12th Commerce Maths 1 Chapter 4 Exercise 4.1 Answers Maharashtra Board

Applications of Derivatives Class 12 Commerce Maths 1 Chapter 4 Exercise 4.1 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 4 Applications of Derivatives Ex 4.1 Questions and Answers.

Std 12 Maths 1 Exercise 4.1 Solutions Commerce Maths

Question 1.
Find the equations of tangent and normal to the following curves at the given point on it:
(i) y = 3x2 – x + 1 at (1, 3)
Solution:
y = 3x2 – x + 1
∴ \(\frac{d y}{d x}=\frac{d}{d x}\) (3x2 – x + 1)
= 3 × 2x – 1 + 0
= 6x – 1
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}\) = 6(1) – 1
= 5
= slope of the tangent at (1, 3).
∴ the equation of the tangent at (1, 3) is
y – 3 = 5(x – 1)
∴ y – 3 = 5x – 5
∴ 5x – y – 2 = 0.
The slope of the normal at (1, 3) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,3)}}=-\frac{1}{5}\)
∴ the equation of the normal at (1, 3) is
y – 3 = \(-\frac{1}{5}\)(x – 1)
∴ 5y – 15 = -x + 1
∴ x + 5y – 16 = 0
Hence, the equations of the tangent and normal are 5x – y – 2 = 0 and x + 5y – 16 = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

(ii) 2x2 + 3y2 = 5 at (1, 1)
Solution:
2x2 + 3y2 = 5
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1 Q 1 (ii)
= slope of the tangent at (1, 1)
∴ the equation of the tangent at (1, 1) is
y – 1 = \(\frac{-2}{3}\)(x – 1)
∴ 3y – 3 = -2x + 2
∴ 2x + 3y – 5 = 0.
The slope of normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{\text {at }(1,1)}}=\frac{-1}{\left(\frac{-2}{3}\right)}=\frac{3}{2}\)
∴ the equation of the normal at (1, 1) is
y – 1 = \(\frac{3}{2}\)(x – 1)
∴ 2y – 2 = 3x – 3
∴ 3x – 2y – 1 = 0
Hence, the equations of the tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively.

(iii) x2 + y2 + xy = 3 at (1, 1)
Solution:
x2 + y2 + xy = 3
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1 Q 1 (iii)
= slope of the tangent at (1, 1)
the equation of the tangent at (1, 1) is
y – 1= -1(x – 1)
∴ y – 1 = -x + 1
∴ x + y = 2
The slope of the normal at (1, 1) = \(\frac{-1}{\left(\frac{d y}{d x}\right)_{a t(1,1)}}\)
= \(\frac{-1}{-1}\)
= 1
∴ the equation of the normal at (1, 1) is y – 1 = 1(x – 1)
∴ y – 1 = x – 1
∴ x – y = 0
Hence, the equations of tangent and normal are x + y = 2 and x – y = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 2.
Find the equations of the tangent and normal to the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = x2 + 5 where the tangent is parallel to the line 4x – y + 1 = 0.
Differentiating y = x2 + 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(x2 + 5) = 2x + 0 = 2x
\(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=2 x_{1}\)
= slope of the tangent at (x1, y1)
Let m1 = 2x1
The slope of the line 4x – y + 1 = 0 is
m2 = \(\frac{-4}{-1}\) = 4
Since, the tangent at P(x1, y1) is parallel to the line 4x – y + 1 = 0,
m1 = m2
∴ 2x1 = 4
∴ x1 = 2
Since, (x1, y1) lies on the curve y = x2 + 5, y1 = \(x_{1}^{2}\) + 5
∴ y1 = (2)2 + 5 = 9 ……[x1 = 2]
∴ the coordinates of the point are (2, 9) and the slope of the tangent = m1 = m2 = 4.
∴ the equation of the tangent at (2, 9) is
y – 9 = 4(x – 2)
∴ y – 9 = 4x – 8
∴ 4x – y + 1 = 0
Slope of the normal = \(\frac{-1}{m_{1}}=-\frac{1}{4}\)
∴ the equation of the normal at (2, 9) is
y – 9 = \(-\frac{1}{4}\)(x – 2)
∴ 4y – 36 = -x + 2
∴ x + 4y – 38 = 0
Hence, the equations of tangent and normal are 4x – y + 1 = 0 and x + 4y – 38 = 0 respectively.

Maharashtra Board 12th Commerce Maths Solutions Chapter 4 Applications of Derivatives Ex 4.1

Question 3.
Find the equations of the tangent and normal to the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Solution:
Let P(x1, y1) be the point on the curve y = 3x2 – 3x – 5 where the tangent is parallel to the line 3x – y + 1 = 0.
Differentiating y = 3x2 – 3x – 5 w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}\)(3x2 – 3x – 5)
= 3 × 2x – 3 × 1 – 0
= 6x – 3
∴ \(\left(\frac{d y}{d x}\right)_{\text {at }\left(x_{1}, y_{1}\right)}=6 x_{1}-3\)
= slope of the tangent at (x1, y1)
Let m1 = 6x1 – 3
The slope of the line 3x – y + 1 = 0
m2 = \(\frac{-3}{-1}\) = 3
Since, the tangent at P(x1, y1) is parallel to the line 3x – y + 1 = 0,
m1 = m2
∴ 6x1 – 3 = 3
∴ 6x1 = 6
∴ x1 = 1
Since, (x1, y1) lies on the curve y = 3x2 – 3x – 5,
\(y_{1}=3 x_{1}{ }^{2}-3 x_{1}-5\), where x1 = 1
= 3(1)2 – 3(1) – 5
= 3 – 3 – 5
= -5
∴ the coordinates of the point are (1, -5) and the slope of the tangent = m1 = m2 = 3.
∴ the equation of the tangent at (1, -5) is
y – (-5) = 3(x – 1)
∴ y + 5 = 3x – 3
∴ 3x – y – 8 = 0
Slope of the normal = \(-\frac{1}{m_{1}}=-\frac{1}{3}\)
∴ the equation of the normal at (1, -5) is
y – (-5) = \(-\frac{1}{3}\)(x – 1)
∴ 3y + 15 = -x + 1
∴ x + 3y + 14 = 0
Hence, the equations of tangent and normal are 3x – y – 8 = 0 and x + 3y + 14 = 0 respectively.

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Differentiation Class 12 Commerce Maths 1 Chapter 3 Miscellaneous Exercise 3 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 3 Differentiation Miscellaneous Exercise 3 Questions and Answers.

Std 12 Maths 1 Miscellaneous Exercise 3 Solutions Commerce Maths

(I) Choose the correct alternative:

Question 1.
If y = (5x3 – 4x2 – 8x)9, then \(\frac{d y}{d x}\) = ___________
(a) 9(5x3 – 4x2 – 8x)8 (15x2 – 8x – 8)
(b) 9(5x3 – 4x2 – 8x)9 (15x2 – 8x – 8)
(c) 9(5x3 – 4x2 – 8x)8 (5x2 – 8x – 8)
(d) 9(5x3 – 4x2 – 8x)9 (5x2 – 8x – 8)
Answer:
(a) 9(5x3 – 4x2 – 8x)8 (15x2 – 8x – 8)

Question 2.
If y = \(\sqrt{x+\frac{1}{x}}\), then \(\frac{d y}{d x}\) = ?
(a) \(\frac{x^{2}-1}{2 x^{2} \sqrt{x^{2}+1}}\)
(b) \(\frac{1-x^{2}}{2 x^{2} \sqrt{x^{2}+1}}\)
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
(d) \(\frac{1-x^{2}}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Answer:
(c) \(\frac{x^{2}-1}{2 x \sqrt{x} \sqrt{x^{2}+1}}\)
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 I Q2

Question 3.
If y = \(e^{\log x}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{e^{\log x}}{x}\)
(b) \(\frac{1}{x}\)
(c) 0
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{e^{\log x}}{x}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 4.
If y = 2x2 + 22 + a2, then \(\frac{d y}{d x}\) = ?
(a) x
(b) 4x
(c) 2x
(d) -2x
Answer:
(b) 4x

Question 5.
If y = 5x . x5, then \(\frac{d y}{d x}\) = ?
(a) 5x . x4(5 + log 5)
(b) 5x . x5(5 + log 5)
(c) 5x . x4(5 + x log 5)
(d) 5x . x5(5 + x log 5)
Answer:
(c) 5x . x4(5 + x log 5)

Question 6.
If y = \(\log \left(\frac{e^{x}}{x^{2}}\right)\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{2-x}{x}\)
(b) \(\frac{x-2}{x}\)
(c) \(\frac{e-x}{ex}\)
(d) \(\frac{x-e}{ex}\)
Answer:
(b) \(\frac{x-2}{x}\)
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 I Q6

Question 7.
If ax2 + 2hxy + by2 = 0, then \(\frac{d y}{d x}\) = ?
(a) \(\frac{(a x+h y)}{(h x+b y)}\)
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)
(c) \(\frac{(a x-h y)}{(h x+b y)}\)
(d) \(\frac{(2 a x+h y)}{(h x+3 b y)}\)
Answer:
(b) \(\frac{-(a x+h y)}{(h x+b y)}\)

Question 8.
If x4 . y5 = (x + y)(m+1) and \(\frac{d y}{d x}=\frac{y}{x}\) then m = ?
(a) 8
(b) 4
(c) 5
(d) 20
Answer:
(a) 8
Hint:
If xp . yq = (x + y)p+q, then \(\frac{d y}{d x}=\frac{y}{x}\)
∴ m + 1 = 4 + 5 = 9
∴ m = 8.

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 9.
If x = \(\frac{e^{t}+e^{-t}}{2}\), y = \(\frac{e^{t}-e^{-t}}{2}\) then \(\frac{d y}{d x}\) = ?
(a) \(\frac{-y}{x}\)
(b) \(\frac{y}{x}\)
(c) \(\frac{-x}{y}\)
(d) \(\frac{x}{y}\)
Answer:
(d) \(\frac{x}{y}\)
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 I Q9

Question 10.
If x = 2at2, y = 4at, then \(\frac{d y}{d x}\) = ?
(a) \(-\frac{1}{2 a t^{2}}\)
(b) \(\frac{1}{2 a t^{3}}\)
(c) \(\frac{1}{t}\)
(d) \(\frac{1}{4 a t^{3}}\)
Answer:
(c) \(\frac{1}{t}\)

(II) Fill in the blanks:

Question 1.
If 3x2y + 3xy2 = 0 then \(\frac{d y}{d x}\) = …………
Answer:
-1
Hint:
3x2y + 3xy2 = 0
∴ 3xy(x + y) = 0
∴ x + y = 0
∴ y = -x
∴ \(\frac{d y}{d x}\) = -1

Question 2.
If xm . yn = (x+y)(m+n) then \(\frac{d y}{d x}=\frac{\ldots \ldots}{x}\)
Answer:
y

Question 3.
If 0 = log(xy) + a then \(\frac{d y}{d x}=\frac{-y}{\ldots . .}\)
Answer:
x

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 4.
If x = t log t and y = tt then \(\frac{d y}{d x}\) = …………
Answer:
y
Hint:
x = t log t = log tt = log y
∴ 1 = \(\frac{1}{y} \cdot \frac{d y}{d x}\)
∴ \(\frac{d y}{d x}\) = y

Question 5.
If y = x . log x then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{1}{x}\)

Question 6.
If y = [log(x)]2 then \(\frac{d^{2} y}{d x^{2}}\) = …………..
Answer:
\(\frac{2(1-\log x)}{x^{2}}\)
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 II Q6

Question 7.
If x = y + \(\frac{1}{y}\) then \(\frac{d y}{d x}\) = …………
Answer:
\(\frac{y^{2}}{y^{2}-1}\)
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 II Q7

Question 8.
If y = eax, then x.\(\frac{d y}{d x}\) = …………
Answer:
axy

Question 9.
If x = t . log t, y = tt then \(\frac{d y}{d x}\) = …………
Answer:
y

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 10.
If y = \(\left(x+\sqrt{x^{2}-1}\right)^{m}\) then \(\sqrt{\left(x^{2}-1\right)} \frac{d y}{d x}\) = ………
Answer:
my
Hint:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 II Q10

(III) State whether each of the following is True or False:

Question 1.
If f’ is the derivative of f, then the derivative of the inverse of f is the inverse of f’.
Answer:
False

Question 2.
The derivative of loga x, where a is constant is \(\frac{1}{x \cdot \log a}\).
Answer:
True

Question 3.
The derivative of f(x) = ax, where a is constant is x . ax-1
Answer:
False

Question 4.
The derivative of a polynomial is polynomial.
Answer:
True

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 5.
\(\frac{d}{d x}\left(10^{x}\right)=x \cdot 10^{x-1}\)
Answer:
False

Question 6.
If y = log x, then \(\frac{d y}{d x}=\frac{1}{x}\).
Answer:
True

Question 7.
If y = e2, then \(\frac{d y}{d x}\) = 2e.
Answer:
False

Question 8.
The derivative of ax is ax. log a.
Answer:
True

Question 9.
The derivative of xm . yn = (x + y)(m+n) is \(\frac{x}{y}\)
Answer:
False

(IV) Solve the following:

Question 1.
If y = (6x3 – 3x2 – 9x)10, find \(\frac{d y}{d x}\)
Solution:
Given y = (6x3 – 3x2 – 9x)10
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q1

Question 2.
If y = \(\sqrt[5]{\left(3 x^{2}+8 x+5\right)^{4}}\), find \(\frac{d y}{d x}\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 3.
If y = [log(log(log x))]2, find \(\frac{d y}{d x}\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q3.1

Question 4.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = 25 + 30x – x2.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q4

Question 5.
Find the rate of change of demand (x) of a commodity with respect to its price (y) if y = \(\frac{5 x+7}{2 x-13}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q5

Question 6.
Find \(\frac{d y}{d x}\) if y = xx.
Solution:
y = xx
∴ log y = log xx = x log x
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q6

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 7.
Find \(\frac{d y}{d x}\) if y = \(2^{x^{x}}\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q7

Question 8.
Find \(\frac{d y}{d x}\), if y = \(\sqrt{\frac{(3 x-4)^{3}}{(x+1)^{4}(x+2)}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q8
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q8.1

Question 9.
Find \(\frac{d y}{d x}\) if y = xx + (7x – 1)x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q9
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q9.1

Question 10.
If y = x3 + 3xy2 + 3x2y, find \(\frac{d y}{d x}\).
Solution:
y = x3 + 3xy2 + 3x2y
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q10

Question 11.
If x3 + y2 + xy = 7, find \(\frac{d y}{d x}\).
Solution:
x3 + y2 + xy = 7
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q11

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 12.
If x3y3 = x2 – y2, find \(\frac{d y}{d x}\).
Solution:
x3y3 = x2 – y2
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q12

Question 13.
If x7 . y9 = (x + y)16, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q13
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q13.1

Question 14.
If xa . yb = (x + y)a+b, then show that \(\frac{d y}{d x}=\frac{y}{x}\).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q14

Question 15.
Find \(\frac{d y}{d x}\) if x = 5t2, y = 10t.
Solution:
x = 5t2, y = 10t
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q15

Question 16.
Find \(\frac{d y}{d x}\) if x = e3t, y = \(e^{\sqrt{t}}\).
Solution:
x = e3t, y = \(e^{\sqrt{t}}\)
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q16

Question 17.
Differentiate log(1 + x2) with respect to ax.
Solution:
Let u = log(1 + x2) and v = ax
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q17

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 18.
Differentiate e(4x+5) with resepct to 104x.
Solution:
Let u = e(4x+5) and v = 104x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q18

Question 19.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = log x.
Solution:
y = log x
Differentiating w.r.t. x, we get
\(\frac{d y}{d x}=\frac{d}{d x}(\log x)=\frac{1}{x}\)
Differentiating again w.r.t. x, we get
\(\frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(\frac{1}{x}\right)=-\frac{1}{x^{2}}\)

Question 20.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = 2at, x = at2.
Solution:
x = at2, y = 2at
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q20

Question 21.
Find \(\frac{d^{2} y}{d x^{2}}\), if y = x2 . ex
Solution:
y = x2 . ex
Differentiating w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q21
= ex (2x + 2 + x2 + 2x)
= ex (x2 + 4x + 2).

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3

Question 22.
If x2 + 6xy + y2 = 10, then show that \(\frac{d^{2} y}{d x^{2}}=\frac{80}{(3 x+y)^{3}}\).
Solution:
x2 + 6xy + y2 = 10 ……..(1)
Differentiating both sides w.r.t. a, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q22
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q22.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q22.2

Question 23.
If ax2 + 2hxy + by2 = 0, then show that \(\frac{d^{2} y}{d x^{2}}\) = 0.
Solution:
ax2 + 2hxy + by2 = 0 ……..(1)
∴ ax2 + hxy + hxy + by2 = 0
∴ x(ax + hy) + y(hx + by) = 0
∴ x(ax + hy) = -y(hx + by)
∴ \(\frac{a x+h y}{h x+b y}=-\frac{y}{x}\) …….(2)
Differentiating (1) w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Miscellaneous Exercise 3 IV Q23

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 3 Exercise 3.6 Answers Maharashtra Board

Differentiation Class 12 Commerce Maths 1 Chapter 3 Exercise 3.6 Answers Maharashtra Board

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Std 12 Maths 1 Exercise 3.6 Solutions Commerce Maths

1. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = √x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 I Q1

Question 2.
y = x5
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 I Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6

Question 3.
y = x-7
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 I Q3

2. Find \(\frac{d^{2} y}{d x^{2}}\) if,

Question 1.
y = ex
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 II Q1

Question 2.
y = e(2x+1)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6

Question 3.
y = elog x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.6 II Q3

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 3 Exercise 3.5 Answers Maharashtra Board

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Std 12 Maths 1 Exercise 3.5 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
x = at2, y = 2at
Solution:
x = at2, y = 2at
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 I Q1

Question 2.
x = 2at2, y = at4
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 I Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5

Question 3.
x = e3t, y = e(4t+5)
Solution:
x = e3t, y = e(4t+5)
Differentiating x and y w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 I Q3

2. Find \(\frac{d y}{d x}\) if:

Question 1.
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\)
Solution:
x = \(\left(u+\frac{1}{u}\right)^{2}\), y = \((2)^{\left(u+\frac{1}{u}\right)}\) ……(1)
Differentiating x and y w.r.t. u, we get,
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 II Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 II Q1.1

Question 2.
x = \(\sqrt{1+u^{2}}\), y = log(1 + u2)
Solution:
x = \(\sqrt{1+u^{2}}\), y = log(1 + u2) ……(1)
Differentiating x and y w.r.t. u, we get,
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5

Question 3.
Differentiate 5x with respect to log x.
Solution:
Let u = 5x and v = log x
Then we want to find \(\frac{d u}{d v}\)
Differentiating u and v w.r.t. x, we get
\(\frac{d u}{d x}=\frac{d}{d x}\left(5^{x}\right)=5^{x} \cdot \log 5\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 II Q3

3. Solve the following:

Question 1.
If x = \(a\left(1-\frac{1}{t}\right)\), y = \(a\left(1+\frac{1}{t}\right)\), then show that \(\frac{d y}{d x}\) = -1
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q1

Question 2.
If x = \(\frac{4 t}{1+t^{2}}\), y = \(3\left(\frac{1-t^{2}}{1+t^{2}}\right)\), then show that \(\frac{d y}{d x}=-\frac{9 x}{4 y}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q2.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q2.2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5

Question 3.
If x = t . log t, y = tt, then show that \(\frac{d y}{d x}\) – y = 0.
Solution:
x = t log t
Differentiating w.r.t. t, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.5 III Q3.1

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 3 Exercise 3.4 Answers Maharashtra Board

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Std 12 Maths 1 Exercise 3.4 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
√x + √y = √a
Solution:
√x + √y = √a
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 I Q1

Question 2.
x3 + y3 + 4x3y = 0
Solution:
x3 + y3 + 4x3y = 0
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 I Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4

Question 3.
x3 + x2y + xy2 + y3 = 81
Solution:
x3 + x2y + xy2 + y3 = 81
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 I Q3

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y.ex + x.ey = 1
Solution:
y.ex + x.ey = 1
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 II Q1

Question 2.
xy = e(x-y)
Solution:
xy = e(x-y)
∴ log xy = log e(x-y)
∴ y log x = (x – y) log e
∴ y log x = x – y …..[∵ log e = 1]
∴ y + y log x = x
∴ y(1 + log x) = x
∴ y = \(\frac{x}{1+\log x}\)
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4

Question 3.
xy = log(xy)
Solution:
xy = log (xy)
∴ xy = log x + log y
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 II Q3

3. Solve the following:

Question 1.
If x5 . y7 = (x + y)12, then show that \(\frac{d y}{d x}=\frac{y}{x}\)
Solution:
x5 . y7 = (x + y)12
∴ log(x5 . y7) = log(x + y)12
∴ log x5 + log y7 = log(x + y)12
∴ 5 log x + 7 log y = 12 log (x + y)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 III Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 III Q1.1

Question 2.
If log(x + y) = log(xy) + a, then show that \(\frac{d y}{d x}=\frac{-y^{2}}{x^{2}}\)
Solution:
log (x + y) = log (xy) + a
∴ log(x + y) = log x + log y + a
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 III Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4

Question 3.
If ex + ey = e(x+y), then show that \(\frac{d y}{d x}=-e^{y-x}\).
Solution:
ex + ey = e(x+y) ……….(1)
Differentiating both sides w.r.t. x, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 III Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.4 III Q3.1

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 3 Exercise 3.3 Answers Maharashtra Board

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Std 12 Maths 1 Exercise 3.3 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(x^{x^{2 x}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 I Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 I Q1.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 I Q1.2

Question 2.
y = \(x^{e^{x}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 I Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3

Question 3.
y = \(e^{x^{x}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 I Q3

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(\left(1+\frac{1}{x}\right)^{x}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 II Q1

Question 2.
y = (2x + 5)x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3

Question 3.
y = \(\sqrt[3]{\frac{(3 x-1)}{(2 x+3)(5-x)^{2}}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 II Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 II Q3.1

3. Find \(\frac{d y}{d x}\) if:

Question 1.
y = (log x)x + xlog x
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 III Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 III Q1.1

Question 2.
y = xx + ax
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 III Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3

Question 3.
y = \(10^{x^{x}}+10^{x^{10}}+10^{10^{x}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 III Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.3 III Q3.1

12th Commerce Maths Digest Pdf

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Std 12 Maths 1 Exercise 3.2 Solutions Commerce Maths

1. Find the rate of change of demand (x) of a commodity with respect to price (y) if:

Question 1.
y = 12 + 10x + 25x2
Solution:
Given y = 12 + 10x + 25x2
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 I Q1
Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{1}{10+50 x}\)

Question 2.
y = 18x + log(x – 4)
Solution:
Given y = 18x + log (x – 4)
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 I Q2
Hence, the rate of change of demand (x) with respect to price (y) \(=\frac{d x}{d y}=\frac{x-4}{18 x-71}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2

Question 3.
y = 25x + log(1 + x2)
Solution:
Given y = 25x + log(1 + x2)
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 I Q3
Hence, the rate of change of demand (x) with respect to price (y) \(\frac{d x}{d y}=\frac{1+x^{2}}{25 x^{2}+2 x+25}\)

2. Find the marginal demand of a commodity where demand is x and price is y.

Question 1.
y = xe-x + 7
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 II Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 II Q1.1

Question 2.
y = \(\frac{x+2}{x^{2}+1}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2

Question 3.
y = \(\frac{5 x+9}{2 x-10}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.2 II Q3

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Std 12 Maths 1 Exercise 3.1 Solutions Commerce Maths

1. Find \(\frac{d y}{d x}\) if,

Question 1.
y = \(\sqrt{x+\frac{1}{x}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 I Q1
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 I Q1.1

Question 2.
y = \(\sqrt[3]{a^{2}+x^{2}}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 I Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1

Question 3.
y = (5x3 – 4x2 – 8x)9
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 I Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 I Q3.1

2. Find \(\frac{d y}{d x}\) if:

Question 1.
y = log(log x)
Solution:
Given y = log(log x)
Let u = log x
Then y = log u
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 II Q1

Question 2.
y = log(10x4 + 5x3 – 3x2 + 2)
Solution:
Given y = log(10x4 + 5x3 – 3x2 + 2)
Let u = 10x4 + 5x3 – 3x2 + 2
Then y = log u
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 II Q2

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1

Question 3.
y = log(ax2 + bx + c)
Solution:
Given y = log(ax2 + bx + c)
Let u = ax2 + bx + c
Then y = log u
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 II Q3

3. Find \(\frac{d y}{d x}\) if:

Question 1.
y = \(e^{5 x^{2}-2 x+4}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 III Q1

Question 2.
y = \(a^{(1+\log x)}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 III Q2
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 III Q2.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1

Question 3.
y = \(5^{(x+\log x)}\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 3 Differentiation Ex 3.1 III Q3

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Std 12 Maths 1 Miscellaneous Exercise 2 Solutions Commerce Maths

(I) Choose the correct alternative.

Question 1.
If AX = B, where A = \(\left[\begin{array}{cc}
-1 & 2 \\
2 & -1
\end{array}\right]\), B = \(\left[\begin{array}{l}
1 \\
1
\end{array}\right]\) then X = ___________
(a) \(\left[\begin{array}{l}
\frac{3}{5} \\
\frac{3}{7}
\end{array}\right]\)
(b) \(\left[\begin{array}{l}
\frac{7}{3} \\
\frac{5}{3}
\end{array}\right]\)
(c) \(\left[\begin{array}{l}
1 \\
1
\end{array}\right]\)
(d) \(\left[\begin{array}{l}
1 \\
2
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{l}
1 \\
1
\end{array}\right]\)

Question 2.
The matrix \(\left[\begin{array}{lll}
8 & 0 & 0 \\
0 & 8 & 0 \\
0 & 0 & 8
\end{array}\right]\) is ___________
(a) identity matrix
(b) scalar matrix
(c) null matrix
(d) diagonal matrix
Answer:
(b) scalar matrix

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 3.
The matrix \(\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]\) is ___________
(a) identity matrix
(b) diagonal matrix
(c) scalar matix
(d) null matrix
Answer:
(d) null matrix

Question 4.
If A = \(\left[\begin{array}{lll}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a
\end{array}\right]\), then |adj A| = ___________
(a) a12
(b) a9
(c) a6
(d) a-3
Answer:
(c) a6
Hint:
adj A = \(\left[\begin{array}{ccc}
a^{2} & 0 & 0 \\
0 & a^{2} & 0 \\
0 & 0 & a^{2}
\end{array}\right]\)
∴ |adj A| = a2(a4 – 0) = a6

Question 5.
Adjoint of \(\left[\begin{array}{ll}
2 & -3 \\
4 & -6
\end{array}\right]\) is ___________
(a) \(\left[\begin{array}{ll}
-6 & 3 \\
-4 & 2
\end{array}\right]\)
(b) \(\left[\begin{array}{cc}
6 & 3 \\
-4 & 2
\end{array}\right]\)
(c) \(\left[\begin{array}{cc}
-6 & -3 \\
4 & 2
\end{array}\right]\)
(d) \(\left[\begin{array}{cc}
-6 & 3 \\
4 & -2
\end{array}\right]\)
Answer:
(a) \(\left[\begin{array}{ll}
-6 & 3 \\
-4 & 2
\end{array}\right]\)

Question 6.
If A = diag. [d1, d2, d3, …, dn], where d1 ≠ 0, for i = 1, 2, 3, …….., n, then A-1 = ___________
(a) diag [1/d1, 1/d2, 1/d3, …, 1/dn]
(b) D
(c) I
(d) O
Answer:
(a) diag [1/d1, 1/d2, 1/d3, …, 1/dn]

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 7.
If A2 + mA + nI = O and n ≠ 0, |A| ≠ 0, then A-1 = ___________
(a) \(\frac{-1}{m}\)(A + nI)
(b) \(\frac{-1}{n}\)(A + mI)
(c) \(\frac{-1}{m}\)(I + mA)
(d) (A + mnI)
Answer:
(b) \(\frac{-1}{n}\)(A + mI)
Hint:
A2 + mA + nI = 0
∴ (A2 + mA + nI) . A-1 = 0 . A-1
∴ A(AA-1) + m(AA-1) + nIA-1 = 0
∴ AI + mI + nA-1 = 0
∴ nA-1 = -A – mI
∴ A-1 = \(\frac{-1}{n}\)(A + mI)

Question 8.
If a 3 × 3 matrix B has its inverse equal to B, then B2 = ___________
(a) \(\left[\begin{array}{lll}
0 & 1 & 1 \\
0 & 1 & 0 \\
1 & 0 & 1
\end{array}\right]\)
(b) \(\left[\begin{array}{lll}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 0 & 1
\end{array}\right]\)
(c) \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{array}\right]\)
(d) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Answer:
(d) \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
Hint:
B-1 = B
∴ B2 = B.B-1 = I

Question 9.
If A = \(\left[\begin{array}{cc}
\alpha & 4 \\
4 & \alpha
\end{array}\right]\) and |A3| = 729 then α = ___________
(a) ±3
(b) ±4
(c ) ±5
(d) ±6
Answer:
(c ) ±5
Hint:
|A|= \(\left|\begin{array}{ll}
\alpha & 4 \\
4 & \alpha
\end{array}\right|\) = α2 – 16
∴ |A3| = |A|3 = (α2 – 16)3 = 729
∴ α2 – 16 = 9
∴ α2 = 25
∴ α = ±5

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 10.
If A and B square matrices of order n × n such that A2 – B2 = (A – B)(A + B), then which of the following will be always true?
(a) AB = BA
(b) either A or B is a zero matrix
(c) either of A and B is an identity matrix
(d) A = B
Answer:
(a) AB = BA
Hint:
A2 – B2 = (A – B)(A + B)
∴ A2 – B2 = A2 + AB – BA – B2
∴ 0 = AB – BA
∴ AB = BA

Question 11.
If A = \(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\) then A-1 = ___________
(a) \(\left[\begin{array}{rr}
3 & -5 \\
1 & 2
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
3 & -5 \\
-1 & 2
\end{array}\right]\)
(c) \(\left[\begin{array}{rr}
3 & 5 \\
-1 & 2
\end{array}\right]\)
(d) \(\left[\begin{array}{ll}
3 & -5 \\
1 & -2
\end{array}\right]\)
Answer:
(b) \(\left[\begin{array}{rr}
3 & -5 \\
-1 & 2
\end{array}\right]\)

Question 12.
If A is a 2 × 2 matrix such that A(adj A) = \(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]\), then |A| = ___________
(a) 0
(b) 5
(c) 10
(d) 25
Answer:
(b) 5
Hint:
A(adj A) = |A|.I

Question 13.
If A is a non-singular matrix, then det(A-1) = ___________
(a) 1
(b) 0
(c) det(A)
(d) 1/det(A)
Answer:
(d) 1/det(A)
Hint:
AA-1 = I
∴ |A|.|A-1| = 1
∴ |A-1| = \(\frac{1}{|\mathrm{~A}|}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 14.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-3 & -1
\end{array}\right]\), B = \(\left[\begin{array}{rr}
-1 & 0 \\
1 & 5
\end{array}\right]\) then AB = ___________
(a) \(\left[\begin{array}{rr}
1 & -10 \\
1 & 20
\end{array}\right]\)
(b) \(\left[\begin{array}{rr}
1 & 10 \\
-1 & 20
\end{array}\right]\)
(c) \(\left[\begin{array}{ll}
1 & 10 \\
1 & 20
\end{array}\right]\)
(d) \(\left[\begin{array}{rr}
1 & 10 \\
-1 & -20
\end{array}\right]\)
Answer:
(c) \(\left[\begin{array}{ll}
1 & 10 \\
1 & 20
\end{array}\right]\)

Question 15.
If x + y + z = 3, x + 2y + 3z = 4, x + 4y + 9z = 6, then (y, z) = ___________
(a) (-1, 0)
(b) (1, 0)
(c) (1, -1)
(d) (-1, 1)
Answer:
(b) (1, 0)

(II) Fill in the blanks:

Question 1.
A = \(\left[\begin{array}{l}
3 \\
1
\end{array}\right]\) is ___________ matrix.
Answer:
column

Question 2.
Order of matrix \(\left[\begin{array}{lll}
2 & 1 & 1 \\
5 & 1 & 8
\end{array}\right]\) is ___________
Answer:
2 × 3

Question 3.
If A = \(\left[\begin{array}{ll}
4 & x \\
6 & 3
\end{array}\right]\) is a singular matrix, then x is ___________
Answer:
2

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 4.
Matrix B = \(\left[\begin{array}{ccc}
0 & 3 & 1 \\
-3 & 0 & -4 \\
p & 4 & 0
\end{array}\right]\) is a skew-symmetric, then value of p is ___________
Answer:
-1

Question 5.
If A = [aij]2×3 and B = [bij]m×1, and AB is defined, then m = ___________
Answer:
3

Question 6.
If A = \(\left[\begin{array}{cc}
3 & -5 \\
2 & 5
\end{array}\right]\), then cofactor of a12 is ___________
Answer:
-2

Question 7.
If A = [aij]m×m is non-singular matrix, then A-1 = \(\frac{1}{\ldots \ldots}\) adj (A).
Answer:
|A|

Question 8.
(AT)T = ___________
Answer:
A

Question 9.
If A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right]\) and A-1 = \(\left[\begin{array}{cc}
1 & -1 \\
x & 2
\end{array}\right]\), then x = ___________
Answer:
-1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 10.
If a1x+ b1y = c1 and a2x + b2y = c2, then matrix form is \(\left[\begin{array}{cc}
\cdots & \ldots \\
\cdots & \ldots
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
\ldots \\
\cdots
\end{array}\right]\)
Answer:
\(\left[\begin{array}{ll}
a_{1} & b_{1} \\
a_{2} & b_{2}
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
c_{1} \\
c_{2}
\end{array}\right]\)

(III) State whether each of the following is True or False:

Question 1.
Single element matrix is row as well as a column matrix.
Answer:
True

Question 2.
Every scalar matrix is unit matrix.
Answer:
False

Question 3.
A = \(\left[\begin{array}{ll}
4 & 5 \\
6 & 1
\end{array}\right]\) is non-singular matrix.
Answer:
True

Question 4.
If A is symmetric, then A = -AT.
Answer:
False

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 5.
If AB and BA both exist, then AB = BA.
Answer:
False

Question 6.
If A and B are square matrices of same order, then (A + B)2 = A2 + 2AB + B2.
Answer:
False

Question 7.
If A and B are conformable for the product AB, then (AB)T = ATBT.
Answer:
False

Question 8.
Singleton matrix is only row matrix.
Answer:
False

Question 9.
A = \(\left[\begin{array}{cc}
2 & 1 \\
10 & 5
\end{array}\right]\) is invertible matrix.
Answer:
False

Question 10.
A(adj A) = |A| I, where I is the unit matrix.
Answer:
True.

(IV) Solve the following:

Question 1.
Find k, if \(\left[\begin{array}{ll}
7 & 3 \\
5 & k
\end{array}\right]\) is a singular matrix.
Solution:
Let A = \(\left[\begin{array}{ll}
7 & 3 \\
5 & k
\end{array}\right]\)
Since, A is singular matrix, |A| = 0
∴ \(\left|\begin{array}{ll}
7 & 3 \\
5 & k
\end{array}\right|\) = 0
∴ 7k – 15 = 0
∴ k = \(\frac{15}{7}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 2.
Find x, y, z if \(\left[\begin{array}{lll}
2 & x & 5 \\
3 & 1 & z \\
y & 5 & 8
\end{array}\right]\) is a symmetric matrix.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q2
By equality of matrices,
x = 3, y = 5 and z = 5.

Question 3.
If A = \(\left[\begin{array}{ll}
1 & 5 \\
7 & 8 \\
9 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
2 & 4 \\
1 & 5 \\
-8 & 6
\end{array}\right]\), C = \(\left[\begin{array}{cc}
-2 & 3 \\
1 & -5 \\
7 & 8
\end{array}\right]\) then show that (A + B) + C = A + (B + C).
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q3.1
From (1) and (2),
(A + B) + C = A + (B + C).

Question 4.
If A = \(\left[\begin{array}{ll}
2 & 5 \\
3 & 7
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 7 \\
-3 & 0
\end{array}\right]\), find the matrix A – 4B + 7I, where I is the unit matrix of order 2.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q4

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 5.
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 + 2BT)T = AT + 2B
(ii) (3A – 5BT)T = 3AT – 5B
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q5
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q5.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q5.2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q5.3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q5.4

Question 6.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 6 \\
1 & 2 & 3
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-3 & 2 & -1 \\
-2 & 1 & 0
\end{array}\right]\) thenshow that AB and BA are both singular matrices.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q6
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q6.1
∴ BA is also a singular matrix.
Hence, AB and BA are both singular matrices.

Question 7.
If A = \(\left[\begin{array}{ll}
3 & 1 \\
1 & 5
\end{array}\right]\), B = \(\left[\begin{array}{cc}
1 & 2 \\
5 & -2
\end{array}\right]\), verify |AB| = |A| |B|.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q7

Question 8.
If A = \(\left[\begin{array}{cc}
2 & -1 \\
-1 & 2
\end{array}\right]\), then show that A2 – 4A + 3I = 0.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q8

Question 9.
If A = \(\left[\begin{array}{cc}
-3 & 2 \\
2 & 4
\end{array}\right]\), B = \(\left[\begin{array}{ll}
1 & a \\
b & 0
\end{array}\right]\) and (A + B)(A – B) = A2 – B2, find a and b.
Solution:
(A + B)(A – B) = A2 – B2
∴ A2 – AB + BA – B2 = A2 – B2
∴ -AB + BA = 0
∴ AB = BA
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q9
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q9.1
By equality of matrices,
-3 + 2b = -3 + 2a ……..(1)
-3a = 2 + 4a ……..(2)
2 + 4b = -3b ……..(3)
2a = 2b ……..(4)
From (2), 7a = -2
∴ a = \(\frac{-2}{7}\)
From (3), 7b = -2
∴ b = \(\frac{-2}{7}\)
These values of a and b also satisfy equations (1) and (4).
Hence, a = \(\frac{-2}{7}\) and b = \(\frac{-2}{7}\)

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 10.
If A = \(\left[\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right]\), then find A3.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q10

Question 11.
Find x, y, z if \(\left\{5\left[\begin{array}{ll}
0 & 1 \\
1 & 0 \\
1 & 1
\end{array}\right]-\left[\begin{array}{cc}
2 & 1 \\
3 & -2 \\
1 & 3
\end{array}\right]\right\}\left[\begin{array}{l}
2 \\
1
\end{array}\right]=\left[\begin{array}{c}
x-1 \\
y+1 \\
2 z
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q11
∴ By equality of matrices, we get
x – 1 = 0 ∴ x = 1
y + 1 = 6 ∴ y = 5
2z = 10 ∴ z = 5

Question 12.
If A = \(\left[\begin{array}{cc}
2 & -4 \\
3 & -2 \\
0 & 1
\end{array}\right]\), B = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 1 & 0
\end{array}\right]\) then showthat(AB)T = BTAT.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q12
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q12.1

Question 13.
If A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\), then reduce it to unit matrix by row tranformation.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q13
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q13.1

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 14.
Two farmers Shantaram and Kantaram cultivate three crops rice, wheat, and groundnut. The sale (in ₹) of these crops by both the farmers for the month of April and May 2016 is given below:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q14
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:
The given information can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q14.1
(i) The total sale in ₹ for two months of each farmer for each crop can be obtained by the addition A + B.
Now, A + B
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q14.2
∴ total sale in ₹ for two months of each farmer for each crop is given by
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q14.3
Hence, the total sale for Shantaram are ₹ 33000 for Rice, ₹ 28000 for Wheat, ₹ 24000 for Groundnut, and for Kantaram are ₹ 39000 for Rice, ₹ 31500 for Wheat, ₹ 24000 for Groundnut.
(ii) The increase in sales from April to May for every crop of each farmer can be obtained by the subtraction of A from B.
Now, B – A
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q14.4
Hence, the increase in sales from April to May of Shantam is ₹ 3000 in Rice, ₹ 2000 in Wheat, nothing in Groundnut and of Kantaram are ₹ 3000 in Rice, ₹ 1500 in Wheat, ₹ 8000 in Groundnut.

Question 15.
Check whether following matrices are invertible or not:
(i) \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
Then |A| = \(\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right|\)
= 1 – 0
= 1 ≠ 0
∴ A is a non-singular matrix.
Hence, A-1 exists.

(ii) \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right]\)
Then |A| = \(\left|\begin{array}{ll}
1 & 1 \\
1 & 1
\end{array}\right|\)
= 1 – 1
= 0
∴ A is a singular matrix.
Hence, A-1 does not exist.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

(iii) \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
Then |A| = \(\left|\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right|\)
= 3(5 – 0) – 4(5 – 0) + 3(4 – 1)
= 15 – 20 + 9
= 4 ≠ 0
∴ A is a non-singular matrix.
Hence, A-1 exists.

(iv) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
\end{array}\right]\)
Then |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & 4 & 5 \\
2 & 4 & 6
\end{array}\right|\)
= 1(24 – 20) – 2(12 – 10) + 3(8 – 8)
= 4 – 4 + 0
= 0
∴ A is a singular matrix.
Hence, A-1 does not exist.

Question 16.
Find inverse of the following matrices (if they exist) by elementary transformation:
(i) \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (i).1

(ii) \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution:

(iii) \(\left[\begin{array}{ccc}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iii).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

(iv) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iv).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q16 (iv).2

Question 17.
Find the inverse of \(\left[\begin{array}{lll}
3 & 1 & 5 \\
2 & 7 & 8 \\
1 & 2 & 5
\end{array}\right]\) by adjoint method.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q17
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q17.1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q17.2

Question 18.
Solve the following equations by method of inversion:
(i) 4x – 3y – 2 = 0, 3x – 4y + 6 = 0
Solution:
The given equations are
4x – 3y = 2
3x – 4y = -6
These equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (i).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (i).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (i).3
By equality of matrices,
x = \(\frac{26}{7}\), y = \(\frac{30}{7}\) is the required solution.

(ii) x + y – z = 2, x – 2y + z = 3 and 2x – y – 3z = -1
Solution:
The given equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (ii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (ii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (ii).3
By equality of matrices,
x = 3, y = 1, z = 2 is the required solution.

(iii) x – y + z = 4, 2x + y – 3z = 0 and x + y + z = 2
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (iii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (iii).3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q18 (iii).4

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 19.
Solve the following equations by method of reduction:
(i) 2x + y = 5, 3x – 5y = -3
Solution:
The given equation can be written in matrix form as:
\(\left[\begin{array}{ll}
2 & 1 \\
3 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{r}
5 \\
-3
\end{array}\right]\)
By R2 – 5R1, we get
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (i)
By equality of matrices,
2x + y = 5 …….(1)
-7x = -28 ……(2)
From (2), x = 4
Substituting x = 4 in (1), we get
2(4) + y = 5
∴ y = -3
Hence, x = 4 and y = -3 is the required solution.

(ii) x + 2y + z = 3, 3x – y + 2z = 1 and 2x – 3y + 3z = 2
Solution:
The given equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (ii)
By equality of matrices,
x + 2y + z = 3 …….(1)
-7y – z = -8 …….(2)
2z = 4 …….(3)
From (3), z = 2
Substituting z = 2 in (2), we get
-7y – 2 = -8
∴ -7y = -6
∴ y = \(\frac{6}{7}\)
Substituting y = \(\frac{6}{7}\), z = 2 in (1), we get
x + 2(\(\frac{6}{7}\)) + 2 = 3
x = 3 – 2 – \(\frac{12}{7}\) = \(\frac{-5}{7}\)
Hence, x = \(\frac{-5}{7}\), y = \(\frac{6}{7}\) and z = 2 is the required solution.

(iii) x – 3y + z = 2, 3x + y + z = 1 and 5x + y + 3z = 3.
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (iii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q19 (iii).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2

Question 20.
The sum of three numbers is 6. If we multiply the third number by 3 and add it to the second number, we get 11. By adding first and third numbers we get a number that is double the second number. Use this information and find a system of linear equations. Find the three numbers using matrices.
Solution:
Let the three numbers be x, y, and z.
According to the given condition,
x + y + z = 6
3z + y = 11, i.e. y + 3z = 11
and x + z = 2y, i.e. x – 2y + z = 0
Hence, the system of linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Miscellaneous Exercise 2 IV Q20
By equality of matrices,
x + y + z = 6 …….(1)
y + 3z = 11 ………(2)
-3y = -6 ………(3)
From (3), y = 2
Substituting y = 2 in (2), we get
2 + 3z = 11
∴ 3z = 9
∴ z = 3
Substituting y = 2, z = 3 in (1), we get
x + 2 + 3 = 6
∴ x = 1
∴ x = 1, y = 2, z = 3
Hence, the required numbers are 1, 2 and 3.

12th Commerce Maths Digest Pdf

12th Commerce Maths 1 Chapter 2 Exercise 2.6 Answers Maharashtra Board

Matrices Class 12 Commerce Maths 1 Chapter 2 Exercise 2.6 Answers Maharashtra Board

Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 2 Matrices Ex 2.6 Questions and Answers.

Std 12 Maths 1 Exercise 2.6 Solutions Commerce Maths

Question 1.
Solve the following equations by the method of inversion:
(i) x + 2y = 2, 2x + 3y = 3
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (i).2
By equality of matrices,
x = 0, y = 1 is the required solution.

(ii) 2x + y = 5, 3x + 5y = -3
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (ii).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

(iii) 2x – y + z = 1, x + 2y + 3z = 8 and 3x + y – 4z = 1
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).3
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).4
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iii).5
By equality of matrices,
x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 1, x – y + z = 2 and x + y – z = 3
Solution:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).1
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).2
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q1 (iv).3

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 2.
Express the following equations in matrix form and solve them by method of reduction:
(i) x + 3y = 2, 3x + 5y = 4.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (i)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (i).1
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

(ii) 3x – y = 1, 4x + y = 6.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (ii)
By equality of matrices,
12x – 4y = 4 …..(1)
7y = 14 …..(2)
From (2), y = 2
Substituting y = 2 in (1), we get
12x – 8 = 4
∴ 12x = 12
∴ x = 1
Hence, x = 1, y = 2 is the required solution.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

(iii) x + 2y + z = 8, 2x + 3y – z = 11 and 3x – y – 2z = 5.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iii)
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iii).1
By equality of matrices,
x + 2y + z = 8 ……(1)
-y – 3z = -5 …….(2)
16z = 16 ……….(3)
From (3), z = 1
Substituting z = 1 in (2), we get
-y – 3= -5
∴ y = 2
Substituting y = 2, z = 1 in (1), we get
x + 4 + 1 = 8
∴ x = 3
Hence, x = 3, y = 2, z = 1 is the required solution.

(iv) x + y + z = 1, 2x + 3y + 2z = 2 and x + y + 2z = 4.
Solution:
The given equations can be written in the matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q2 (iv)
By equality of matrices,
x + y + z = 1 ……(1)
y = 0
z = 3
Substituting y = 0, z = 3 in (1), we get
x + 0 + 3 = 1
∴ x = -2
Hence, x = -2, y = 0, z = 3 is the required solution.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 3.
The total cost of 3 T.V. and 2 V.C.R. is ₹ 35000. The shopkeeper wants a profit of ₹ 1000 per T.V. and ₹ 500 per V.C.R. He sells 2 T.V. and 1 V.C.R. and he gets total revenue of ₹ 21500. Find the cost and selling price of T.V. and V.C.R.
Solution:
Let the cost of each T.V. be ₹ x and each V.C.R. be ₹ y.
Then the total cost of 3 T.V. and 2 V.C.R. is ₹ (3x + 2y) which is given to be ₹ 35000.
∴ 3x + 2y = 35000
The shopkeeper wants a profit of ₹ 1000 per T.V. and ₹ 500 per V.C.R.
The selling price of each T.V. is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. and 1 V.C.R is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21500.
∴ 2(x + 1000) + (y + 500) = 21500
∴ 2x + 2000 + y + 500 = 21500
∴ 2x + y = 19000
Hence, the system of linear equations is
3x + 2y = 35000
2x + y = 19000
The equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q3
By equality of matrices,
-x = -3000 …….(1)
2x + y = 19000 ……….(2)
From (1), x = 3000
Substituting x = 3000 in (2), we get
2(3000) + y = 19000
∴ y = 19000 – 6000 = 13000
Hence, the cost price of one T.V. is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. is ₹ 4000 and of one V.C.R. is ₹ 13500.

Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6

Question 4.
The sum of the cost of one Economics book, one Cooperation book, and one Account book is ₹ 420. The total cost of an Economic book, 2 Cooperation books, and an Account book is ₹ 480. Also, the total cost of an Economic book, 3 Cooperation books, and 2 Account books is ₹ 600. Find the cost of each book.
Solution:
Let the cost of 1 Economic book, 1 Cooperation book and 1 Account book be ₹ x, ₹ y and ₹ z respectively.
Then, from the given information
x + y + z = 420
x + 2y + z = 480
x + 3y + 2z = 600
These equations can be written in matrix form as:
Maharashtra Board 12th Commerce Maths Solutions Chapter 2 Matrices Ex 2.6 Q4
By equality of matrices,
x + y + z = 420 …….(1)
y = 60
2y + z = 180 ………(2)
Substituting y = 60 in (2), we get
2(60) + z = 180
∴ z = 180 – 120 = 60
Substituting y = 60, z = 60 in (1), we get
x + 60 + 60 = 420
∴ x = 420 – 120 = 300
Hence, the cost of each Economic book is ₹ 300, each Cooperation book is ₹ 60 and each Account book is ₹ 60.

12th Commerce Maths Digest Pdf