 8 \\
3 & 1 & 1
\end{array}\right|\)
= 2(2 – 8) – (-1)(1 – 24) + 1(1 – 6)
= 2(-6) + 1(-23) + 1(-5)
= -12 – 23 – 5
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 1 are the solutions of the given equations.

(ii) Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
p + q = \(\frac{3}{2}\)
i.e., 2p + 2q = 3
i.e., 2p + 2q + 0 = 3
q + r = \(\frac{5}{6}\)
i.e., 6q + 6r = 5,
i.e., 0p + 6q + 6r = 5
r + p = \(\frac{4}{3}\)
i.e., 3r + 3p = 4,
i.e., 3p + 0q + 3r = 4
D = \(\left|\begin{array}{lll}
2 & 2 & 0 \\
0 & 6 & 6 \\
3 & 0 & 3
\end{array}\right|\)
= 2(18 – 0) -2(0 – 18) + 0
= 2(18) – 2(-18)
= 36 + 36
= 72 ≠ 0

Dp = \(\left|\begin{array}{lll}
3 & 2 & 0 \\
5 & 6 & 6 \\
4 & 0 & 3
\end{array}\right|\)
= 3(18 – 0) – 2(15 – 24) + 0
= 3(18) – 2(-9)
= 54 + 18
= 72

Dq = \(\left|\begin{array}{lll}
2 & 3 & 0 \\
0 & 5 & 6 \\
3 & 4 & 3
\end{array}\right|\)
= 2(15 – 24) – 3(0 – 18) + 0
= 2(-9) – 3(-18)
= -18 + 54
= 36

Dr = \(\left|\begin{array}{lll}
2 & 2 & 3 \\
0 & 6 & 5 \\
3 & 0 & 4
\end{array}\right|\)
= 2(24 – 0) – 2(0 – 15) + 3(0 – 18)
= 2(24) – 2(-15) + 3(-18)
= 48 + 30 – 54
= 24
By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

(iii) Given equations are
2x + 3y + 3z = 5
x – 2y + z = -4
3x – y – 2z = 3
D = \(\left|\begin{array}{ccc}
2 & 3 & 3 \\
1 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 2(4 + 1) – 3(-2 – 3) + 3(-1 + 6)
= 2(5) – 3(-5) + 3(5)
= 10 + 15 + 15
= 40 ≠ 0

Dx = \(\left|\begin{array}{ccc}
5 & 3 & 3 \\
-4 & -2 & 1 \\
3 & -1 & -2
\end{array}\right|\)
= 5(4 + 1) – 3(8 – 3) + 3(4 + 6)
= 5(5) – 3(5) + 3(10)
= 25 – 15 + 30
= 40

Dy = \(\left|\begin{array}{ccc}
2 & 5 & 3 \\
1 & -4 & 1 \\
3 & 3 & -2
\end{array}\right|\)
= 2(8 – 3) – 5(-2 – 3) + 3(3 + 12)
= 2(5) – 5(-5) + 3(15)
= 10 + 25 + 45
= 80

Dz = \(\left|\begin{array}{ccc}
2 & 3 & 5 \\
1 & -2 & -4 \\
3 & -1 & 3
\end{array}\right|\)
= 2(-6 – 4) – 3(3 + 12) + 5(-1 + 6)
= 2(-10) – 3(15) + 5(5)
= -20 -45 + 25
= -40
By Cramer’s Rule,

∴ x = 1, y = 2 and z = -1 are the solutions of the given equations.

(iv) Given equations are
x – y + 2z = 7
3x + 4y – 5z = 5
2x – y + 3z = 12
D = \(\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 4 & -5 \\
2 & -1 & 3
\end{array}\right|\)
= 1(12 – 5) – (-1)(9 + 10) + 2(-3 – 8)
= 1(7) + 1(19) + 2(-11)
= 7 + 19 – 22
= 4 ≠ 0

Dx = \(\left|\begin{array}{ccc}
7 & -1 & 2 \\
5 & 4 & -5 \\
12 & -1 & 3
\end{array}\right|\)
= 7(12 – 5) – (-1)(15 + 60) + 2(-5 – 48)
= 7(7) + 1(75) + 2(-53)
= 49 + 75 – 106
= 18

Dy = \(\left|\begin{array}{ccc}
1 & 7 & 2 \\
3 & 5 & -5 \\
2 & 12 & 3
\end{array}\right|\)
= 1(15 + 60) – 7(9 + 10) + 2(36 – 10)
= 1(75) – 7(19) + 2(26)
= 75 – 133 + 52
= -6

Dz = \(\left|\begin{array}{ccc}
1 & -1 & 7 \\
3 & 4 & 5 \\
2 & -1 & 12
\end{array}\right|\)
= 1(48 + 5) – (-1)(36 – 10) + 7(-3 – 8)
= 1(53) + 1(26) + 7(-11)
= 53 + 26 – 77
= 2
By Cramer’s Rule,

∴ x = \(\frac{9}{2}\), y = \(\frac{-3}{2}\) and z = \(\frac{1}{2}\) are the solutions of the given equations.

Question 10.
Find the value of k, if the following equations are consistent.
(i) (k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
(ii) 3x + y – 2 = 0, kx + 2y – 3 = 0 and 2x – y = 3
(iii) (k – 2)x + (k – 1)y = 17, (k – 1)x +(k – 2)y = 18 and x + y = 5
Solution:
(i) Given equations are
(k + 1)x + (k – 1)y + (k – 1) = 0
(k – 1)x + (k + 1)y + (k – 1) = 0
(k – 1)x + (k – 1)y + (k + 1) = 0
Since these equations are consistent,

⇒ 2(2k + 2 + 2k – 2) – 0 + (k – 1) (4 – 0) = 0
⇒ 2(4k) + (k – 1)4 = 0
⇒ 8k + 4k – 4 = 0
⇒ 12k – 4 = 0
⇒ k = \(\frac{4}{12}=\frac{1}{3}\)

(ii) Given equations are
3x + y – 2 = 0
kx + 2y – 3 = 0
2x – y = 3, i.e., 2x – y – 3 = 0.
Since these equations are consistent,
\(\left|\begin{array}{rrr}
3 & 1 & -2 \\
k & 2 & -3 \\
2 & -1 & -3
\end{array}\right|=0\)
⇒ 3(-6 – 3) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ 3(-9) – 1(-3k + 6) – 2(-k – 4) = 0
⇒ -27 + 3k – 6 + 2k + 8 = 0
⇒ 5k – 25 = 0
⇒ k = 5

(iii) Given equations are
(k – 2)x + (k – 1)y = 17
⇒ (k – 2)x + (k – 1)y – 17 = 0
(k – 1)x + (k – 2)y = 18
⇒ (k – 1)x + (k – 2)y – 18 = 0
x + y = 5
⇒ x + y – 5 = 0
Since these equations are consistent,

⇒ -1(-5k + 10 + 18) – 1(-5k + 5 + 18) + 1(k – 1 – k + 2) = 0
⇒ -1(-5k + 28) – 1(-5k + 23) + 1(1) = 0
⇒ 5k – 28 + 5k – 23 + 1 = 0
⇒ 10k – 50 = 0
⇒ k = 5

Question 11.
Find the area of triangle whose vertices are
(i) A(-1, 2), B(2, 4), C(0, 0)
(ii) P(3, 6), Q(-1, 3), R(2, -1)
(iii) L(1, 1), M(-2, 2), N(5, 4)
Solution:
(i) Here, A(x₁, y₁) = A(-1, 2)
B(x₂, y₂) = B(2, 4)
C(x₃, y₃) = C(0, 0)

Since area cannot be negative,
A(ΔABC) = 4 sq.units

(ii) Here, P(x₁, y₁) = P(3, 6)
Q(x₂, y₂) = Q(-1, 3)
R(x₃, y₃) = R(2, -1)

A(ΔPQR) = \(\frac{25}{2}\) sq.units

(iii) Here, L(x₁, y₁) = L(1, 1)
M(x₂, y₂) = M(-2, 2)
N(x₃, y₃) = N(5, 4)

Since area cannot be negative,
A(ΔLMN) = \(\frac{13}{2}\) sq.units

Question 12.
Find the value of k,
(i) if the area of a triangle is 4 square units and vertices are P(k, 0), Q(4, 0), R(0, 2).
(ii) if area of triangle is \(\frac{33}{2}\) square units and vertices are L(3, -5), M(-2, k), N(1, 4).
Solution:
(i) Here, P(x₁, y₁) = P(k, 0)
Q(x₂, y₂) = Q(4, 0)
R(x₃, y₃) = R(0, 2)
A(ΔPQR) = 4 sq.units

(ii) Here, L(x₁, y₁) = L(3, -5), M(x₂, y₂) = M(-2, k), N(x₃, y₃) = N(1, 4)
A(ΔLMN) = \(\frac{33}{2}\) sq. units
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
\(\pm \frac{33}{2}=\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
-2 & k & 1 \\
1 & 4 & 1
\end{array}\right|\)
⇒ \(\pm \frac{33}{2}=\frac{1}{2}\) [3(k – 4) – (-5) (-2 – 1) + 1 (-8 – k)]
⇒ ±33 = 3k – 12 – 15 – 8 – k
⇒ ±33 = 2k – 35
⇒ 2k – 35 = 33 or 2k – 35 = -33
⇒ 2k = 68 or 2k = 2
⇒ k = 34 or k = 1

Question 13.
Find the area of quadrilateral whose vertices are A(0, -4), B(4, 0), C(-4,0), D (0, 4).
Solution:
A(0, -4), B(4, 0), C(-4, 0), D(0, 4)


∴ A(ABDC) = A(ΔABC) + A(ΔBDC)
= 16 + 16
= 32 sq.units

Question 14.
An amount of ₹ 5000 is put into three investments at the rate of interest of 6%, 7%, and 8% per annum respectively. The total annual income is ₹ 350. If the combined income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment.
Solution:
Let the amount of each investment be ₹ x, ₹ y and ₹ z.
According to the given conditions,
x + y + z = 5000,
6% x + 7% y + 8% z = 350




∴ The amounts of investments are ₹ 1750, ₹ 1500, and ₹ 1750.

Question 15.
Show that the lines x – y = 6, 4x – 3y = 20 and 6x + 5y + 8 = 0 are concurrent. Also, find the point of concurrence.
Solution:
Given equations of the lines are
x – y = 6, i.e., x – y – 6 = 0 ……(i)
4x – 3y = 20, i.e., 4x – 3y – 20 = 0 …..(ii)
6x + 5y + 8 = 0 ……(iii)
The given lines will be concurrent, if

= 1(-24 + 100) – (-1) (32 + 120) – 6(20 + 18)
= 1(76) + 1(152) – 6(38)
= 76 + 152 – 228
= 0
∴ The given lines are concurrent.
To find the point of concurrence, solve any two equations.
Multiplying (i) by 5, we get
5x – 5y – 30 = 0 …….(iv)
Adding (iii) and (iv), we get
11x – 22 = 0
∴ x = 2
Substituting x = 2 in (i), we get
2 – y – 6 = 0
∴ y = -4
∴ The point of concurrence is (2, -4).

Question 16.
Show that the following points are collinear using determinants:
(i) L(2, 5), M(5, 7), N(8, 9)
(ii) P(5,1), Q(1, -1), R(11, 4)
Solution:
(i) Here, L(x₁, y₁) = L(2, 5)
M(x₂, y₂) = M(5, 7)
N(X₃ y₃) = N(8, 9)
If A(ΔLMN) = 0, then the points L, M, N are collinear.

∴ The points L, M, N are collinear.

(ii) Here, P(x₁, y₁) = P(5, 1)
Q(x₂, y₂) = Q(1, -1)
R(x₃, y₃) = R(11, 4)
If A(ΔPQR) = 0, then the points P, Q, R are collinear.

∴ The points P, Q, R are collinear.

Balbharti Maharashtra State Board Class 11 Maths Solutions Pdf Chapter 4 Determinants and Matrices Ex 4.3 Questions and Answers.

Maharashtra State Board 11th Maths Solutions Chapter 4 Determinants and Matrices Ex 4.3

Question 1.
Solve the following linear equations by using Cramer’s Rule.
x+y + z = 6, x – y + z = 2,.x + 2y – z = 2
x + y – 2z = -10,
2x +y – 3z = -19, 4x + 6y + z = 2
x + z = 1, y + z = 1, x + y = 4
\(\frac{-2}{x}-\frac{1}{y}-\frac{3}{z}\) = 3, \(\frac{2}{x}-\frac{3}{y}+\frac{1}{z}\) = -13 and \(\frac{2}{x}-\frac{3}{z}\) = -11
Solution:
Given equations are
x + y + z = 6,
x – y + z = 2,
x + 2y – z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
1 2 -1 = 1(1 -2) – 1(-1 – 1) + 1(2 + 1)
= 1 (-1)-1 (-2)+ 1(3)
= -1 + 2 + 3
= 4 ≠ 0

D_x = \(\left|\begin{array}{ccc}
6 & 1 & 1 \\
2 & -1 & 1 \\
2 & 2 & -1
\end{array}\right|\)
= 6(1 – 2) – 1(-2 – 2) + 1(4 + 2)
= 6(-1) -1 (-4) + 1(6)
= -6 + 4 + 6
= 4

D_y = \(\left|\begin{array}{ccc}
1 & 6 & 1 \\
1 & 2 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= 1(-2 – 2) – 6(-l – 1) + 1(2 = 1 (- 4) – 6 (- 2) + 1(0)
= -4+12 + 0 = 8

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & -1 & 2 \\
1 & 2 & 2
\end{array}\right|\)
= l(-2 – 4) – 1(2 – 2) + 6(2 + 1)
= l(-6)-l(0) + 6(3)
= -6 + 0+18 = 12

By Cramer’s Rule,

∴ x = 1, y = 2 and z = 3 are the solutions of the given equations.

ii. Given equations are x+y- 2z = -10,
2x + y – 3z = -19,
Ax + 6y + z = 2.
D = \(\left|\begin{array}{ccc}
1 & 1 & -2 \\
2 & 1 & -3 \\
4 & 6 & 1
\end{array}\right|\)
= 1(1 + 18)- 1(2+ 12)-2(12-4)
= 1(19)-1(14)-2(8)
= 19-14-16 = -11 ≠ 0

D_x = \(\left|\begin{array}{ccc}
-10 & 1 & -2 \\
-19 & 1 & -3 \\
2 & 6 & 1
\end{array}\right|\)
= -10(1 + 18) – 1(-19 + 6) – 2(- 114 – 2)
= -10(19)- 1(-13) -2(-l 16)
= -190+ 13 + 232 = 55

D_y = \(\left|\begin{array}{ccc}
1 & -10 & -2 \\
2 & -19 & -3 \\
4 & 2 & 1
\end{array}\right|\)
= 1(-19 + 6) – (-10)(2 + 12) – 2(4 + 76) = 1(-13) + 10(14) – 2(80)
= -13 + 140-160 = -33

D_z = \(\left|\begin{array}{ccc}
1 & 1 & -10 \\
2 & 1 & -19 \\
4 & 6 & 2
\end{array}\right|\)
= 1(2+ 114)-1(4+ 76)-10(12-4)
= 1(116)-1(80)-10(8)
= 116-80-80 .
= -44
By Cramer’s Rule,

∴ x = -5, y = 3 and z = 4 are the solutions of the given equations.
[Note: The question has been modified]

iii. Given equations are
x + z = 1, i.e.,x + 0y + z = 1,
y + z = 1, i.e., 0x + y + z = 1,
x + y = 4, i.e., x + y + 0z = 4.
D = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(0 – 1)
= 1(-1)+1(-1)
= -1-1 = -2 ≠ 0
D_x = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 0
\end{array}\right|\)
= 1(0 – 1) – 0 + 1(1 -4) = l(-l)+l(-3)
= -1 – 3
= -4

D_y = \(\left|\begin{array}{lll}
1 & 1 & 1 \\
0 & 1 & 1 \\
1 & 4 & 0
\end{array}\right|\)
= 1(0 – 4) – 1(0 – 1) + 1(0 – 1)
= 1(-4) – 1(-1) + 1(-1)
= -4 + 1 – 1
= -4

D_z = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 1 & 4
\end{array}\right|\)
= 1(4 – 1) – 0 + 1(0 – 1)
= 1(3) + 1(-1)
= 3 – 1
= 2

By Cramer’s Rule,

∴ x = 2, y = 2 and z = -1 are the solutions of the given equations.

Let \(\frac{1}{x}\) = p, \(\frac{1}{y}\) = q, \(\frac{1}{z}\) = r
∴ The given equations become
-2p – q – 3r = 3, i.e., 2p + q + 3r = -3,
2p-3q + r = -13,
2p – 3r = -11, i.e., 2p + 0q – 3r = -11.
D = \(\left|\begin{array}{ccc}
2 & 1 & 3 \\
2 & -3 & 1 \\
2 & 0 & -3
\end{array}\right|\)
= 2(9 – 0) – 1(-6 – 2) + 3(0 + 6)
= 2(9) – 1(-8) + 3(6)
= 18 + 8 + 18
= 44 ≠ 0

D_P = \(\left|\begin{array}{ccc}
-3 & 1 & 3 \\
-13 & -3 & 1 \\
-11 & 0 & -3
\end{array}\right|\)
= -3(9 – 0) – 1(39 + 11) + 3(0 – 33)
= -3(9) – 1(50) + 3(-33)
= -27 – 50 – 99
= -176

D_q = \(\left|\begin{array}{ccc}
2 & -3 & 3 \\
2 & -13 & 1 \\
2 & -11 & -3
\end{array}\right|\)
= 2(39 + 11)- (-3)(-6 – 2) + 3(-22 + 26)
= 2(50) + 3(-8) + 3(4)
= 100 – 24 + 12
= 88

D_r = \(\left|\begin{array}{ccc}
2 & 1 & -3 \\
2 & -3 & -13 \\
2 & 0 & -11
\end{array}\right|\)
= 2(33 – 0) – 1(-22 + 26) – 3(0 + 6)
= 2(33) – 1(4) – 3(6)
= 66 – 4 – 18
= 44

By Cramer’s Rule,

∴ x = \(\frac/{-1}{4}\), y = \(\frac{1}{2}\) z = 1 are the solutions of the given equations.

Question 2.
The sum of three numbers is 15. If the second number is subtracted from the sum of first and third numbers, then we get 5. When the third number is subtracted from the sum of twice the first number and the second number, we get 4. Find the three numbers.
Solution:
Let the three numbers be x, y and z.
According to the given conditions, x + y + z = 15,
x + z-y = 5, i.e., x – y + z = 5,
2x + y – z = 4.
D = \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -1 & 1 \\
2 & 1 & -1
\end{array}\right|\)
= 1(1 – 1) – 1(-1 – 2) + 1(1 + 2)
= 1(0) – 1(-3) + 1(3)
= 0 + 3 + 3
= 6 ≠ 0

D_x = \(\left|\begin{array}{ccc}
15 & 1 & 1 \\
5 & -1 & 1 \\
4 & 1 & -1
\end{array}\right|\)
= 15(1 – 1) – 1(-5 – 4) + 1(5 + 4)
= 15(0) – 1(-9) + 1(9)
= 0 + 9 + 9
= 18

D_y = \(\left|\begin{array}{ccc}
1 & 15 & 1 \\
1 & 5 & 1 \\
2 & 4 & -1
\end{array}\right|\)
= 1(-5 – 4) – 15(-1 – 2) + 1(4 – 10)
= 1(-9) – 15(-3) + 1(-6)
= -9 + 45 – 6 = 30

D_z = \(\left|\begin{array}{ccc}
1 & 1 & 15 \\
1 & -1 & 5 \\
2 & 1 & 4
\end{array}\right|\)
= 1(-4 – 5) – 1(4 – 10) + 15(1 + 2)
= 1(-9) – 1(-6) + 15(3)
= -9 + 6 + 45
= 42
By Cramer’s Rule,

∴ The three numbers are 3, 5 and 7.

Question 3.
Examine the consistency of the following equations.
i. 2x – y + 3 = 0, 3x + y – 2 = 0, 11x + 2y – 3 = 0
ii. 2x + 3y – 4 = 0, x + 2y = 3, 3x + 4y + 5 = 0
iii. x + 2y – 3 = 0,7x + 4y – 11 = 0,2x + 4y – 6 = 0
Solution:
i. Given equations are 2x – y + 3 = 0,
3x + y – 2 = 0,
11x + 2y – 3 = 0.
D = \(\left|\begin{array}{ccc}
2 & -1 & 3 \\
3 & 1 & -2 \\
11 & 2 & -3
\end{array}\right|\)
= 2(-3 + 4) – (-l)(-9 + 22) + 3(6-11)
= 2(1)+1(13)+ 3(-5)
= 2 + 13-15 = 0
∴ The given equations are consistent.

ii. Given equations are 2x + 3y – 4 = 0,
x + 2y = 3, i.e., x + 2y – 3 = 0,
3x + 4y + 5 = 0.
\(\left|\begin{array}{ccc}
2 & 3 & -4 \\
1 & 2 & -3 \\
3 & 4 & 5
\end{array}\right|\)
= 2(10 + 12) – 3(5 + 9) – 4(4 – 6)
= 2 (22) – 3(14) – 4(-2)
= 44 – 42 + 8
= 10 ≠ 0
∴ The given equations are not consistent.

iii. Given equations are x + 2y – 3 =
7x + 4y – 11 =0,
2x + 4y – 6 = 0.
\(\left|\begin{array}{ccc}
1 & 2 & -3 \\
7 & 4 & -11 \\
2 & 4 & -6
\end{array}\right|\)
= 1(-24 + 44) – 2(-42 + 22) – 3(28 – 8)
= 1(20) – 2(-20) – 3(20)
= 20 + 40 – 60
= 0
∴ The given equations are consistent.

Question 4.
Find k, if the following equations are consistent.
i. 2x + 3y-2 = 0,2x + 4y-k = 0,x-2j + 3k = 0
ii. kx + 3,y +1 = 0, x + 2y+1 = 0, x + y = 0
Solution:
i. Given equations are 2x + 3y – 2 = 0,
2x + 4y – k = 0,
x – 2y + 3k = 0.
Since these equations are consistent,
\(\left|\begin{array}{ccc}
2 & 3 & -2 \\
2 & 4 & -k \\
1 & -2 & 3 k
\end{array}\right|\) = 0
∴ 2(12k – 2k) – 3(6k + k) – 2(- 4 – 4) = 0
∴ 2(10k) – 3(7k) – 2(- 8) = 0
∴ 20k – 21k + 16 = 0
∴ k = 16

Given equations are are
kx + 3y + 1 = 0,
x + 2y +1=0,
x + y = 0, i.e., x + y + 0 = 0.
Since these equations are consistent,
\(\left|\begin{array}{lll}
k & 3 & 1 \\
1 & 2 & 1 \\
1 & 1 & 0
\end{array}\right|\) = 0
∴ k(0 – 1) – 3(0 – 1) + 1(1 – 2) = 0
∴ k(-1) – 3(-1) + 1(-1) = 0
∴ -k + 3 – 1 = 0
∴ k = 2.

Question 5.
Find the area of triangle whose vertices are
i. A (5,8), B (5,0), C (1,0)
ii. P(3/2, 1), Q(4,2), R(4, -1/2)
iii. M (0, 5), N (- 2, 3), T (1, – 4)
Solution:
i. Here, A(x₁, y₁) ≡ A(5, 8), B(x₂, y₂) = B(5, 0), C(x₃, y₃) = C(1,0)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABC) = \(\frac{1}{2}\left|\begin{array}{lll}
5 & 8 & 1 \\
5 & 0 & 1 \\
1 & 0 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[5(0 – 0) – 8(5 – 1) + 1(0 – 0)]
= \(\frac{1}{2}\)[0 – 8(4) + 0]
= \(\frac{1}{2}\)(-32)
= -16
Since area cannot be negative,
A(ΔABC) = 16 sq. units

ii. Here, P(x₁, y₁) ≡ P(3/2, 1), Q(x₂, y₂) ≡ Q(4, 2), R(x₃, y₃) ≡ R(4,-\(\frac{1}{2}\) )


Since area cannot be negative
A(ΔPQR) = 25/8 sq. units

iii. Here, M(x₁, y₁) ≡ M(0, 5), N(x₂, y₂) ≡ N(-2, 3)
T(x₃, y₃) ≡ T(1, -4)
Area of a triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ A(ΔMNT) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & 5 & 1 \\
-2 & 3 & 1 \\
1 & -4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [ 0 – 5 (-2 -1) + 1 (8 – 3)]
= \(\frac{1}{2}\)[-5 (-3) + 1(5)]
= \(\frac{1}{2}\) (15 + 5)
= \(\frac{1}{2}\) (20)
= 10 sq. units

Question 6.
Find the area of quadrilateral whose vertices are A (- 3,1), B (- 2, – 2), C (1,4), D (3, – 1).
Solution:
A(-3, 1), B(-2, -2), C(l, 4), D(3, -1)

A(□ ABDC) = A(ΔABD) + A(ΔADC)
Area of triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
A(ΔABD) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
-2 & -2 & 1 \\
3 & -1 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3 (-2 + 1) – 1(-2 – 3) + 1(2 + 6)
= \(\frac{1}{2}\) [-3(-1) – 1(-5) + 1(8)]
= \(\frac{1}{2}\) (3 + 5 + 8)
= \(\frac{1}{2}\) (16)
∴ A(ΔABD) = 8 sq. units
A(ΔADC) = \(\frac{1}{2}\left|\begin{array}{ccc}
-3 & 1 & 1 \\
3 & -1 & 1 \\
1 & 4 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [-3(-1-4) – 1(3 – 1) + 1(12 + 1)]
= \(\frac{1}{2}\) [-3(-5) – 1(2) + 1(13)]
= \(\frac{1}{2}\) [15 – 2 + 13]
= \(\frac{1}{2}\) (26)
∴ A(ΔADC) = 13 sq. units
∴ A(□ ABDC) = A(ΔABD) + A(ΔADC)
= 8 + 13
= 21 sq. units

Question 7.
Find the value of k, if the area of triangle whose vertices are P (k, 0), Q (2,2), R (4,3) is \(\frac{3}{2}\) sq. units.
Solution:
Here, P(x₁, y₁) ≡ P(k, 0), Q(x₂, y₂) ≡ Q(2, 2), R(x₃, y₃) ≡ R(4,3)
∴ A(ΔPQR) = \(\frac{3}{2}\) sq. units
Area if triangle = \(\frac{1}{2}\left|\begin{array}{lll}
x_{1} & y_{1} & 1 \\
x_{2} & y_{2} & 1 \\
x_{3} & y_{3} & 1
\end{array}\right|\)
∴ \(\pm \frac{3}{2}=\frac{1}{2}\left|\begin{array}{lll}
k & 0 & 1 \\
2 & 2 & 1 \\
4 & 3 & 1
\end{array}\right|\)
∴ ± \( [k(2 – 3) – 0 + 1 (6 – 8)]
∴ ± [latex\frac{3}{2}=\frac{1}{2}\) (-k -2)
∴ ± 3 = -k – 2
∴ 3 = -k – 2 or -3 = -k – 2
∴ k = -5 or k = 1

Question 8.
Examine the collinearity of the following set of points:
i. A (3, – 1), B (0, – 3), C (12, 5)
ii. P (3, – 5), Q (6,1), R (4, 2)
iii. L(0,1/2), M(2,-1), N(-4, 7/2)
Solution:
i. Here, A(x₁, y₁) ≡ A(3, -1), B(x₂, y₂) ≡ B(0, -3), C(x₃, y₃) ≡ C(12, 5)
If A(∆ABC) = 0, then the points A, B, C are collinear.
∴ A(∆ABC) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -1 & 1 \\
0 & -3 & 1 \\
12 & 5 & 1
\end{array}\right|\)
= \(\frac{1}{2}\)[3(-3 – 5) – (-1) (0 – 12) + 1(0 + 36)]
= \(\frac{1}{2}\)[3(-8)+ 1(-12)+ 1(36)]
= \(\frac{1}{2}\)(-24 – 12 + 36)
= 0
∴ The points A, B, C are collinear.

ii. Here, P(x₁, y₁) ≡ P(3, -5), Q(x₂, y₂) ≡ Q(6, 1), R(x₃, y₃) ≡ R(4,2)
∴ If A(∆PQR) = 0, then the points P,Q, R are collinear
∴ A(∆PQR) = \(\frac{1}{2}\left|\begin{array}{ccc}
3 & -5 & 1 \\
6 & 1 & 1 \\
4 & 2 & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [3(1-2) – (-5)(6 – 4) + 1(12 – 4)]
= \(\frac{1}{2}\) [3(-1) + 5(2) + 1(8)]
= \(\frac{1}{2}\)(-3 + 10 + 8)= \(\frac{15}{2}\) ≠ 0
∴ The points P, Q, R are non-collinear.

iii. Here, L(x₁, y₁) ≡ L(0,1/2), M(x₂, y₂) ≡ M(2, -1), N(x₃, y₃) ≡ N(-4, 7/2)
If A(∆LMN) = 0, then the points L, M, N are collinear.
∴ A(∆LMN) = \(\frac{1}{2}\left|\begin{array}{ccc}
0 & \frac{1}{2} & 1 \\
2 & -1 & 1 \\
-4 & \frac{7}{2} & 1
\end{array}\right|\)
= \(\frac{1}{2}\) [0 – ]\(\frac{1}{2}\) (2 + 4) + 1(7 – 4)]
= \(\frac{1}{2}\)[ –\(\frac{1}{2}\) (6) + 1(3)]
= \(\frac{1}{2}\) (-3 + 3) = 0
∴ The points L, M, N are collinear.