Class 12

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Exercise 3.2 Questions And Answers Maharashtra Board

Question 1.
Find the Cartesian co-ordinates of the point whose polar co-ordinates are:
(i) \(\left(\sqrt{2}, \frac{\pi}{4}\right)\)
Solution:
Here, r = \(\sqrt {2}\) and θ = \(\frac{\pi}{4}\)
Let the cartesian coordinates be (x, y)
Then, x = rcosθ = \(\sqrt {2}\)cos\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
y = rsinθ = \(\sqrt {2}\)sin\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1
∴ the cartesian coordinates of the given point are (1, 1).

(ii) \(\left(4, \frac{\pi}{2}\right)\)
Solution:

(iii) \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\)
Solution:
Here, r = \(\frac{3}{4}\) and θ = \(\frac{3 \pi}{4}\)
Let the cartesian coordinates be (x, y)

(iv) \(\left(\frac{1}{2}, \frac{7 \pi}{3}\right)\)
Solution:
Here, r = \(\frac{1}{2}\) and θ = \(\frac{7 \pi}{4}\)
Let the cartesian coordinates be (x, y)


∴ the cartesian coordinates of the given point are \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)

Question 2.
Find the of the polar co-ordinates point whose Cartesian co-ordinates are.
(i) \((\sqrt{2}, \sqrt{2})\)
Solution:
Here x = \(\sqrt {2}\) and y = \(\sqrt {2}\)
∴ the point lies in the first quadrant.
Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (\(\sqrt {2}\) )² + (\(\sqrt {2}\) )² = 2 + 2 = 4
∴ r = 2 … [∵ r > 0]
cos θ = \(\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
and sin θ = \(\frac{y}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)
∴ tan θ = 1
Since the point lies in the first quadrant and
0 ≤ θ ≤ 2π, tan θ = 1 = tan\(\frac{\pi}{4}\)
∴ θ = \(\frac{\pi}{4}\)
∴ the polar coordinates of the given point are \(\left(2, \frac{\pi}{4}\right)\).

(ii) \(\left(0, \frac{1}{2}\right)\)
Solution:
Here x = 0 and y = \(\frac{1}{2}\)
the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)
Then, r² = x² + y² = (0)² + \(\left(\frac{1}{2}\right)^{2}=0+\frac{1}{4}=\frac{1}{4}\)
∴ r = \(\frac{1}{2}\) …[∵ r > 0]
cosθ = \(\frac{x}{r}=\frac{0}{(1 / 2)}\) = 0
and sin θ = \(\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}\) = 1
Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π
cosθ = 0 = cos\(\frac{\pi}{2}\) and sinθ = 1 = sin\(\frac{\pi}{2}\)
∴ θ = \(\frac{\pi}{2}\)
∴ the polar coordinates of the given point are \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\).

(iii) \((1,-\sqrt{3})\)
Solution:
Here x = 1 and y = \(-\sqrt{3}\)
∴ the point lies in the fourth quadrant.
Let the polar coordinates be (r, θ).
Then, r² = x² + y² = (1)² + (\(-\sqrt {3}\) )² = 1 + 3 = 4
∴ r = 2 … [∵ r > 0]

∴ the polar coordinates of the given point are \(\left(2, \frac{5 \pi}{3}\right)\).

(iv) \(\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\)
Solution:

Question 3.
In ∆ABC, if ∠A = 45º, ∠B = 60º then find the ratio of its sides.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\)
∴ \(\frac{a}{b}=\frac{\sin A}{\sin B}\) and \(\frac{b}{c}=\frac{\sin B}{\sin C}\)
∴ a : b : c = sinA : sinB : sinC
Given ∠A = 45° and ∠B = 60°
∵ ∠A + ∠B + ∠C = 180°
∴ 45° + 60° + ∠C = 180°
∴ ∠C = 180° – 105° = 75°

Question 4.
In ∆ABC, prove that sin \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)=\left(\frac{\boldsymbol{b}-\boldsymbol{c}}{a}\right)\) cos \(\frac{A}{2}\).
Solution:
By the sine rule,

Question 5.
With usual notations prove that 2 \(\left\{a \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{A}{2}\right\}\) = a – b + c.
Solution:

Question 6.
In ∆ABC, prove that a³sin(B – C) + b³sin(C – A) + c³sin(A – B) = 0
Solution:
By the sine rule,
\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a³sin (B – C) + b³sin (C – A) + c³sin (A – B)
= a³(sin B cos C – cos B sin C) + b³(sinCcos A – cos C sin A) + c³(sinAcosB – cos A sin B)

= \(\frac{1}{2 k}\) [a²(a² + b² – c²) – a²(a² + c² – b²) + b²(b² + c² – a²) – b²(a² + b² – c²) + c²(c² + a² – b²) – c²(b² + c² – a²)]
= \(\frac{1}{2 k}\) [a⁴ + a²b² – a²c² – a⁴ – a²c² + a²b² + b⁴ + b²c² – a²b² – a²b² – b⁴ + b²c² + c⁴ + a²c² – b²c² – b²c² – c⁴ + a²c²]
= \(\frac{1}{2 k}\)(0) = 0 = RHS.

Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a², b², c² are also in A.P
Solution:
By the sine rule,
\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) =\(\frac{\sin \mathrm{C}}{c}\) = k
∴ sin A = ka, sin B = kb, sin C = kc …(1)
Now, cot A, cotB, cotC are in A.P.
∴ cotC – cotB = cotB – cot A
∴ cotA + cotC = 2cotB

Question 8.
In ∆ABC, if a cos A = b cos B then prove that the triangle is right angled or an isosceles traingle.
Solution:
By the sine rule,
\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = k
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sin A cos A = k sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B ∴ sin 2A – sin 2B = 0
∴ 2 cos (A + B)∙sin (A -B) = 0
∴ 2cos (π – C)∙sin(A – B) = 0 … [∵ A + B + C = π]
∴ -2 cos C∙sin (A – B) = 0
∴ cos C = 0 OR sin(A -B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.

Question 9.
With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a² + b² + c².
Solution:
LHS = 2 (bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
= 2bc \(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\) + 2ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) + 2ab\(\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\) …(By cosine rule]
= b² + c² – a² + c² + a² – b² + a² + b² – c² = a² + b² + c² = RHS.

Question 10.
In △ABC, if a = 18, b = 24, c = 30 then find the values of
(i) cos A
Solution:
Given : a = 18, b = 24 and c = 30
∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36

(ii) sin\(\frac{A}{2}\)
Solution:

(iii) cos\(\frac{A}{2}\)
Solution:

(iv) tan\(\frac{A}{2}\)
Solution:

(v) A(△ABC)
Solution:

(iv) sin A.
Solution:

Question 11.
In △ABC prove that (b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan\(\frac{B}{2}\) = (a + b – c) tan\(\frac{C}{2}\).
Solution:
(b + c – a) tan \(\frac{A}{2}\)

Question 12.
In △ABC prove that sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\) = \(\frac{[A(\triangle A B C)]^{2}}{a b c s}\)
Solution:
LHS = sin \(\frac{A}{2}\)∙sin \(\frac{B}{2}\)∙sin \(\frac{C}{2}\)

Class 12 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 12 Maths 1
• Exercise 3.2 Class 12 Maths 1
• Exercise 3.3 Class 12 Maths 1
• Miscellaneous Exercise 3 Class 12 Maths 1
• Exercise 4.1 Class 12 Maths 1
• Exercise 4.2 Class 12 Maths 1
• Exercise 4.3 Class 12 Maths 1
• Miscellaneous Exercise 4 Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.1 Questions and Answers.

12th Maths Part 1 Trigonometric Functions Exercise 3.1 Questions And Answers Maharashtra Board

Question 1.
Find the principal solutions of the following equations :
(i) cos θ= \(\frac{1}{2}\)
Solution:
We know that, cos\(\frac{\pi}{3}\) = \(\frac{1}{2}\) and cos (2π – θ) = cos θ
∴ cos\(\frac{\pi}{3}\) = cos(2π – \(\frac{\pi}{3}\)) = cos\(\frac{5 \pi}{3}\)
∴ cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\) = \(\frac{1}{2}\), where
0 < \(\frac{\pi}{3}\) < 2π and 0 < \(\frac{5 \pi}{3}\) < 2π
∴ cos θ = \(\frac{1}{2}\) gives cos θ = cos\(\frac{\pi}{3}\) = cos\(\frac{5 \pi}{3}\)
∴ θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)
Hence, the required principal solutions are
θ = \(\frac{\pi}{3}\) and θ = \(\frac{5 \pi}{3}\)

(ii) sec θ = \(\frac{2}{\sqrt{3}}\)
Solution:

(iii) cot θ = \(\sqrt {3}\)
Solution:
The given equation is cot θ = \(\sqrt {3}\) which is same as tan θ = \(\frac{1}{\sqrt{3}}\).
We know that,

Hence, the required principal solution are
θ = \(\frac{\pi}{6}\) and θ = \(\frac{7 \pi}{6}\).

(iv) cot θ = 0.
Solution:

Question 2.
Find the principal solutions of the following equations:
(i) sinθ = \(-\frac{1}{2}\)
Solution:
We know that,
sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\) and sin (π + θ) = -sinθ,
sin(2π – θ) = -sinθ

Hence, the required principal solutions are
θ = \(\frac{7\pi}{6}\) and θ = \(\frac{11 \pi}{6}\).

(ii) tanθ = -1
Solution:
We know that,
tan\(\frac{\pi}{4}\) = 1 and tan(π – θ) = -tanθ,
tan(2π – θ) = -tanθ

Hence, the required principal solutions are
θ = \(\frac{3\pi}{4}\) and θ = \(\frac{7 \pi}{4}\).

(iii) \(\sqrt {3}\) cosecθ + 2 = 0.
Solution:

Question 3.
Find the general solutions of the following equations :
(i) sinθ = \(\frac{1}{2}\)
Solution:
(i) The general solution of sin θ = sin ∝ is
θ = nπ + (-1 )ⁿ∝, n ∈ Z
Now, sinθ = \(\frac{1}{2}\) = sin\(\frac{\pi}{6}\) …[∵ sin\(\frac{\pi}{6}\) = \(\frac{1}{2}\)]
∴ the required general solution is
θ = nπ + (-1)ⁿ\(\frac{\pi}{6}\), n ∈ Z.

(ii) cosθ = \(\frac{\sqrt{3}}{2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
Now, cosθ = \(\frac{\sqrt{3}}{2}\) = cos\(\frac{\pi}{6}\) …[∵ cos\(\frac{\pi}{6}\) = \(\frac{\sqrt{3}}{2}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{6}\), n ∈ Z.

(iii) tanθ = \(\frac{1}{\sqrt{3}}\)
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, tan θ = \(\frac{1}{\sqrt{3}}\) = tan\(\frac{\pi}{6}\) …[tan\(\frac{\pi}{6}\) = \(\frac{1}{\sqrt{3}}\)]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{6}\) , n ∈ Z.

(iv) cotθ = 0.
Solution:
The general solution of tan θ = tan ∝ is
θ = nπ + ∝, n ∈ Z
Now, cot θ = 0 ∴ tan θ does not exist
∴ tanθ = tan\(\frac{\pi}{2}\) [∵ tan\(\frac{\pi}{2}\) does not exist]
∴ the required general solution is
θ = nπ + \(\frac{\pi}{2}\), n ∈ Z.

Question 4.
Find the general solutions of the following equations:
(i) secθ = \(\sqrt {2}\)
Solution:
The general solution of cos θ = cos ∝ is
θ = nπ ± ∝, n ∈ Z.
Now, secθ = \(\sqrt {2}\) ∴ cosθ = \(\frac{1}{\sqrt{2}}\)
∴ cosθ = cos\(\frac{\pi}{4}\) ….[cos\(\frac{\pi}{4}\) = \(\frac{1}{\sqrt{2}}\)]
∴ the required general solution is
θ = 2nπ ± \(\frac{\pi}{4}\), n ∈ Z.

(ii) cosecθ = –\(\sqrt {2}\)
Solution:
The general solution of sinθ = sin∝ is

(iii) tanθ = -1
Solution:
The general solution of tanθ = tan∝ is

Question 5.
Find the general solutions of the following equations :
(i) sin 2θ = \(\frac{1}{2}\)
Solution:
The general solution of sin θ = sin ∝ is
θ = nπ + (-1)ⁿ∝, n ∈ Z

(ii) tan \(\frac{2 \theta}{3}\) = \(\sqrt {3}\)
Solution:
The general solution of tan θ = tan ∝ is

(iii) cot 4θ = -1
Solution:
The general solution of tan θ = tan ∝ is

Question 6.
Find the general solutions of the following equations :
(i) 4 cos²θ = 3
Solution:
The general solution of cos²θ = cos² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 cos²θ = 3

(ii) 4 sin²θ = 1
Solution:
The general solution of sin²θ = sin² ∝ is
θ = nπ ± ∝, n ∈ Z
Now, 4 sin²θ = 3

(iii) cos 4θ = cos 2θ
Solution:
The general solution of cos θ = cos ∝ is
θ = 2nπ ± ∝, n ∈ Z
∴ the general solution of cos 4θ = cos 2θ is given by
4θ = 2nπ ± 2θ, n ∈ Z
Taking positive sign, we get
4θ = 2nπ + 2θ, n ∈ Z
∴ 2θ = 2nπ, n ∈ Z
∴ θ = nπ, n ∈ Z
Taking negative sign, we get
4θ = 2nπ – 2θ, n ∈ Z
∴ 6θ = 2nπ, n ∈ Z
∴ θ = \(\frac{n \pi}{3}\), n ∈ Z
Hence, the required general solution is
θ = \(\frac{n \pi}{3}\), n ∈ Z or ∴ θ = nπ, n ∈ Z.
Alternative Method:
cos 4θ = cos 2θ
∴ cos4θ – cos 20 = 0
∴ -2sin\(\left(\frac{4 \theta+2 \theta}{2}\right)\)∙sin\(\left(\frac{4 \theta-2 \theta}{2}\right)\) = 0
∴ sin3θ∙sinθ = 0
∴ either sin3θ = 0 or sin θ = 0
The general solution of sin θ = 0 is
θ = nπ, n ∈ Z.
∴ the required general solution is given by
3θ = nπ, n ∈ Z or θ = nπ, n ∈ Z
i.e. θ = \(\frac{n \pi}{3}\), n ∈ Z or θ = nπ, n ∈ Z.

Question 7.
Find the general solutions of the following equations :
(i) sinθ = tanθ
Solution:
sin θ = tan θ
∴ sin θ = \(\frac{\sin \theta}{\cos \theta}\)
∴ sin θ cos θ = sin θ
∴ sin θ cos θ – sinθ = 0
∴ sin θ (cos θ – 1) = θ
∴ either sinθ = 0 or cosθ – 1 = 0
∴ either sin θ = 0 or cos θ = 1
∴ either sinθ = 0 or cosθ = cosθ …[∵ cos0 = 1]
The general solution of sinθ = 0 is θ = nπ, n ∈ Z and cos θ = cos ∝ is θ = 2nπ ± ∝, where n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = 2nπ ± 0, n ∈ Z
∴ θ = nπ, n ∈ Z or θ = 2nπ, n ∈ Z.

(ii) tan³θ = 3tanθ
Solution:
tan³θ = 3tanθ
∴ tan³θ – 3tanθ = 0
∴ tan θ (tan²θ – 3) = 0
∴ either tan θ = 0 or tan²θ – 3 = 0
∴ either tanθ = 0 or tan²θ = 3
∴ either tan θ = 0 or tan²θ = (\(\sqrt {3}\) )³
∴ either tan θ = 0 or tan²θ = (tan\(\frac{\pi}{3}\))³ …[tan\(\frac{\pi}{3}\) = \(\sqrt {3}\)]
∴ either tanθ = 0 or tan²θ = tan²\(\frac{\pi}{3}\)
The general solution of
tanθ = 0 is θ = nπ, n ∈ Z and
tan²θ = tan²∝ is θ = nπ ± ∝, n ∈ Z.
∴ the required general solution is given by
θ = nπ, n ∈ Z or θ = nπ ± \(\frac{\pi}{3}\), n ∈ Z.

(iii) cosθ + sinθ = 1.
Solution:
cosθ + sinθ = 1

Question 8.
Which of the following equations have solutions ?
(i) cos 2θ = -1
Solution:
cos 2θ = -1
Since -1 ≤ cos θ ≤ 1 for any θ,
cos 2θ = -1 has solution.

(ii) cos²θ = -1
Solution:
cos²θ = -1
This is not possible because cos²θ ≥ 0 for any θ.
∴ cos²θ = -1 does not have any solution.

(iii) 2 sinθ = 3
Solution:
2 sin θ = 3 ∴ sin θ = \(\frac{3}{2}\)
This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

(iv) 3 tanθ = 5
Solution:
3tanθ = 5 ∴ tanθ = \(\frac{5}{3}\)
This is possible because tan θ is any real number.
∴ 3tanθ = 5 has solution.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 3.1 Class 12 Maths 1
• Exercise 3.2 Class 12 Maths 1
• Exercise 3.3 Class 12 Maths 1
• Miscellaneous Exercise 3 Class 12 Maths 1
• Exercise 4.1 Class 12 Maths 1
• Exercise 4.2 Class 12 Maths 1
• Exercise 4.3 Class 12 Maths 1
• Miscellaneous Exercise 4 Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2B Questions and Answers.

12th Maths Part 1 Matrices Miscellaneous Exercise 2B Questions And Answers Maharashtra Board

I. Choose the correct answer from the given alternatives in each of the following questions:

Question 1.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\), adj = \(\left(\begin{array}{ll}
4 & a \\
-3 & b
\end{array}\right)\) then the values of a and b are,
(a) a = – 2, b = 1
(b) a = 2, b = 4
(c) a = 2, b = –1
(d) a = 1, b = –2
Solution:
(a) a = – 2, b = 1

Question 2.
The inverse of \(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\) is

Solution:
\(\left(\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right)\)

Question 3.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
2 & 1
\end{array}\right)\) and A(adj A) = k 1, then the value of k is
(a) 1
(b) -1
(c) 0
(d) -3
Solution:
(d) -3 [Hint : A(adj A) = |A| ∙ I]

Question 4.
If A = \(\left(\begin{array}{ll}
2 & -4 \\
3 & 1
\end{array}\right)\), then the adjoint of matrix A is

Solution:\(\left(\begin{array}{ll}
1 & 4 \\
-3 & 2
\end{array}\right)\)

Question 5.
If A = \(\left(\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right)\) and A(adj A) = kI, then the value of k is
(a) 2
(b) -2
(c) 10
(d) -10
Solution:
(b) -2

Question 6.
If A = \(\left(\begin{array}{rr}
\lambda & 1 \\
-1 & -\lambda
\end{array}\right)\), then A^-1 does not exist if λ = ………..
(a) 0
(b) ± 1
(c) 2
(d) 3
Solution:
(b) ± 1

Question 7.
If A = \(\left[\begin{array}{ll}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\) then A^-1 = ….

Solution:
\(\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
-\sin \alpha & \cos \alpha
\end{array}\right]\)

Question 8.
If F (∝) = \(\left[\begin{array}{ccc}
\cos \alpha & -\sin \alpha & 0 \\
\sin \alpha & \cos \alpha & 0 \\
0 & 0 & 1
\end{array}\right]\) where ∝ ∈ R then [F(∝)]^-1 is =
(a) F(-∝)
(b) F(∝^-1)
(c) F(2∝)
(d) None of these
Solution:
(a) F(-∝)

Question 9.
The inverse of A = \(\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]\)
(a) I
(b) A
(c) A’
(d) -I
Solution:
(b) A

Question 10.
The inverse of a symmetric matrix is
(a) Symmetric
(b) Non-symmetric
(c) Null matrix
(d) Diagonal matrix
Solution:
(a) Symmetric

Question 11.
For a 2 × 2 matrix A, if A(adjA) = \(\left(\begin{array}{ll}
10 & 0 \\
0 & 10
\end{array}\right)\) then determinant A equals
(a) 20
(b) 10
(c) 30
(d) 40
Solution:
(b) 10

Question 12.
If A² = \(-\frac{1}{2}\left[\begin{array}{cc}
1 & -4 \\
-1 & 2
\end{array}\right]\) then A =

Solution:
\(-\frac{1}{2}\left[\begin{array}{cc}
2 & 4 \\
1 & 1
\end{array}\right]\)

II. Solve the following equations by the methods of inversion.

(i) 2x – y = -2 , 3x + 4y = 5
Solution:
The given equations can be written in the matrix form as :



By equality of matrices,
x = \(-\frac{5}{11}\), y = \(\frac{12}{11}\) is the required solution.

(ii) x + y + z = 1, 2x + 3y + 2z = 2 and ax + ay + 2az = 4, a ≠ 0.
Solution:
The given equations can be written in the matrix form as :

= 1(6a – 2a) – 1(4a – 2a) + 1(2a – 3a)
= 4a – 2a – a = a ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I

(iii) 5x – y +4z = 5, 2x + 3y + 5z = 2 and 5x – 2y + 6z = -1
Solution:
The given equations can be written in the matrix form as :

= 5(18 + 10) + 1 (12 – 25) + 4( -4 – 15)
= 140 – 13 – 76 = 51 #0
∴ A^-1 exists.
Now, we have to find the cofactor matrix


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B
∴ X = \(\frac{1}{51}\left[\begin{array}{rrr}
28 & -2 & -17 \\
13 & 10 & -17 \\
-19 & 5 & 17
\end{array}\right]\left[\begin{array}{r}
5 \\
2 \\
-1
\end{array}\right]\)

By equality of matrices,
x = 3, y = 2, z = -2 is the required solution.

(iv) 2x + 3y = -5, 3x + y = 3
Solution:

(v) x + y + z = -1, y + z = 2 and x + y – z = 3
Solution:
The given equations can be written in the matrix form as :

= 1(-1 – 1) – 1 (0 – 1) + 1(0 – 1)
= -2 + 1 – 1 = -2 ≠ 0 ∴ A^-1 exists.
Consider AA^-1 = I


Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B

∴ by equality of the matrices, x= -3, y = 4, z = -2 is the required solution.

Question 2.
Express the following equation in matrix from and solve them by the method of reduction.
(i) x – y + z = 1, 2x – y = 1, 3x + 3y – 4z = 2
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x – y + z = 1 ……(1)
y – 2z = -1 …..(2)
5z = 5 ….(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y – 2 = -1 ∴ y = 1
Substituting y = 1, z = 1 in (1), we get,
x – 1 + 1 = 1
∴ x = 1
Hence, x = 1, y = 1, z = 1 is the required solution.

(ii) x + y = 1, y + z = \(\frac{5}{3}\), z + x = \(\frac{4}{3}\).
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
x + y = 1 ……(1)
y + z = \(\frac{5}{3}\) …(2)
2z = 2 ……..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
y + 1 = \(\frac{5}{3}\) ∴ y = \(\frac{2}{3}\)
Substituting y = \(\frac{2}{3}\) in (1), we get,
x + \(\frac{2}{3}\) = 1 ∴ x = \(\frac{1}{3}\)
Hence, x = \(\frac{1}{3}\), y = \(\frac{2}{3}\), z = 1 is the required solution.

(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 :

∴ \(\left[\begin{array}{r}
x+2 y+3 z \\
0-5 y-5 z \\
0+0-8 z
\end{array}\right]\) = \(\left[\begin{array}{r}
8 \\
-15 \\
-8
\end{array}\right]\)
By equality of matrices,
x + 2y + 3z = 8 …..(1)
-5y – 5z = -15 ….(2)
-8z = -8 …..(3)
From (3), z = 1
Substituting z = 1 in (2), we get,
-5y – 5 = -15
-5y = -10
∴ y = 2
Substituting y = 2, z = 1 in (1), we get,
x + 4 + 3 = 8 ∴ x = 1
Hence, x = 1, y = 2, z = 1 is the required solution.

(iv) x + y + z = 6, 3x – y + 3z =10 and 5x + 5y – 4z = 3.
Solution:

(v) 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 :

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.

(vi) x + 3y + 2z = 6, 3x – 2y + 5z =5 and 2x – 3y + 6z = 7.
Solution:
The given equations can be written in the matrix form as :


By equality of matrices,
x + 3y + 2z = 6 …(1)
y + \(\frac{3}{2}\) z = 4 …(2)
\(\frac{31}{2}\)z = 31 …..(3)
From (3), z = 2
Substituting z = 2 in (2), we get,
y + \(\frac{3}{2}\)z = 4
y + \(\frac{3}{2}\)(2) = 4
y + 3 = 4
y = 1
Substituting y = 1, z = 2 in (2), we get,
x + 3y + 2z = 6
x + 3(1) + 2(2) = 6
x + 3 + 4 = 6
x = -1
Hence, x = -1, y = 1, z = 2 is the required solution.

Question 3.
The sum of three numbers is 6. If we multiply third number by 3 and add it to the second number we get 11. By adding first and the third numbers we get a number which 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 conditions,
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 the linear equations is
x + y + z = 6
y + 3z = 11
x – 2y + z = 0
These equations can be written in the matrix form as :

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
Put 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.

Question 4.
The cost of 4 pencils, 3 pens and 2 books is ₹ 150. The cost of 1 pencil, 2 pens and 3 books is ₹ 125. The cos of 6 pencils, 2 pens and 3 books is ₹ 175. Fild the cost of each item by using Matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 book be ₹x, ₹ y, ₹ z respectively.
According to the given conditions,
4x + 3y + 2z = 150
x + 2y + 3z = 125
6x + 2y + 3z = 175
The equations can be written in matrix form as :

By equality of matrices,
x + 2y + 3z = 125 …(1)
-5y – 10z = -350 …(2)
5z = 125 …(3)
From (3), z = 25
Substituting z = 25 in (2), we get
-5y – 10(25) = -350
∴ -5y = -350 + 250 = -100
∴ y = 20
Substituting y = 20, z = 25 in (1), we get
x + 2(20) + 3(25) = 125
∴ x = 125 – 40 – 75 = 10
∴ x = 10, y = 20, z = 25
Hence, the cost of 1 pencil is ₹ 10, 1 pen is ₹ 20 and 1 book is ₹ 25.

Question 5.
The sum of three numbers is 6. Thrice the third number when added to the first number, gives 7. On adding three times first number to the sum of second and third number, we get 12. Find the three numbers by using Matrices.
Solution:
Let the numbers be x, y and z.
According to the given conditions,
x + y + z = 6
3z + x = 7, i.e., x + 3z = 7
and 3x + y + z = 12
Hence, the system of linear equations is
x + y + z = 6
x + 3z = 7
3x + y + z = 12
These equations can be written in matrix form as :

By equality of matrices,
x + y + z = 6 …(1)
-y + 2z = 1 …(2)
-3y = -5 …(3)
From (3), y = \(\frac{5}{3}\)
Substituting y = \(\frac{5}{3}\) in (2), we get,
–\(\frac{5}{3}\) + 2z = 1
∴ 2z = 1 + \(\frac{5}{3}\) = \(\frac{8}{3}\)
∴ z = \(\frac{4}{3}\)
Substituting y =\(\frac{5}{3}\), z = \(\frac{5}{3}\) in (1), we get,
x + \(\frac{5}{3}+\frac{4}{3}\) = 6
∴ x = 3
∴ x = 3, y = \(\frac{5}{3}\), z = \(\frac{4}{3}\)
Hence, the required numbers are 3, \(\frac{5}{3}\) and \(\frac{4}{3}\).

Question 6.
The sum of three numbers is 2. If twice the second number is added to the sum of first and third number, we get 1 adding five times the first number to the sum of second and third we get 6. Find the three numbers by using matrices.
Solution:
Let the three numbers be x, y and z.
According to the question,
x + y + 2
x + 2y + z = 1
5x + y + z = 6
The given system of equations can be written in matrix form as follows:

Question 7.
An amount of ₹ 5000 is invested in three types of investments, at interest rates 6%, 7%, 8% per annum respectively. The total annual income from this investment is ₹ 350. If the total annual income from the first two investments is ₹ 70 more than the income from the third, find the amount of each investment using the matrix method.
Solution:
Let the amounts in three investments by ₹ x, ₹ y, and ₹ z respectively.
Then x + y + z = 5000
Since the rate of interest in these investments are 6%, 7% and 8% respectively, the annual income of the three investments are \(\frac{6 x}{100}\), \(\frac{7 y}{100}\) and \(\frac{8 z}{100}\) respectively.
According to the given conditions,
\(\frac{6 x}{100}+\frac{7 y}{100}+\frac{8 z}{100}\) = 350
i.e. 6x + 7y + 8z = 35000
Also, \(\frac{6 x}{100}+\frac{7 y}{100}\) = \(\frac{8 z}{100}\) + 70
i.e. 6x + 7y – 8z = 7000
Hence, the system of linear equation is
x + y + z = 5000
6x + 7y + 8z = 35000
6x + 7y – 8z = 7000
These equations can be written in matrix form as :


By equality of matrices,
x + y + z = 5000 …(1)
y + 2z = 5000 …(2)
-16z = -28000 ….(3)
From (3), z = 1750
Substituting z = 1750 in (2), we get,
y + 2(1750) = 5000
∴ y = 5000 – 3500 = 1500
Substituting y = 1500, z = 1750 in (1), we get,
x + 1500 + 1750 = 5000
∴ x = 5000 – 3250 = 1750
∴ x = 1750, y = 1500, z = 1750
Hence, the amounts of the three investments are ₹ 1750, ₹ 1500 and ₹ 1750 respectively.

Question 8.
The sum of the costs of one ook each of Mathematics, Physics and Chemistry is ₹ 210. Total cost of a mathematics book, 2 physics books, and a chemistry book is ₹ 240 Also the total cost of a Mathematics book, 3 physics book and chemistry books is Rs. 300/-. Find the cost of each book, using Matrices.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Miscellaneous Exercise 2A Questions and Answers.

12th Maths Part 1 Matrices Miscellaneous Exercise 2A Questions And Answers Maharashtra Board

Question 1.
If A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\) then reduce it to I₃ by using column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right|\)
= 1(1 – 0) – 0 + 0 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By C₁ – 2C₂, we get, A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
-3 & 3 & 1
\end{array}\right]\)
By C₁ + 3C₃ and C₂ – 3C₃, we get,
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 2.
If A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\), then reduce it to I₃ by using row transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right|\)
= 2 (0 – 1) – 1(1 – 1) + 3 (1 – 0)
= -2 – 0 + 3 = 1 ≠ 0
∴ A is a non-singular matrix.
Hence, the required transformation is possible.
Now, A = \(\left[\begin{array}{lll}
2 & 1 & 3 \\
1 & 0 & 1 \\
1 & 1 & 1
\end{array}\right]\)
By R₁ – R₂, we get,

By R₁ – R₃ and By R₂ – R₃, we get
A ~ \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\) = I₃.

Question 3.
Check whether the 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.

(iii) \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
1 & 2 \\
3 & 3
\end{array}\right|\) = 3 – 6 = -3 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(iv) \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{ll}
2 & 3 \\
10 & 15
\end{array}\right|\) = 30 – 30 = 0.
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(v) \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{cc}
\sec \theta & \tan \theta \\
\tan \theta & \sec \theta
\end{array}\right|\)
= sec²θ – tan²θ = 1 ≠ 0.
∴ A is a non-singular matrix.
Hence, A^-1 exist.

(vii) \(\left[\begin{array}{lll}
3 & 4 & 3 \\
1 & 1 & 0 \\
1 & 4 & 5
\end{array}\right]\)
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 exist.

(viii) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
2 & -1 & 3 \\
1 & 2 & 3
\end{array}\right|\)
= 1 (-3 -6) – 2 (6 – 3) + 3 (4 + 1)
= -9 – 6 + 15 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

(ix) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right]\)
Then, |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
3 & 4 & 5 \\
4 & 6 & 8
\end{array}\right|\)
= 1(32 – 30) – 2(24 – 20) + 3(18 – 16)
= 2 – 8 + 6 = 0
∴ A is a singular matrix.
Hence, A^-1 does not exist.

Question 4.
Find AB, if A = \(\left[\begin{array}{ccc}
1 & 2 & 3 \\
1 & -2 & -3
\end{array}\right]\) and B = \(\left[\begin{array}{cc}
1 & -1 \\
1 & 2 \\
1 & -2
\end{array}\right]\) Examine whether AB has inverse or not.
Solution:

∴ A is a non-singular matrix.
Hence, (AB)^-1 exist.

Question 5.
If A = \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\) is a nonsingular matrix then find A^-1 by elementary row transformations.
Hence, find the inverse of \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\)
Solution:
Since A is a non-singular matrix, then find A^-1 by using elementary row transformations.
We write AA^-1 = I

Comparing \(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) with \(\left[\begin{array}{lll}
x & 0 & 0 \\
0 & y & 0 \\
0 & 0 & z
\end{array}\right]\),
we get, x = 2, y = 1, z = -1
∴ \(\frac{1}{x}\) = \(\frac{1}{2}\), \(\frac{1}{y}\) = \(\frac{1}{1}\) = 1, \(\frac{1}{z}\) = \(\frac{1}{-1}\) = -1
\(\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right]\) is \(\left(\begin{array}{rrr}
\frac{1}{2} & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1
\end{array}\right)\).

Question 6.
if A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) and X is a 2 × 2 matrix such that AX = I , then find X.
Solution:
We will reduce the matrix A to the identity matrix by using row transformations. During this pro¬cess, I will be converted to the matrix X.
We have AX = I.

Question 7.
Find the inverse of each of the following matrices (if they exist).
(i) \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & -1 \\
2 & 3
\end{array}\right|\) = 3 + 2 = 5 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ii) \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
1 & -1
\end{array}\right|\) = -2 – 1 = -3 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iii) \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right|\) = 7 – 6 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(iv) \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & -3 \\
5 & 7
\end{array}\right|\) = 14 + 15 = 29 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(v) \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right|\) = 8 – 7 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vi) \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
3 & -10 \\
2 & -7
\end{array}\right|\) = -21 + 20 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(vii) \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right|\)
= 2(4 + 6) +3(4 – 9) + 3(-4 – 6)
= 20 – 15 – 30 = -25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(viii) \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right|\)
= 1(0 + 25) + 3(0 + 10) + 2(-15 – 0)
= 25 + 30 -30
= 25 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

(ix) \(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A =\(\left[\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right|\)
= 2(3 – 0) – 0 – 1(5 – 0)
= 6 – 0 – 5 = 1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)

(x) \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
∴ A^-1 = \(\left[\begin{array}{lll}
1 & 2 & -2 \\
0 & -2 & 1 \\
-1 & 3 & 0
\end{array}\right]\)
= 1\(\left|\begin{array}{ll}
-2 & 1 \\
3 & 0
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & 1 \\
-1 & 1
\end{array}\right|\) – 2\(\left|\begin{array}{ll}
0 & -2 \\
-1 & 3
\end{array}\right|\)
|A| = 1(0 – 3) – 2(0 + 1) – 2(0 – 2)
= -3 – 2 + 4
= -1 ≠ 0
∴ A^-1 exists.
We have
AA^-1 = I


∴ A^-1 = \(\left[\begin{array}{lll}
3 & 6 & 2 \\
1 & 2 & 1 \\
2 & 5 & 2
\end{array}\right]\)

Question 8.
Find the inverse of A = \(\left[\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right]\) by
(i) elementary row transformations
Solution:
|A| = \(\left|\begin{array}{ccc}
\cos \theta & -\sin \theta & 0 \\
\sin \theta & \cos \theta & 0 \\
0 & 0 & 1
\end{array}\right|\)
= cosθ (cosθ – 0) + sinθ (sinθ – 0) + 0
= cos²θ + sin²θ = 1 ≠ 0
∴ A^-1 exists.
(i) Consider AA^-1 = I

(ii) elementary column transformations
Solution:
Consider A^-1A = I

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




From (1) and (2), (AB)^-1 = B^-1 ∙ A^-1.

Question 10.
If A = \(\left[\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right]\), then show that A^-1 = \(\frac{1}{6}\)(A – 5I)
Solution:
|A| = \(\left|\begin{array}{ll}
4 & 5 \\
2 & 1
\end{array}\right|\) = 4 – 10 = -6 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 11.
Find matrix X such that AX = B, where A = \(\left[\begin{array}{ll}
1 & 2 \\
-1 & 3
\end{array}\right]\) and B = \(\left[\begin{array}{ll}
0 & 1 \\
2 & 4
\end{array}\right]\)
Solution:
AX = B

Question 12.
Find X, if AX = B where A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\) and B = \(\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\).
Solution:
AX = B

Question 13.
If A = \(\left[\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right]\), B = \(\left[\begin{array}{ll}
4 & 1 \\
3 & 1
\end{array}\right]\) and C = \(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\) then find matrix X such that AXB = C.
Solution:
AXB = C
∴ \(\left(\begin{array}{ll}
1 & 1 \\
1 & 2
\end{array}\right)(\mathrm{XB})\) =\(\left[\begin{array}{ll}
24 & 7 \\
31 & 9
\end{array}\right]\)
First we perform the row transformations.

Question 14.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by adjoint method.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]_3×3 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = \(\left|\begin{array}{ll}
1 & 5 \\
4 & 7
\end{array}\right|\) = 7 – 20 = -13
A₁₂ = (-1)¹⁺²M₁₂ = \(\left|\begin{array}{ll}
1 & 5 \\
2 & 7
\end{array}\right|\) = -(7 – 10) = 3

Question 15.
Find the inverse of \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by adjoint method.
Solution:
where A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\)
|A| = 1(2 – 6) – 0(0 – 3) + 1(0 – 2)
|A| = -4 – 2
|A| = -6 ≠ 0
∴ A^-1 exists.
First we have to find the cofactor matrix
= [A_ij]3×3, where A_ij = (-1)^i+jM_ij

Question 16.
Find A^-1 by adjoint method and by elementary transformations if A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right]\)
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
-1 & 1 & 2 \\
1 & 2 & 4
\end{array}\right|\)
= 1(4 – 4) – 2(-4 – 2) + 3(-2 – 1)
= 0 + 12 – 9 = 3 ≠ 0
∴ A^-1 exists.
A^-1by adjoint method :
We have to find the cofactor matrix
= [A_ij]_3×3, where A_ij = (-1)^i+j M_ij

Question 17.
Find the inverse of A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) by elementary column transformations.
Solution:
|A| = \(\left|\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right|\)
= 1 (2 – 6) – 0 + 1 (0 – 2)
= -4 – 2= -6 ≠ 0
∴ A^-1 exists.
Consider A^-1A = I
∴ A^-1\(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) = \(\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\)
By C₃ – C₁, we get,

Question 18.
Find the inverse of \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\) by elementary row transformations.
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right|\)
= 1(7 – 20) – 2(7 – 10) + 3(4 – 2)
= -13 + 6 + 6 = -1 ≠ 0
∴ A^-1 exists.
Consider AA^-1 = I

Question 19.
Show with usual notations that for any matrix A = [a_ij]_3×3
(i) a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃ = 0
Solution:
A = [a_ij]_3×3 = \(\left[\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right]\)
(i) A₂₁ = (-1)²⁺¹M₂₁ = \(-\left|\begin{array}{ll}
a_{12} & a_{13} \\
a_{32} & a_{33}
\end{array}\right|\)
= -(a₁₂a₃₃ – a₁₃a₃₂)
= -a₁₂a₃₃ + a₁₃a₃₂
A₂₂ = (-1)²⁺²M₂₂ = \(\left|\begin{array}{ll}
a_{11} & a_{13} \\
a_{31} & a_{33}
\end{array}\right|\)
= a₁₁a₃₃ – a₁₃a₃₁
A₂₃ = (-1)²⁺³M₂₃ = \(-\left|\begin{array}{ll}
a_{11} & a_{12} \\
a_{31} & a_{32}
\end{array}\right|\)
= -(a₁₁a₃₂ – a₁₂a₃₁)
= -a₁₁a₃₂+ a₁₂a₃₁
∴ a₁₁A₂₁ + a₁₂A₂₂ + a₁₃A₂₃
= a₁₁(-a₁₂₃₃ + a₁₃a₃₂) + a₁₂(a₁₁a₃₃ – a₁₃a₃₁) + a₁₃(-a₁₁a₃₂ + a₁₂a₃₁)
= -a₁₁a₁₂a₃₃ + a₁₁a₁₃a₃₂ + a₁₁a₁₂a₃₃ – a₁₂a₁₃a₃₁ – a₁₁a₁₃a₃₂ + a₁₂a₁₃a₃₁
= 0

(ii) a₁₁A₁₁ + a₁₂A₁₂ + a₁₃A₁₃ = |A|
Solution:

Question 20.
If A = \(\left[\begin{array}{lll}
1 & 0 & 1 \\
0 & 2 & 3 \\
1 & 2 & 1
\end{array}\right]\) and B = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
1 & 1 & 5 \\
2 & 4 & 7
\end{array}\right]\), then find a matrix X such that XA= B.
Solution:
Consider XA = B

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.3 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.3 Questions And Answers Maharashtra Board

Question 1.
Solve the following equations by the inversion method.
(i) x + 2y = 2, 2x + 3y = 3
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 2 \\
2 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
This is of the form AX = B, where

∴ A^-1 = \(\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\)
Now, premultiply AX = B by A^-1, we get,
A^-1(AX) = A^-1B
∴ (A^-1A)X = A^-1B
∴ IX = A^-1B
∴ X = \(=\left[\begin{array}{rr}
-3 & 2 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
2 \\
3
\end{array}\right]\)
∴ \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(=\left[\begin{array}{r}
-6+6 \\
4-3
\end{array}\right]\) = \(=\left[\begin{array}{l}
0 \\
1
\end{array}\right]\)
By equality of matrices,
x = 0, y = 1 is the required solution.

(ii) x + y = 4, 2x – y = 5
Solution:
x + y = 4, 2x – y = 5
The given equations can be written in the matrix form as:
\(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
4 \\
5
\end{array}\right]\)
This is of the form AX = B ⇒ X ⇒ A^-1B
A = \(\left[\begin{array}{cc}
1 & 1 \\
2 & -1
\end{array}\right]\)
|A| = -1 – 2 = -3 ≠ 0

By equality of matrices.
x = 3, y = 1

(iii) 2x + 6y = 8, x + 3y = 5
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]=\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
This is of the form AX = B, where
A = \(\left[\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right]\), X = \(\left[\begin{array}{l}
x \\
y
\end{array}\right]\) and B = \(\left[\begin{array}{l}
8 \\
5
\end{array}\right]\)
Let us find A^-1.
|A| = \(\left|\begin{array}{ll}
2 & 6 \\
1 & 3
\end{array}\right|\) = 6 – 6 = 0
∴ A^-1 does not exist.
Hence, x and y do not exist.

Question 2.
Solve the following equations by reduction method.
(i) 2x + y = 5, 3x + 5y = -3
Solution:
The given equations can be written in the matrix form as :

By equality of matrices,
2x + y = 5 …(1)
7y = -21 …(2)
From (2), y = -3
Substituting y = -3 in (1), we get,
2x – 3 = 5
∴ 2x = 8 ∴ x = 4
Hence, x = 4, y = -3 is the required solution.

(ii) x + 3y = 2, 3x + 5y = 4.
Solution:
The given equations can be written in the matrix form as :
\(\left[\begin{array}{ll}
1 & 3 \\
3 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left[\begin{array}{l}
2 \\
4
\end{array}\right]\)
By R₂ – 3R₁, we get
\(\left[\begin{array}{rr}
1 & 3 \\
0 & -4
\end{array}\right]\left[\begin{array}{l}
x \\
y
\end{array}\right]\) = \(\left(\begin{array}{r}
2 \\
-2
\end{array}\right)\)
∴ \(\left[\begin{array}{l}
x+3 \\
0-4 y
\end{array}\right]\) = \(\left[\begin{array}{r}
2 \\
-2
\end{array}\right]\)
By equality of matrices,
x + 3y = 2 …(1)
-4y = -2
From (2), y = \(\frac{1}{2}\)
Substituting y = \(\frac{1}{2}\) in (1), we get,
x + \(\frac{3}{2}\) = 2
∴ x = 2 – \(\frac{3}{2}=\frac{1}{2}\)
Hence, x = \(\frac{1}{2}\), y = \(\frac{1}{2}\) is the required solution.

(iii) 3x – y = 1, 4x + y = 6
Solution:
The given equations can be written in the matrix form as :

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.

(iv) 5x + 2y = 4, 7x + 3y = 5
Solution:
5x + 2y = 4 ………..(1)
7x + 3y = 5 …………(2)
Multiplying Eq. (1) with 7 and Eq. (2) with 5

Put y = -3 into Eq. (1)
5x + 2y = 4
5x + 2(-3) = 4
5x – 6 = 4
5x = 4 + 6
5x = 10
x = \(\frac{10}{5}\)
x = 2
Hence, x = 2, y = -3 is the required solution.

Question 3.
The cost of 4 pencils, 3 pens and 2 erasers is ₹ 60. The cost of 2 pencils, 4 pens and 6 erasers is ₹ 90, whereas the cost of 6 pencils, 2 pens and 3 erasers is ₹ 70. Find the cost of each item by using matrices.
Solution:
Let the cost of 1 pencil, 1 pen and 1 eraser be ₹ x, ₹ y and ₹ z respectively.
Then, from the given conditions,
4x + 3y + 2z = 60
2x + 4y + 6z = 90, i.e., x + 2y + 3z = 45
6x + 2y + 3z = 70
These equations can be written in the matrix form as :

By equality of matrices,
x + 2y + 3z = 45 …….(1)
– 5y – 10z = – 120 …….(2)
5z = 40
From (3), z = 8
Substituting z = 8 in (2), we get,
– 5y – 80 = -120
∴ – 5y = -40 ∴ y = 8
Substituting y = 8, z = 8 in (1), we get,
x + 16 + 24 = 45
∴ x + 40 = 45 ∴ x = 5
∴ x = 5, y = 8, z = 8
Hence, the cost is ₹ 5 for a pencil, ₹ 8 for a pen and ₹ 8 for an eraser.

Question 4.
If three numbers are added, their sum is 2. If 2 times the second number is subtracted from the sum of first and third numbers, we get 8 and if three times the first number is added to the sum of second and third numbers, we get 4. Find the numbers using matrices.
Solution:
Let the three numbers be x, y and z. According to the given conditions,
x + y + z = 2
x + z – 2y = 8, i.e., x – 2y + 2 = 8
and y + z + 3x = 4, i.e., 3x + y + z = 4
Hence, the system of linear equations is
x + y + z = 2
x – 2y + z = 8
3x + y + z = 4
These equations can be written in the matrix form as :

By equality of matrices,
x + y + z = 2 ……(1)
-3y = 6 ……(2)
– 2y – 2z = -2 ……..(3)
From (2), y = -2
Substituting y = -2 in (3), we get,
-2(-2) – 2z = -2
∴ -2z = -6 ∴ z = 3
Substituting y = -2, z = 3 in (1), we get,
x – 2 + 3 = 2 ∴ x = 1
Hence, the required numbers are 1, -2 and 3.

Question 5.
The total cost of 3 T.V. sets and 2 V.C.R.s is ₹ 35000. The shop-keeper wants profit of ₹ 1000 per television and ₹ 500 per V.C.R. He can sell 2 T. V. sets and 1 V.C.R. and get the total revenue as ₹ 21,500. Find the cost price and the selling price of a T.V. sets and a V.C.R.
Solution:
Let the cost of each T.V. set be ₹ x and each V.C.R. be ₹ y. Then the total cost of 3 T.V. sets and 2 V.C.R.’s is ₹ (3x + 2y) which is given to be ₹ 35,000.
∴ 3x + 2y = 35000
The shopkeeper wants profit of ₹ 1000 per T.V. set and of ₹ 500 per V.C.R.
∴ the selling price of each T.V. set is ₹ (x + 1000) and of each V.C.R. is ₹ (y + 500).
∴ selling price of 2 T.V. set and 1 V.C.R. is
₹ [2(x + 1000) + (y + 500)] which is given to be ₹ 21,500.
∴ 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
These equations can be written in the matrix form as :

By equality of matrices,
2x + y = 19000 ……….(1)
-x = -3000 ……….(2)
From (2), x = 3000
Substituting x = 3000 in (1), we get,
2(3000) + y = 19000
∴ y = 13000
∴ the cost price of one T.V. set is ₹ 3000 and of one V.C.R. is ₹ 13000 and the selling price of one T.V. set is ₹ 4000 and of one V.C.R. is ₹ 13500.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.2 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.2 Questions And Answers Maharashtra Board

Question 1.
Find the co-factors of the elements of the following matrices
(i) \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
-1 & 2 \\
-3 & 4
\end{array}\right]\)
Here, a₁₁ = -11, M₁₁ = 4
∴ A₁₁ = (-1)¹⁺¹(4) = 4
a₁₂ = 2, M₁₂ = -3
∴ A₁₂ = (-1)¹⁺²(- 3) = 3
a₂₁ = – 3, M₂₁ = -2
∴ A₂₁ = (- 1)²⁺¹(2) = -2
a₂₂ = 4, M₂₂ = -1
∴ A₂₂ = (-1)²⁺²(-1) = -1.

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
The co-factor of a_ij is given by A_ij = (-1)^i+jM_ij

Question 2.
Find the matrix of co-factors for the following matrices
(i) \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rr}
1 & 3 \\
4 & -1
\end{array}\right]\)
Here, a₁₁ = 1, M₁₁ = -1
∴ A₁₁ = (-1)¹⁺¹(-1) = -1
a₁₂ = 3, M₁₂ = 4
∴ A₁₂ = (-1)¹⁺²(4) = -4
a₂₁ = 4, M₂₁ = 3
∴ A₂₁ = (-1)²⁺¹(3) = -3
a₂₂ = -1, M₂₂ = 1
∴ A₂₂ = (-1)²⁺¹(1) = 1
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left(\begin{array}{rr}
-1 & -4 \\
-3 & 1
\end{array}\right)\)

(ii) \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & 0 & 2 \\
-2 & 1 & 3 \\
0 & 3 & -5
\end{array}\right]\)



A₁₁ = -14, A₁₂ = -10, A₁₃ = -6,
A₂₁ = 6, A₂₂ = -5, A₂₃ = -3,
A₃₁ = -2, A₃₂ = -7, A₃₃ = 1.
∴ the co-factor matrix
= \(\left[\begin{array}{lll}
A_{11} & A_{12} & A_{13} \\
A_{21} & A_{22} & A_{23} \\
A_{31} & A_{32} & A_{33}
\end{array}\right]\) = \(\left[\begin{array}{rrr}
-14 & -10 & -6 \\
6 & -5 & -3 \\
-2 & -7 & 1
\end{array}\right]\)

Question 3.
Find the adjoint of the following matrices.
(i) \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
3 & 5
\end{array}\right]\)
Here, a₁₁ = 2, M₁₁= 5
∴ A₁₁ = (-1)¹⁺¹(5) = 5
a₁₂ = -3, M₁₂ = 3
∴ A₁₂ = (-1)¹⁺²(3) = -3
a₂₁ = 3, M₂₁ = -3
∴ A A₂₁ = (-1)²⁺¹(-3) = 3
a₂₂ = 5, M₂₂ = 2
∴ A₂₂ = (-1)²⁺¹ = 2
∴ the co-factor matrix = \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\)
= \(\left[\begin{array}{rr}
5 & -3 \\
3 & 2
\end{array}\right]\)
∴ adj A = \(\left(\begin{array}{rr}
5 & 3 \\
-3 & 2
\end{array}\right)\)

(ii) \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
-2 & 3 & 5 \\
-2 & 0 & -1
\end{array}\right]\)
Solution:


A₁₁ = -3, A₁₂ = -12, A₁₃ = 6,
A₂₁ = -1, A₂₂ = 3, A₂₃ = 2,
A₃₁ = -11, A₃₂ = -9, A₃₃ = 1
∴ the co-factor matrix = \(\left[\begin{array}{lll}
\mathrm{A}_{11} & \mathrm{~A}_{12} & \mathrm{~A}_{15} \\
\mathrm{~A}_{21} & \mathrm{~A}_{22} & \mathrm{~A}_{23} \\
\mathrm{~A}_{31} & \mathrm{~A}_{32} & \mathrm{~A}_{33}
\end{array}\right]\)
= \(\left[\begin{array}{rrr}
-3 & -12 & 6 \\
-1 & 3 & 2 \\
-11 & -9 & 1
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{rrr}
-3 & -1 & -11 \\
-12 & 3 & -9 \\
6 & 2 & 1
\end{array}\right]\)

Question 4.
If A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\), verify that A (adj A) = (adj A) A = | A | ∙ I
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right]\)




From (1), (2) and (3), we get,
A(adj A) = (adj A)A = |A|∙I.
Note: This relation is valid for any non-singular matrix A.

Question 5.
Find the inverse of the following matrices by the adjoint method
(i) \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right]\)
∴ |A| = \(\left|\begin{array}{ll}
-1 & 5 \\
-3 & 2
\end{array}\right|\) = -2 + 15 = 13 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix
= [A_ij]_2×2, where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 2
A₁₂ = (-1)¹⁺²M₁₂ = -(-3) = 3
A₂₁ = (-1)²⁺¹M₂₁ = -5
A₂₂ = (-1)²⁺²M₂₂ = -1
Hence, the co-factor matrix

(ii) \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -2 \\
4 & 3
\end{array}\right]\)
|A| = \(\) = 6 + 8 = 14 ≠ 0
∴ A^-1 exist
First we have to find the co-factor matrix
= [A_ij] _2×2 where A_ij = (-1)^i+jM_ij
Now, A₁₁ = (-1)¹⁺¹M₁₁ = 3
A₁₂ = (-1)¹⁺²M = -4
A₂₁ = (-2)²⁺¹M₂₁ = (-2) = 2
A₂₂ = (-1)²⁺²M₂₂ = 2
Hence the co-factor matrix
= \(\left[\begin{array}{ll}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{array}\right]\) = \(\left[\begin{array}{cc}
3 & -4 \\
2 & 2
\end{array}\right]\)
∴ adj A = \(\left[\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right]\)
∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A) = \(\frac{1}{14}\left(\begin{array}{cc}
3 & 2 \\
-4 & 2
\end{array}\right)\)

(iii) \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
1 & 0 & 0 \\
3 & 3 & 0 \\
5 & 2 & -1
\end{array}\right]\)



∴ A^-1 = \(\frac{1}{3}\left[\begin{array}{rrr}
3 & 0 & 0 \\
-3 & 1 & 0 \\
9 & 2 & -3
\end{array}\right]\)

(iv) \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
∴ |A| = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 2 & 4 \\
0 & 0 & 5
\end{array}\right]\)
= 1(10 – 0) – 0 + 0
= 1(10) – 0 + 0
= 10 ≠ 0
∴ A^-1 exists.
First we have to find the co-factor matrix


∴ A^-1 = \(\frac{1}{|\mathrm{~A}|}\) (adj A)
= \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)
∴ A^-1 = \(\frac{1}{10}\left(\begin{array}{rrr}
10 & -10 & 2 \\
0 & 5 & -4 \\
0 & 0 & 2
\end{array}\right)\)

Question 6.
Find the inverse of the following matrices
(i) \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
1 & 2 \\
2 & -1
\end{array}\right]\)

(ii) \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{cc}
2 & -3 \\
-1 & 2
\end{array}\right]\)

∴ A^-1 = \(\left(\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right)\)

(iii) \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{lll}
0 & 1 & 2 \\
1 & 2 & 3 \\
3 & 1 & 1
\end{array}\right]\)

(iv) \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution:
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 2 Matrices Ex 2.1 Questions and Answers.

12th Maths Part 1 Matrices Exercise 2.1 Questions And Answers Maharashtra Board

Question 1.
Apply the given elementary transformation on each of the following matrices.
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\), R₁ ↔ R₂
Solution:
A = \(\left[\begin{array}{cc}
1 & 0 \\
-1 & 3
\end{array}\right]\)
By R₁ ↔ R₂, we get,
A ~ \(\left[\begin{array}{rr}
-1 & 3 \\
1 & 0
\end{array}\right]\)

Question 2.
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\), R₁ → R₁ → R₂
Solution:
B = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 5 & 4
\end{array}\right]\),
R₁ → R₁ → R₂ gives,
B ~ \(\left[\begin{array}{rrr}
-1 & -6 & -1 \\
2 & 5 & 4
\end{array}\right]\)

Question 3.
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\), C₁ ↔ C₂; B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\), R₁ ↔ R₂. What do you observe?
Solution:
A = \(\left[\begin{array}{ll}
5 & 4 \\
1 & 3
\end{array}\right]\)
By C₁ ↔ C₂, we get,
A ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(1)
B = \(\left[\begin{array}{ll}
3 & 1 \\
4 & 5
\end{array}\right]\)
By R₁ ↔ R₂, we get,
B ~ \(\left[\begin{array}{ll}
4 & 5 \\
3 & 1
\end{array}\right]\) …(2)
From (1) and (2), we observe that the new matrices are equal.

Question 4.
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\), 2C₂
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\), -3R₁
Find the addition of the two new matrices.
Solution:
A = \(\left[\begin{array}{ccc}
1 & 2 & -1 \\
0 & 1 & 3
\end{array}\right]\)
By 2C₂, we get,
A ~ \(\left[\begin{array}{rrr}
1 & 4 & -1 \\
0 & 2 & 3
\end{array}\right]\)
B = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
2 & 4 & 5
\end{array}\right]\)
By -3R₁, we get,
B ~ \(\left[\begin{array}{rrr}
-3 & 0 & -6 \\
2 & 4 & 5
\end{array}\right]\)
Now, addition of the two new matrices

Question 5.
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\), 3R₃ and then C₃ + 2C₂.
Solution:
A = \(\left[\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right]\)
By 3R₃, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
9 & 9 & 3
\end{array}\right]\)
By C₃ + 2C₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
9 & 9 & 3+2(9)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)

Question 6.
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\), C₃ + 2C₂ and then 3R₃. What do you conclude from Ex. 5 and Ex. 6 ?
Solution:
A = \(\left(\begin{array}{rrr}
1 & -1 & 3 \\
2 & 1 & 0 \\
3 & 3 & 1
\end{array}\right)\)
By C₃ + 2C₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 3+2(-1) \\
2 & 1 & 0+2(1) \\
3 & 3 & 1+2(3)
\end{array}\right)\)
∴ A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
3 & 3 & 7
\end{array}\right)\)
By 3R₃, we get
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 1 \\
2 & 1 & 2 \\
9 & 9 & 21
\end{array}\right)\)
We conclude from Ex. 5 and Ex. 6 that the matrix remains same by interchanging the order of the elementary transformations. Hence, the transformations are commutative.

Question 7.
Use suitable transformation on \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\) into an upper triangular matrix.
Solution:
Let A = \(\left[\begin{array}{ll}
1 & 2 \\
3 & 4
\end{array}\right]\)
By R₂ – 3R₁, we get,
A ~ \(\left[\begin{array}{rr}
1 & 2 \\
0 & -2
\end{array}\right]\)
This is an upper triangular matrix.

Question 8.
Convert \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\) into an identity matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rr}
1 & -1 \\
2 & 3
\end{array}\right]\)
By R₂ – 2R₁, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 5
\end{array}\right]\)
By \(\left(\frac{1}{5}\right)\)R₂, we get,
A ~ \(\left[\begin{array}{rr}
1 & -1 \\
0 & 1
\end{array}\right]\)
By R₁ + R₂, we get,
A ~ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\)
This is an identity matrix.

Question 9.
Transform \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\) into an upper triangular matrix by suitable row transformations.
Solution:
Let A = \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
2 & 1 & 3 \\
3 & 2 & 4
\end{array}\right]\)
By R₂ – 2R₁ and R₃ – 3R₁, we get
A ~ \(\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 5 & -2
\end{array}\right]\)
By R₃ – \(\left(\frac{5}{3}\right)\)R₂, we get,
A ~ \(\left(\begin{array}{rrr}
1 & -1 & 2 \\
0 & 3 & -1 \\
0 & 0 & -\frac{1}{3}
\end{array}\right)\)
This is an upper triangular matrix.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 2.1 Class 12 Maths 1
• Exercise 2.2 Class 12 Maths 1
• Exercise 2.3 Class 12 Maths 1
• Miscellaneous Exercise 2A Class 12 Maths 1
• Miscellaneous Exercise 2B Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Miscellaneous Exercise 1 Questions and Answers.

12th Maths Part 1 Mathematical Logic Miscellaneous Exercise 1 Questions And Answers Maharashtra Board

Question 1.
Select and write the correct answer from the given alternatives in each of the following questions:
i) If p ∧ q is false and p ∨ q is true, the ________ is not true.
(A) p ∨ q
(B) p ↔ q
(C) ~p ∨ ~q
(D) q ∨ ~p
Solution:
(b) p ↔ q.

(ii) (p ∧ q) → r is logically equivalent to ________.
(A) p → (q → r)
(B) (p ∧ q) → ~r
(C) (~p ∨ ~q) → ~r
(D) (p ∨ q) → r
Solution:
(a) p → (q → r) [Hint: Use truth table.]

(iii) Inverse of statement pattern (p ∨ q) → (p ∧ q) is ________.
(A) (p ∧ q) → (p ∨ q)
(B) ~(p ∨ q) → (p ∧ q)
(C) (~p ∧ ~q) → (~p ∨ ~q)
(D) (~p ∨ ~q) → (~p ∧ ~q)
Solution:
(c) (~p ∧ ~q) → (~p ∨ ~ q)

(iv) If p ∧ q is F, p → q is F then the truth values of p and q are ________.
(A) T, T
(B) T, F
(C) F, T
(D) F, F
Solution:
(b) T, F

(v) The negation of inverse of ~p → q is ________.
(A) q ∧ p
(B) ~p ∧ ~q
(C) p ∧ q
(D) ~q → ~p
Solution:
(a) q ∧ p

(vi) The negation of p ∧ (q → r) is ________.
(A) ~p ∧ (~q → ~r)
(B) p ∨ (~q ∨ r)
(C) ~p ∧ (~q → ~r)
(D) ~p ∨ (~q ∧ ~r)
Solution:
(d) ~p ∨ (q ∧ ~r)

(vii) If A = {1, 2, 3, 4, 5} then which of the following is not true?
(A) Ǝ x ∈ A such that x + 3 = 8
(B) Ǝ x ∈ A such that x + 2 < 9
(C) Ɐ x ∈ A, x + 6 ≥ 9
(D) Ǝ x ∈ A such that x + 6 < 10
Solution:
(c) Ǝ x ∈ A, x + 6 ≥ 9.

Question 2.
Which of the following sentences are statements in logic? Justify. Write down the truth
value of the statements :
(i) 4! = 24.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(ii) π is an irrational number.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(iii) India is a country and Himalayas is a river.
Solution:
It is a statement which is false, hence its truth value is ‘F’. ….[T ∧ F ≡ F]

(iv) Please get me a glass of water.
Solution:
It is an imperative sentence, hence it is not a statement.

(v) cos²θ – sin²θ = cos2θ for all θ ∈ R.
Solution:
It is a statement which is true, hence its truth value is ‘T’.

(vi) If x is a whole number the x + 6 = 0.
Solution:
It is a statement which is false, hence its truth value is ‘F’.

Question 3.
Write the truth values of the following statements :
(i) \(\sqrt {5}\) is an irrational but \(3\sqrt {5}\) is a complex number.
Solution:
Let p : \(\sqrt {5}\) is an irrational.
q : \(3\sqrt {5}\) is a complex number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F]

(ii) Ɐ n ∈ N, n² + n is even number while n² – n is an odd number.
Solution:
Let p : Ɐ n ∈ N, n² + n is an even number.
q : Ɐ n ∈ N, n² – n is an odd number.
Then the symbolic form of the given statement is p ∧ q.
The truth values of p and q are T and F respectively.
∴ the truth value of p ∧ q is F. … [T ∧ F ≡ F].

(iii) Ǝ n ∈ N such that n + 5 > 10.
Solution:
Ǝ n ∈ N, such that n + 5 > 10 is a true statement, hence its truth value is T.
(All n ≥ 6, where n ∈ N, satisfy n + 5 > 10).

(iv) The square of any even number is odd or the cube of any odd number is odd.
Solution:
Let p : The square of any even number is odd.
q : The cube of any odd number is odd.
Then the symbolic form of the given statement is p ∨ q.
The truth values of p and q are F and T respectively.
∴ the truth value of p ∨ q is T. … [F ∨ T ≡ T].

(v) In ∆ ABC if all sides are equal then its all angles are equal.
Solution:
Let p : ABC is a triangle and all its sides are equal.
q : Its all angles are equal.
Then the symbolic form of the given statement is p → q
If the truth value of p is T, then the truth value of q is T.
∴ the truth value of p → q is T. … [T → T ≡ T].

(vi) Ɐ n ∈ N, n + 6 > 8.
Solution:
Ɐ n ∈ N, 11 + 6 > 8 is a false statement, hence its truth value is F.
{n = 1 ∈ N, n = 2 ∈ N do not satisfy n + 6 > 8).

Question 4.
If A = {1, 2, 3, 4, 5, 6, 7, 8, 9}, determine the truth value of each of the following statement :
(i) Ǝ x ∈ A such that x + 8 = 15.
Solution:
True

(ii) Ɐ x ∈ A, x + 5 < 12.
Solution:
False

(iii) Ǝ x ∈ A, such that x + 7 ≥ 11.
Solution:
True

(iv) Ɐ x ∈ A, 3x ≤ 25.
Solution:
False

Question 5.
Write the negations of the following :
(i) Ɐ n ∈ A, n + 7 > 6.
Solution:
The negation of the given statements are :
Ǝ n ∈ A, such that n + 7 ≤ 6.
OR Ǝ n ∈ A, such that n + 7 ≯ 6.

(ii) Ǝ x ∈ A, such that x + 9 ≤ 15.
Solution:
Ɐ x ∈ A, x + 9 > 15.

(iii) Some triangles are equilateral triangle.
Solution:
All triangles are not equilateral triangles.

Question 6.
Construct the truth table for each of the following :
(i) p → (q → p)
Solution:

(ii) (~p ∨ ~q) ↔ [~(p ∧ q)]
Solution:

(iii) ~(~p ∧ ~q) ∨ q
Solution:

(iv) [(p ∧ q) ∨ r] ∧ [~r ∨ (p ∧ q)]
Solution:

(v) [(~p ∨ q) ∧ (q → r)] → (p → r)
Solution:

Question 7.
Determine whether the following statement patterns are tautologies contradictions or contingencies :
(i) [(p → q) ∧ ~q)] → ~p
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p → q) ∧ ~q)] → ~p is a tautology.

(ii) [(p ∨ q) ∧ ~p] ∧ ~q
Solution:

All the entries in the last column of the above truth table are F.
∴ [(p ∨ q) ∧ ~p] ∧ ~q is a contradiction.

(iii) (p → q) ∧ (p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are F.
∴ (p → q) ∧ (p ∧ ~q) is a contradiction.

(iv) [p → (q → r)] ↔ [(p ∧ q) → r]
Solution:

All the entries in the last column of the above truth table are T.
∴ [p → (q → r)] ↔ [(p ∧ q) → r] is a tautology.

(v) [(p ∧ (p → q)] → q
Solution:

All the entries in the last column of the above truth table are T.
∴ [(p ∧ (p → q)] → q is a tautology.

(vi) (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p ∧ q) ∨ (~p ∧ q) ∨ (p ∨ ~q) ∨ (~p ∧ ~q) is a tautology.

(vii) [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r
Solution:

The entries in the last column are neither T nor all F.
∴ [(p ∨ ~q) ∨ (~p ∧ q)] ∧ r is a contingency.

(viii) (p → q) ∨ (q → p)
Solution:

All the entries in the last column of the above truth table are T.
∴ (p → q) ∨ (q → p) is a tautology.

Question 8.
Determine the truth values ofp and q in the following cases :
(i) (p ∨ q) is T and (p ∧ q) is T
Solution:

Since p ∨ q and p ∧ q both are T, from the table the truth values of both p and q are T.

(ii) (p ∨ q) is T and (p ∨ q) → q is F
Solution:

Since the truth values of (p ∨ q) is T and (p ∨ q) → q is F, from the table, the truth values of p and q are T and F respectively.

(iii) (p ∧ q) is F and (p ∧ q) → q is T
Solution:

Since the truth values of (p ∧ q) is F and (p ∧ q) → q is T, from the table, the truth values of p and q are either T and F respectively or F and T respectively or both F.

Question 9.
Using truth tables prove the following logical equivalences :
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:

The entries in the columns 3 and 8 are identical.
∴ p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q).

(ii) (p ∧ q) → r ≡ p → (q → r)
Solution:

The entries in the columns 5 and 7 are identical.
∴ (p ∧ q) → r ≡ p → (q → r).

Question 10.
Using rules in logic, prove the following :
(i) p ↔ q ≡ ~ (p ∧ ~q) ∧ ~(q ∧ ~p)
Solution:
By the rules of negation of biconditional,
~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)
∴ ~ [(p ∧ ~ q) ∨ (q ∧ ~p)] ≡ p ↔ q
∴ ~(p ∧ ~q) ∧ ~(q ∧ ~p) ≡ p ↔ q … (Negation of disjunction)
≡ p ↔ q ≡ ~(p ∧ ~ q) ∧ ~ (q ∧ ~p).

(ii) ~p ∧ q ≡ (p ∨ q) ∧ ~p
Solution:
(p ∨ q) ∧ ~ p
≡ (p ∧ ~p) ∨ (q ∧ ~p) … (Distributive Law)
≡ F ∨ (q ∧ ~p) … (Complement Law)
≡ q ∧ ~ p … (Identity Law)
≡ ~p ∧ q …(Commutative Law)
∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(iii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p
Solution:
~ (p ∨ q) ∨ (~p ∧ q)
≡ (~p ∧ ~q) ∨ (~p ∧ q) … (Negation of disjunction)
≡ ~p ∧ (~q ∨ q) … (Distributive Law)
≡ ~ p ∧ T … (Complement Law)
≡ ~ p … (Identity Law)
∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 11.
Using the rules in logic, write the negations of the following :
(i) (p ∨ q) ∧ (q ∨ ~r)
Solution:
The negation of (p ∨ q) ∧ (q ∨ ~ r) is
~ [(p ∨ q) ∧ (q ∨ ~r)]
≡ ~ (p ∨ q) ∨ ~ (q ∨ ~r) … (Negation of conjunction)
≡ (~p ∧ ~q) ∨ [~q ∧ ~(~r)] … (Negation of disjunction)
≡ {~ p ∧ ~q) ∨ (~q ∧ r) … (Negation of negation)
≡ (~q ∧ ~p) ∨ (~q ∧ r) … (Commutative law)
≡ (~ q) ∧ (~ p ∨ r) … (Distributive Law)

(ii) p ∧ (q ∨ r)
Solution:
The negation of p ∧ (q ∨ r) is
~ [p ∧ (q ∨ r)]
≡ ~ p ∨ ~(q ∨ r) … (Negation of conjunction)
≡ ~p ∨ (~q ∧ ~r) … (Negation of disjunction)

(iii) (p → q) ∧ r
Solution:
The negation of (p → q) ∧ r is
~ [(p → q) ∧ r]
≡ ~ (p → q) ∨ (~ r) … (Negation of conjunction)
≡ (p ∧ ~q) ∨ (~ r) … (Negation of implication)

(iv) (~p ∧ q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∧ q) ∨ (p ∧ ~ q) is
~ [(~p ∧ q) ∨ (p ∧ ~q)]
≡ ~(~p ∧ q) ∧ ~ (p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∨ ~q] ∧ [~p ∨ ~(q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~ p ∨ q) … (Negation of negation)

Question 12.
Express the following circuits in the symbolic form. Prepare the switching table :
(i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q: the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is :
(p ∧ q) ∨ (~p) ∨ (p ∧ ~q).

(ii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed.
Then the symbolic form of the given statement is : (p ∨ q) ∧ (p ∨ r).

Question 13.
Simplify the following so that the new circuit has minimum number of switches. Also, draw the simplified circuit.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p: the switch S₁‘ is closed or the switch S₁ is open
~ q: the switch S₂‘ is closed or the switch S₂ is open.
Then the given circuit in symbolic form is :
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~q)
Using the laws of logic, we have,
(p ∧ ~q) ∨ (~p ∧ q) ∨ (~p ∧ ~ q)
= (p ∧ ~q) ∨ [(~p ∧ q) ∨ (~p ∧ ~q) …(By Complement Law)
= (p ∧ ~q) ∨ [~p ∧ (q ∨ ~q)} (By Distributive Law)
= (p ∧ ~q) ∨ (~p ∧ T) …(By Complement Law)
= (p ∧ ~q) ∨ ~ p …(By Identity Law)
= (p ∨ ~p) ∧ (~q ∨ ~p) …(By Distributive Law)
= ~q ∨ ~p …(By Identity Law)
= ~p ∨ ~p …(By Commutative Law)
Hence, the simplified circuit for the given circuit is :

(ii)

Solution:
(ii) Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
s : the switch S₄ is closed
t : the switch S₅ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open
~ s : the switch S₄‘ is closed or the switch S₄ is open
~ t : the switch S₅‘ is closed or the switch S₅ is open.
Then the given circuit in symbolic form is
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~t] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)]
Using the laws of logic, we have,
[(p ∧ q) ∨ ~r ∨ ~s ∨ ~ t] ∧ [(p A q) ∨ (r ∧ s ∧ t)]
= [(p∧ q) ∨ ~(r ∧ s ∧ t)] ∧ [(p ∧ q) ∨ (r ∧ s ∧ t)] … (By De Morgan’s Law)
= (p ∧ q) ∨ [ ~(r ∧ s ∧ t) ∧ (r ∧ s ∧ t)] … (By Distributive Law)
= (p ∧ q) ∨ F … (By Complement Law)
= p ∧ q … (By Identity Law)
Hence, the alternative simplified circuit is :

Question 14.
Check whether the following switching circuits are logically equivalent – Justify.
(A)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
(A) The symbolic form of the given switching circuits are
p ∧ (q ∨ r) and (p ∧ q) ∨ (p ∧ r) respectively.
By Distributive Law, p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)
Hence, the given switching circuits are logically equivalent.

(B)

Solution:
The symbolic form of the given switching circuits are
(p ∨ q) ∧ (p ∨ r) and p ∨ (q ∧ r)
By Distributive Law,
p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)
Hence, the given switching circuits are logically equivalent.

Question 15.
Give alternative arrangement of the switching following circuit, has minimum switches.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed, or the switch S₁ is open
~q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form Of the given circuit is :
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r)
≡ (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) y (p ∧ ~q ∧ r) …(By Commutative Law)
≡ (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Complement Law)
≡ F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [(~p ∨ p) ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) … (By Distributive Law)
≡ [T ∧ (q ∧ r)] ∨ (p ∧ ~q ∧ r) = (q ∧ r) ∨ (p ∧ ~q ∧ r) …(By Complement Law)
≡ (q ∧ r) ∨ (p ∧ ~q ∧ r) … (By Identity Law)
≡ [q ∨ (p ∧ ~q)] ∧ r … (By Distributive Law)
≡ [q ∨ p) ∧ ((q ∨ ~q)] ∧ r … (By Distributive Law)
≡ [(q ∨ p) ∧ T] ∧ r …(By Complement Law)
≡ (q ∨ p) ∧ r … (By Identity Law)
≡ (p ∨ q) ∧ r …(By Commutative Law)
∴ the alternative arrangement of the new circuit with minimum switches is :

Question 16.
Simplify the following so that the new circuit circuit.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given switching circuit is :
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
Using the laws of logic, we have,
(~p ∨ q) ∨ (p ∨ ~q) ∨ (p ∨ q)
≡ (~p ∨ q ∨ p ∨ ~q) ∨ (p ∨ q)
≡ [(~p ∨ p) ∨ (q ∨ ~q)] ∨ (p ∨ q) … (By Commutative Law)
≡ (T ∨ T) ∨ (p ∨ q) … (By Complement Law)
≡ T ∨ (p ∨ q) … (By Identity Law)
≡ T … (By Identity Law)
∴ the current always flows whether the switches are open or closed. So, it is not necessary to use any switch in the circuit.
∴ the simplified form of given circuit is :

Question 17.
Represent the following switching circuit in symbolic form and construct its switching table. Write your conclusion from the switching table.

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open.
Then, the symbolic form of the given switching circuit is : [p ∨ (~ q) ∨ (~ r)] ∧ [p ∨ (q ∧ r)]

From the table, the’ final column’ and the column of p are identical. Hence, the given circuit is equivalent to the simple circuit with only one switch S₁.
the simplified form of the given circuit is :

Class 12 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 12 Maths 1
• Exercise 1.2 Class 12 Maths 1
• Exercise 1.3 Class 12 Maths 1
• Exercise 1.4 Class 12 Maths 1
• Exercise 1.5 Class 12 Maths 1
• Miscellaneous Exercise 1 Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.5 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.5 Questions And Answers Maharashtra Board

Question 1.
Express the following circuits in the symbolic form of logic and write the input-output table.
(i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed or the switch S₁is open
~q : the switch S₂‘ is closed or the switch S₂ is open
~r : the switch S₃‘ is closed or the switch S₃ is open
l : the lamp L is on
(i) The symbolic form of the given circuit is : p ∨ (q ∧ r) = l
l is generally dropped and it can be expressed as : p ∨ (q ∧ r).

(ii)

Solution:
The symbolic form of the given circuit is : (~ p ∧ q) ∨ (p ∧ ~ q).

(iii)

Solution:
The symbolic form of the given circuit is : [p ∧ (~q ∨ r)] ∨ (~q ∧ ~ r).

(iv)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ q ∧ (r ∨ ~p).

(v)

Solution:
The symbolic form of the given circuit is : [p ∨ (~p ∧ ~q)] ∨ (p ∧ q).

(vi)

Solution:
The symbolic form of the given circuit is : (p ∨ q) ∧ (q ∨ r) ∧ (r ∨ p)

Question 2.
Construct the switching circuit of the following :
(i) (~p∧ q) ∨ (p∧ ~r)
Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open
~ r : the switch S₃‘ is closed or the switch S₃ is open.
Then the switching circuits corresponding to the given statement patterns are :

(ii) (p∧ q) ∨ [~p ∧ (~q ∨ p ∨ r)]
Solution:

(iii) [(p ∧ r) ∨ (~q ∧ ~r)] ∧ (~p ∧ ~r)
Solution:

(iv) (p ∧ ~q ∧ r) ∨ [p ∧ (~q ∨ ~r)]
Solution:

(v) p ∨ (~p ) ∨ (~q) ∨ (p ∧ q)
Solution:

(vi) (p ∧ q) ∨ (~p) ∨ (p ∧ ~q)
Solution:

Question 3.
Give an alternative equivalent simple circuits for the following circuits :
(i)

Solution:
(i) Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch Si is open Then the symbolic form of the given circuit is :
p ∧ (~p ∨ q).
Using the laws of logic, we have,
p ∧ (~p ∨ q)
= (p ∧ ~ p) ∨ (p ∧ q) …(By Distributive Law)
= F ∨ (p ∧ q) … (By Complement Law)
= p ∧ q… (By Identity Law)
Hence, the alternative equivalent simple circuit is :

(ii)

Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~q : the switch S₂‘ is closed or the switch S₂ is open
~r : the switch S₃‘ is closed or the switch S₃ is open.
Then the symbolic form of the given circuit is :
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p).
Using the laws of logic, we have
[p ∧ (q ∨ r)] ∨ (~r ∧ ~q ∧ p)
≡ [p ∧ (q ∨ r)] ∨ [ ~(r ∨ q) ∧ p] …. (By De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] … (By Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~(q ∨ r)) … (By Distributive Law)
≡ p ∧ T … (By Complement Law)
≡ p … (By Identity Law)
Hence, the alternative equivalent simple circuit is :

Question 4.
Write the symbolic form of the following switching circuits construct its switching table and interpret it.
i)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is :
(p ∨ ~q) ∨ (~p ∧ q)

Since the final column contains all’ 1′, the lamp will always glow irrespective of the status of switches.

ii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~p : the switch S₁ is closed or the switch S₁ is open.
~q : the switch S₂‘ is closed or the switch S₂ is open.
Then the symbolic form of the given circuit is : p ∨ (~p ∧ ~q) ∨ (p ∧ q)

Since the final column contains ‘0’ when p is 0 and q is ‘1’, otherwise it contains ‘1′.
Hence, the lamp will not glow when S₁ is OFF and S₂ is ON, otherwise the lamp will glow.

iii)

Solution:
Let p : the switch S₁ is closed
q : the switch S₂ is closed
r : the switch S₃ is closed
~q : the switch S₂‘ is closed or the switch S₂ is open
~r: the switch S₃‘ is closed or the switch S₃ is open.
Then the symbolic form of the given circuit is : [p ∨ (~q) ∨ r)] ∧ [p ∨ (q ∧ r)]

From the switching table, the ‘final column’ and the column of p are identical. Hence, the lamp will glow which S₁ is ‘ON’.

Question 5.
Obtain the simple logical expression of the following. Draw the corresponding switching circuit.
(i) p ∨ (q ∧ ~ q)
Solution:
Using the laws of logic, we have, p ∨ (q ∧ ~q)
≡ p ∨ F … (By Complement Law)
≡ p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S₁ is closed
Then the corresponding switching circuit is :

(ii) (~p ∧ q) ∨ (~p ∧ ~q) ∨ (p ∧ ~q)]
Solution:
Using the laws of logic, we have,
(~p ∧ q) ∨ (~p ∨ ~q) ∨ (p ∧ ~q)
≡ [~p ∧ (q ∨ ~q)] ∨ (p ∧ ~ q)… (By Distributive Law)
≡ (~p ∧ T) ∨ (p ∧ ~q) … (By Complement Law)
≡ ~p ∨ (p ∧ ~q) … (By Identity Law)
≡ (~p ∨ p) ∧ (~p ∧~q) … (By Distributive Law)
≡ T ∧ (~p ∧ ~q) … (By Complement Law)
≡ ~p ∨ ~q … (By Identity Law)
Hence, the simple logical expression of the given expression is ~ p ∨ ~q.
Let p : the switch S₁ is closed
q : the switch S₂ is closed
~ p : the switch S₁‘ is closed or the switch S₁ is open
~ q : the switch S₂‘ is closed or the switch S₂ is open,
Then the corresponding switching circuit is :

(iii) [p (∨ (~q) ∨ ~r)] ∧ (p ∨ (q ∧ r)
Solution:
Using the laws of logic, we have,
[p ∨ (~ (q) ∨ (~r)] ∧ [p ∨ (q ∧ r)]
= [p ∨ { ~(q ∧ r)}] ∧ [p ∨ (q ∧ r)] … (By De Morgan’s Law)
= p ∨ [~(q ∧ r) ∧ (q ∧ r) ] … (By Distributive Law)
= p ∨ F … (By Complement Law)
= p … (By Identity Law)
Hence, the simple logical expression of the given expression is p.
Let p : the switch S₁ is closed
Then the corresponding switching circuit is :

(iv) (p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ ~q ∧ r) ∨ (p ∧ q ∧ r)
Question is Modified
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r)∨ (p ∧ q ∧ r)
Solution:
Using the laws of logic, we have,
(p ∧ q ∧ ~p) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r)
= (p ∧ ~p ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Commutative Law)
= (F ∧ q) ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Complement Law)
= F ∨ (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~p ∧ q ∧ r) ∨ (p ∧ q ∧ r) … (By Identity Law)
= (~ p ∨ p) ∧ (q ∧ r) … (By Distributive Law)
= T ∧ (q ∧ r) … (By Complement Law)
= q ∧ r … (By Identity Law)
Hence, the simple logical expression of the given expression is q ∧ r.
Let q : the switch S₂ is closed
r : the switch S₃ is closed.
Then the corresponding switching circuit is :

Class 12 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 12 Maths 1
• Exercise 1.2 Class 12 Maths 1
• Exercise 1.3 Class 12 Maths 1
• Exercise 1.4 Class 12 Maths 1
• Exercise 1.5 Class 12 Maths 1
• Miscellaneous Exercise 1 Class 12 Maths 1

Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.4 Questions and Answers.

12th Maths Part 1 Mathematical Logic Exercise 1.4 Questions And Answers Maharashtra Board

Question 1.
Using rules of negation write the negations of the following with justification.
(i) ~q → p
Solution:
The negation of ~q → p is
~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)

(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)

(iv) (p ∨ ~q) ∧ r
Solution:
The negation of (p ∨ ~ q) ∧ r is
~[(p ∨ ~q) ∧ r] ≡ ~(p ∨ ~q) ∨ ~r … (Negation of conjunction)
≡ [ ~p ∧ ~(~q)] ∨ ~ r… (Negation of disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)

(v) p → (p ∨ ~q)
Solution:
The negation of p → (p ∨ ~q) is
~ [p → (p ∨ ~q)] ≡ p ∧ ~ (p ∧ ~p) … (Negation of implication)
≡ p ∧ [ ~ p ∧ ~ (~(q)] … (Negation of disjunction)
≡ p ∧ (~ p ∧ q) (Negation of negation)

(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)

(vii) (p ∨ ~q) → (p ∧ ~q)
Solution:
The negation of (p ∨ ~q) → (p ∧ ~q) is
~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [ ~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~q) ∧ (~p ∨ q) … (Negation of negation)

(viii) (~ p ∨ ~q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∨ ~q) ∨ (p ∧ ~ q) is
~ [(~p ∨ ~q) ∨ (p ∧ ~ q)]
≡ ~(~p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∧ ~(~q)] ∧ [~p ∨ ~(~q)] … (Negation of disjunction and conjunction)
≡ (p ∧ q) ∧ (~p ∨ q) … (Negation of negation)

Question 2.
Rewrite the following statements without using if .. then.
(i) If a man is a judge then he is honest.
Solution:
Since p → ≡ ~p ∨ q, the given statements can be written as :
A man is not a judge or he is honest.

(ii) It 2 is a rational number then \(\sqrt {2}\) is irrational number.
Solution:
2 is not a rational number or \(\sqrt {2}\) is irrational number.

(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).

Question 3.
Without using truth table prove that :
(i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p ↔ q) ∧ (q ↔ p) … (Biconditional Law)
≡ (~p ∨ q) ∧ (~q ∨ p) … (Conditional Law)
≡ [~p ∧ (~q ∨ p)] ∨ [q ∧ (~q ∨ p)] … (Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] … (Distributive Law)
≡ [(~p ∧ ~q) ∨ F] ∨ [F ∨ (q ∧ p)] … (ComplementLaw)
≡ (~ p ∧ ~ q) ∨ (q ∧ p) … (Identity Law)
≡ (~ p ∧ ~ q) ∨ (p ∧ q) … (Commutative Law)
≡ (p ∧ q) ∨ (~p ∧ ~q) … (Commutative Law)
≡ RHS.

(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.

(iii) (p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~q) ≡ p ∨ q
Solution:
LHS = (p ∧ q) v (~p ∧ q) ∨ (p ∧ ~q)
≡ [(p ∨ ~p) ∧ q] ∨ (p ∧ ~q) … (Distributive Law)
≡ (T ∧ q) ∨ (p ∧ ~q) … (Complement Law)
≡ q ∨ (p ∧ ~q) … (Identity Law)
≡ (q ∨ p) ∧ (q ∨ ~q) … (Distributive Law)
≡ (q ∨ p) ∧ T .. (Complement Law)
≡ q ∨ p … (Identity Law)
≡ p ∨ q … (Commutative Law)
≡ RHS.

(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.

Class 12 Maharashtra State Board Maths Solution 

• Exercise 1.1 Class 12 Maths 1
• Exercise 1.2 Class 12 Maths 1
• Exercise 1.3 Class 12 Maths 1
• Exercise 1.4 Class 12 Maths 1
• Exercise 1.5 Class 12 Maths 1
• Miscellaneous Exercise 1 Class 12 Maths 1