Author name: Prasanna

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-1/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.1

Ex 3.1 Class 12 NCERT Question 1.
In the matrix A = \(\left[\begin{array}{cccc} 2 & 5 & 19 & -7 \\ 35 & -2 & \frac{5}{2} & 12 \\ \sqrt{3} & 1 & -5 & 17 \end{array}\right]\), write
i. the order of the matrix,
ii. the number of elements.
iii. Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃.
Solution:
i. The matrix has 3 rows and 4 columns. Hence the order of A is 3 x 4

ii. Number of elements = 3(4) = 12

iii. a₁₃ = 19,
a₂₁ = 35,
a₃₃ = – 5,
a₂₄ = 12,
a₂₃ = \(\frac { 5 }{ 2 }\)

Class 12 Maths Ex 3.1 NCERT Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
The ordered pairs of natural numbers whose product is 24 are (1, 24), (2, 12), (3, 8), (4, 6), (6,4), (8,3), (12,2) and (24,1). Hence the possible orders are 1 X 24, 2 x 12, 3 x 3, 8, 4 x 6, 6 x 4, 8 x 3, 12 x 2, 24 x 1.
The possible orders with 13 elements are 1 x 13, 13 x 1

Exercise 3.1 Class 12 Maths NCERT  Question 3.
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Solution:
The ordered pairs of natural numbers whose product is 18 are (1, 18), (2, 9), (3, 6), (6, 3), (9, 2) and (18, 1).
Hence the possible orders are 1 x 18, 2 x 9, 3 x 6, 6 x 3, 9 x 2, 1.8 x 1
The possible orders with 5 elements are 1 x 5, 5 x 1.

Class 12 Chapter 3 Exercise 3.1 NCERT Question 4.
Construct a 2 x 2 matrix, A = [a_ij], whose elements are given by
i. a_ij = \(\frac{(i+j)^{2}}{2}\)
ii. a_ij = \(\frac{i}{j}\)
iii. a_ij = \(\frac{(i+2j)^{2}}{2}\)
Solution:

Question 5.
Construct a 3 x 4 matrix, whose elements are given by
i. a_ij = \(\frac{1}{2}\)|-3i + j|
i. a_ij = 2i – j
Solution:
The general form of a 3 x 4 matrix is

ii. The general form of a 3 x 4 matrix is

Question 6.
Find the values of x, y and z from the following equations:
i. \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
ii. \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
iii. \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
Solution:
i. \(\left[\begin{array}{ll}
4 & 3 \\
x & 5
\end{array}\right]=\left[\begin{array}{ll}
y & z \\
1 & 5
\end{array}\right]\)
Comparing the corresponding elements of the two matrices, we get x = 1, y = 4, z = 3.

ii. \(\left[\begin{array}{cc}
x+y & 2 \\
5+z & x y
\end{array}\right]=\left[\begin{array}{cc}
6 & 2 \\
5 & 8
\end{array}\right]\)
Equating the corresponding elements, we get
x + y = 6, 5 + z = 5, xy = 8
Solving we get x = 2, y = 4, z = 0
or x = 4, y = 2, z = 0

iii. \(\left[\begin{array}{c}
x+y+z \\
x+z \\
y+z
\end{array}\right]=\left[\begin{array}{l}
9 \\
5 \\
7
\end{array}\right]\)
⇒ x + y + z = 9 … (1)
x + z = 5 … (2)
y + z = 7 … (3)
(2) + (3) ⇒ x + y + 2z = 12
(1) ⇒ x + y + z = 9
Subtracting we get z = 3
∴ x = 2,y = 4, z = 3

Question 7.
Find the value of a, b, c and d from the equation:
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Solution:
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+d
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
Equating the corresponding elements of the two matrices, we get
a – b = – 1 ………… (1)
2a + c = 5 ……….. (2)
2a – b = 0 ……….. (3)
3c + d = 13 ……….. (4)
(3) – (1) ⇒ a = 1
(3) ⇒ b = 2
(2) ⇒ c = 3
(4) ⇒ d = 4

Question 8.
A = [a]_{m x n} is a square matrix, if
(a) m n
(c) m = n
(d) None of these
Solution:
(c) m = n
For a square matrix
Number of rows = Number of columns
∴ m = n

Question 9.
Which of the given values of x and y make the following pair of matrices equal?
\(\left[\begin{array}{cc}
3 x+7 & 5 \\
y+1 & 2-3 x
\end{array}\right],\left[\begin{array}{cc}
0 & y-2 \\
8 & 4
\end{array}\right]\)
(a) x= \(\frac { -1 }{ 3 }\), y = 7
(b) Not possible to find
(c) y = 7, x = \(\frac { -2 }{ 3 }\)
(d) x = \(\frac { -1 }{ 3 }\), y = \(\frac { – 2 }{ 3 }\)
Solution:
(b) For equal matrices, the corresponding elements are equal. Equating the corre-spondingelements, we get x = \(\frac { -7 }{ 3 }\), y = 7, y = 7 and x = \(\frac { -2 }{ 3 }\)
x has 2 values which is not possible.

Question 10.
The number of all possible matrices of order 3 x 3 with each entry 0 or 1 is:
(a) 27
(b) 18
(c) 81
(d) 512
Solution:
(d) 512
There are 9 elements and each element can be either 0 or 1.
The 9 places can be filled in 29 ways.
∴ Total number of ways = 29 = 512

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-4/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4

Ex 3.4 Class 12 NCERT Solutions Question 1.
\(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
Write
A = IA

Matrices Exercise 3.4 Solutions Question 2.
\(\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\)
Write
A = IA

Question 3.
\(\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}\)
Write
A = IA

Question 4.
\(\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}\)
Write
A = IA

Question 5.
\(\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}\)
Write
A = IA

Question 6.
\(\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}\)
Write
A = IA

Question 7.
\(\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}\)
Write
A = IA

Question 8.
\(\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}\)
Write
A = IA

Question 9.
\(\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}\)
Write
A = IA

Question 10.
\(\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)
Write
A = IA

Question 11.
\(\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}\)
Write
A = IA

Question 12.
\(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\)
To use column transformation write A = AI
\(\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}\) = A\(\left[\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right]\)
Applying C₁ → C₁ + 2C₂
\(\left[\begin{array}{ll} 0 & -3 \\ 0 & 1 \end{array}\right]\) = A\(\left[\begin{array}{ll} 1 & 0 \\ 2 & 1 \end{array}\right]\)

Question 13.
\(\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}\)
Write
A = IA

Question 14.
\(\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
Solution:
Let \(A=\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}\)
Write
A = IA

Question 15.
\(\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right] \)
Solution:

Question 16.
\(\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right] \)
Solution:

Question 17.
\(\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right] \)
Solution:
Row transformation
Let \(A=\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right] \)

Question 18.
Choose the correct answer in the following question:
Matrices A and B will be inverse of each other only if
(a) AB = BA
(b) AB = BA = 0
(c) AB = 0, BA = 1
(d) AB = BA = I
Solution:
Choice (d) is correct
i.e., AB = BA = I

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-2/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.2

Ex 3.2 Class 12 Question 1.
Let
A = \(\left[\begin{array}{ll} 2 & 4 \\ 3 & 2 \end{array}\right]\)
B = \(\left[\begin{array}{ll} 1 & 3 \\ -2 & 5 \end{array}\right]\)
C = \(\left[\begin{array}{ll} -2 & 5 \\ 3 & 4 \end{array}\right]\)
F ind each of the following:
i. A + B
ii. A – B
iii. 3A – C
iv. AB
v. BA
Solution:

Class 12 Maths 3.2 Question 2.
Compute the following
i. \(\left[\begin{array}{cc}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]\)
ii. \(\left[\begin{array}{cc}
a^{2}+b^{2} & b^{2}+c^{2} \\
a^{2}+c^{2} & a^{2}+b^{2}
\end{array}\right]+\left[\begin{array}{cc}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]\)
iii. \(\left[\begin{array}{ccc}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{ccc}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\)
iv. \(\left[\begin{array}{cc}
\cos ^{2} x & \sin ^{2} x \\
\sin ^{2} x & \cos ^{2} x
\end{array}\right]+\left[\begin{array}{cc}
\sin ^{2} x & \cos ^{2} x \\
\cos ^{2} x & \sin ^{2} x
\end{array}\right]\)
Solution:

Exercise 3.2 Class 12 Maths Question 3.
Compute the indicated products.

Solution:

iii. Number of columns of first matrix = Number of rows of second matrix.

Class 12 Maths Ncert Solutions Chapter 3 Matrices Exercise 3.2 Question 4.
If A = \(\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]\), B = \(=\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]\) and C = \(\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right]\) then compute (A + B) and (B – C). Also, verify that A + (B – C) = (A+ B) – C.
Solution:

Class 12 Ex 3.2 Question 5.
If A = \(\left[\begin{array}{ccc}
\frac{2}{\dot{3}} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]\) and B = \(\left[\begin{array}{ccc}
\frac{2}{5} & \frac{3}{5} & 1 \\
\frac{1}{5} & \frac{2}{5} & \frac{4}{5} \\
\frac{7}{5} & \frac{6}{5} & \frac{2}{5}
\end{array}\right]\), then compute 3A – 5B.
Solution:

Question 6.
Simplify cos θ\(\left[\begin{array}{ll}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]\) + sin θ\(\left[\begin{array}{cc}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]\)
Solution:

Question 7.
Find X and Y, if
i. X + Y = \(\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]\) and X – Y = \(\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]\)

ii. 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) and 3X + 2Y = \(\left[\begin{array}{ll}
2 & -2 \\
-1 & 5
\end{array}\right]\)
Solution:

ii. 2X + 3Y = \(\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]\) …… (1)
3X + 2Y = \(\left[\begin{array}{ll}
2 & -2 \\
-1 & 5
\end{array}\right]\) …… (2)

Question 8.
Find X, if Y = \(\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]\) and 2X + Y = \(\left[\begin{array}{cc}
1 & 0 \\
-3 & 2
\end{array}\right]\).
Solution:

Question 9.
Find x and y, if 2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Solution:
2\(\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\) ⇒ \(\left[\begin{array}{ll}
2 & 6 \\
0 & 2x
\end{array}\right]\) + \(\left[\begin{array}{cc}
y & 0 \\
1 & 2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]\) = \(\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]\)
Equating the corresponding elements, we get
2 + y = 5 and 2x + 2 = 8
⇒ y = 3 and 2x = 6 ⇒ x = 3
∴ x = 3, y = 3

Question 10.
Solve the equation for x, y, z and t, if 2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]\) + 3\(\left[\begin{array}{ll}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
Solution:
2\(\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]\) + 3\(\left[\begin{array}{ll}
1 & -1 \\
0 & 2
\end{array}\right]\) = 3\(\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]\)
⇒ \(\left[\begin{array}{ll}
2x & 2z \\
2y & 2t
\end{array}\right]\) + \(\left[\begin{array}{cc}
3 & -3 \\
0 & 6
\end{array}\right]\) = \(\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ \(\left[\begin{array}{cc}
2x+3 & 2z-3 \\
2y & 2t+6
\end{array}\right]\) = \(\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right]\)
⇒ 2x + 3 = 9, 2z – 3 = 15
2y = 12, 2t + 6 = 18 ⇒ x = 3, y = 6, t = 6, z = 9

Question 11.
If x\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\) + y\(\left[\begin{array}{l}
-1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\)
Solution:
x\(\left[\begin{array}{l}
2 \\
3
\end{array}\right]\) + y\(\left[\begin{array}{l}
-1 \\
1
\end{array}\right]\) = \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\) ⇒ \(\left[\begin{array}{l}
2 x \\
3 x
\end{array}\right]+\left[\begin{array}{c}
-y \\
y
\end{array}\right]=\left[\begin{array}{c}
10 \\
5
\end{array}\right]\) ⇒ \(\left[\begin{array}{l}
10 \\
5
\end{array}\right]\)
⇒ 2x – y = 10, 3x + y = 5
Solving, we get x = 3, y = – 4

Question 12.
Given
\(3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix} \)
find the values of x,y,z and w.
Solution:
\(3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix} \)
⇒ \(\begin{bmatrix} 3x & \quad 3y \\ 3z & \quad 3w \end{bmatrix}=\begin{bmatrix} x+4 & \quad 6+x+y \\ -1+z+w & \quad 2w+3 \end{bmatrix}\)
⇒ 3x = x + 4 ⇒ x = 2
and 3y = 6 + x + y ⇒ y = 4
Also, 3w = 2w + 3 ⇒ w = 3
Again, 3z = – 1 + z + w
⇒ 2z = – 1 + 3
⇒ 2z = 2
⇒ z = 1
Hence x = 2 ,y = 4, z = 1, w = 3.

Question 13.
If F(x) = \(\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\) then show that F(x).F(y) = F(x+y)
Solution:

Question 14.
Show that

Solution:

Question 15.
Find A² – 5A + 6I, if A = \(\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\)
Solution:

Question 16.
If A = \(\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\) Prove that A³ – 6A² + 7A + 2I = 0
Solution:

Question 17.
If \(\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\) and I = \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\) find k so that A² = kA – 2I
Solution:

Question 18.
If A = \(\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\) and I is the identity matrix of order 2, show that I + A = (I – A)\(\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]\)
Solution:

Question 19.
A trust has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bond if the trust fund obtains an annual total interest of
(a) Rs 1800
(b) Rs 2000
Solution:
(a) Let ₹ and ₹ y be the amount invested in two bonds such that x + y = 30000 … (1)
Then [x y]\(\left[\begin{array}{l}
5 \% \\
7 \%
\end{array}\right]\) = [1800]
⇒ [x × 5% + y x 7%] = [1800]
⇒ \(\frac{x \times 5}{100}+\frac{y \times 7}{100}\)
⇒ 5x + 7y = 180000 … (2)
(1) x 5 = 5x + 5y = 150000 … (3)
Subtracting (3) from (2), we get
2y = 30000 ⇒ y = 15000
From (1), we get x= 15000
∴ To get an interest of ₹ 1800, the amount to be invested in each bonds are ₹ 15000 and ₹ 15000

(b) Here x + y = 30000 … (4)
[x y]\(\left[\begin{array}{l}
5 \% \\
7 \%
\end{array}\right]\) = 2000
⇒ \(\frac{x \times 5}{100}+\frac{y \times 7}{100}\) = 2000
⇒ 5x + 7y = 200000 … (5)
(4) x 5 ⇒ 5x + 5y = 150000 …. (6)
Subtracting (6) from (5), we get
2y = 50000 ⇒ y = 25000
From (4) we get x = 5000
To get an interest of ₹ 2000, the amount to be invested in each bonds is ₹ 5000 and ₹ 25000.

Question 20.
The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are ₹ 80, ₹ 60 and ₹ 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
Solution:
1 dozen = 12
∴ 10 dozen chemistry books = 10 x 12 = 120
8 dozen physics books = 8 x 12 = 96
10 dozen economics books = 10 x 12 = 120
Total amount the bookshop will receive
= [120 96 120]\(\left[\begin{array}{l} 80 \\ 60 \\ 40 \end{array}\right]\)
= [120 x 80 + 96 x 60 + 120 x 40]
= [9600 + 5760 + 4800] = ₹ 20160
Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k, respectively. Choose the correct answer in Exercises 21 and 22.

Question 21.
The restriction on n, k and p so that PY + WY will be defined are
a. k = 3, p = n
b. k is arbitrary, p = 2
c. p is arbitrary, k = 3
d. k = 2, p = 3
Solution:
Answer:
PY + WY = (P + W)Y
⇒ order of P = order of W
⇒ p k = n x 3
⇒ p = n and k = 3

Question 22.
If n = p, then the order of the matrix 7X – 5Z is
a. p x 2
b. 2 x n
c. n x 3
d. p x n
Solution:
Order of X = 2 x n
Order of Z = 2 x p = 2 x n ∵ (n = p)
∴ Order of 7X – 5Z is order of X or order of Z.

These NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-3-ex-3-3/

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.3

Ex 3.3 Class 12 NCERT Solutions Question 1.
Find the transpose of each of the following matrices:
(i) \(\left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right] \)
(ii) \(\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\)
(iii) \(\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt { 3 } & 5 & 6 \\ 2 & 3 & -1 \end{matrix} \right] \)
Solution:

Question 2.
If \(A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right] \)
then verify that:
(i) (A + B)’ = A’ + B’
(ii) (A – B)’ = A’ – B’
Solution:
i. (A + B)

From (1) and (2), (A + B)’ = A’ + B’

ii. (A – B)

From (1) and (2), (A – B)’ = A’ – B’.

Question 3.
If \(A’=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right] \)
then verify that:
(i) (A+B)’ = A’+B’
(ii) (A-B)’ = A’-B’
Solution:

Question 4.
If \(A’=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix} ,B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} \) then find (A+2B)’
Solution:

Question 5.
For the matrices A and B, verify that (AB)’ = B’A’, where
\((i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \end{matrix} \right] \)
\((ii)\quad A=\left[ \begin{matrix} 0 \\ 1 \\ 2 \end{matrix} \right] ,B=\left[ \begin{matrix} 1 & 5 & 7 \end{matrix} \right] \)
Solution:
\((i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] \)
\(A’=\left[ \begin{matrix} 1 & -4 & 3 \end{matrix} \right] \)

Question 6.
If (i) \(A=\begin{bmatrix} cos\alpha & \quad sin\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix} \) , the verify that A’A = I
If (ii) \(A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -cos\alpha & \quad sin\alpha \end{bmatrix} \), the verify that A’A = I
Solution:

Question 7.
(i) Show that the matrix \(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \) is a symmetric matrix.
(ii) Show that the matrix \(A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \) is a skew-symmetric matrix.
Solution:
(i) \(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \), A’ = \(\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \) = A
A is symmetric matrix, since A’ = A

(ii) \(A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -cos\alpha & \quad sin\alpha \end{bmatrix} \)
A’ = \(\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \) = – 1\(\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right] \)
A’ = – A
A is a skew-symmetric matrix, since A’ = A

Question 8.
For the matrix, \(A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}\)
(i) (A + A’) is a symmetric matrix.
(ii) (A – A’) is a skew-symmetric matrix.
Solution:

Question 9.
Find \(\\ \frac { 1 }{ 2 } (A+A’)\) and \(\\ \frac { 1 }{ 2 } (A-A’)\), when \(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
Solution:

Question 10.
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
(i) \(\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}\)
(ii) \(\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right] \)
(iii) \(\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right] \)
(iv) \(\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}\)
Solution:

Thus A can be represented as a sum of a symmetric and a skew-symmetric matrix.

Thus B can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Thus A can be expressed as the sum of a symmetric and skew-symmetric matrix.

Thus D can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Question 11.
Choose the correct answer in the following questions:
If A, B are symmetric matrices of same order then AB-BA is a
(a) Skew – symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity matrix
Solution:
Now A’ = B, B’ = B
(AB – BA)’ = (AB)’ – (BA)’
= B’A’ – A’B’
= BA – AB
= – (AB – BA)
AB – BA is a skew-symmetric matrix Hence, option (a) is correct.

Question 12.
If \(A=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}\) then A+A’ = I, if the
value of α is
(a) \(\frac { \pi }{ 6 } \)
(b) \(\frac { \pi }{ 3 } \)
(c) π
(d) \(\frac { 3\pi }{ 2 } \)
Solution:
Now

Thus option (b) is correct.

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-miscellaneous-exercise/

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Question 1.
cos^-1(cos\(\frac { 13π }{ 6 }\))
Solution:
The principal value brach of cos^-1 is [0, p]

Question 2.
tan^-1(tan\(\frac { 7π }{ 6 }\))
Solution:
The principal value brach of tan^-1 is (- \(\frac { π }{ 2 }\), \(\frac { π }{ 2 }\))

Question 3.
2sin^-1\(\frac { 3 }{ 5 }\) = tan^-1\(\frac { 24 }{ 7 }\)
Solution:

Question 4.
sin^-1\(\frac { 8 }{ 17 }\) + sin^-1\(\frac { 3 }{ 5 }\) = tan^-1\(\frac { 77 }{ 36 }\)
Solution:

Question 5.
cos^-1\(\frac { 4 }{ 5 }\) + cos^-1\(\frac { 12 }{ 13 }\) = cos^-1\(\frac { 33 }{ 65 }\)
Solution:

Question 6.
cos^-1\(\frac { 12 }{ 13 }\) + sin^-1\(\frac { 3 }{ 5 }\) = sin^-1\(\frac { 56 }{ 65 }\)
Solution:

Question 7.
tan^-1\(\frac { 63 }{ 16 }\) = sin^-1\(\frac { 5 }{ 13 }\) + cos^-1\(\frac { 3 }{ 5 }\)
Solution:

Question 8.
tan^-1\(\frac { 1 }{ 5 }\) + tan^-1\(\frac { 1 }{ 7 }\) + tan^-1\(\frac { 1 }{ 3 }\) + tan^-1\(\frac { 1 }{ 8 }\) = \(\frac { π }{ 4 }\)
Solution:

Question 9.
tan^-1\(\sqrt{x}\) = \(\frac { 1 }{ 2 }\)cos^-1\(\left(\frac{1-x}{1+x}\right), x \in[0,1]\)
Solution:

Question 10.
cot^-1\(\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\) = \(\frac { x }{ 2 }\) x ∈ (0, \(\frac { π }{ 4 }\))
Solution:

Question 11.
tan^-1\(\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)\) = \(\frac { π }{ 4 }\) – \(\frac { 1 }{ 2 }\) cos^-1 x, – \(\frac{1}{\sqrt{2}} \leq x \leq 1\)
Solution:

Question 12.
\(\frac { 9p }{ 8 }\) – \(\frac { 9 }{ 4 }\)sin^-1\(\frac { 1 }{ 3 }\) = \(\frac { 9 }{ 4 }\)sin^-1\(\frac{2 \sqrt{2}}{3}\)
Solution:

Question 13.
2tan^-1(cosx) = tan^-1(2cosecx)
Solution:

Question 14.
tan^-1\(\frac{1-x}{1+x}\) = \(\frac { 1 }{ 2 }\) tan^-1x, (x > 0)
Solution:

Question 15.
sin(tan^-1x), |x| < 1 is equal to
a. \(\frac{x}{\sqrt{1-x^{2}}}\)
b. \(\frac{1}{\sqrt{1-x^{2}}}\)
c. \(\frac{1}{\sqrt{1+x^{2}}}\)
d. \(\frac{x}{\sqrt{1+x^{2}}}\)
Solution:
d. \(\frac{x}{\sqrt{1+x^{2}}}\)

Question 16.
sin^-1(1 – x) – 2sin^-1x = \(\frac { π }{ 2 }\), then x is equal to
a. 0, \(\frac { 1 }{ 2 }\)
b. 1, \(\frac { 1 }{ 2 }\)
c. 0
d. \(\frac { 1 }{ 2 }\)
Solution:
c. 0

But x ≠ \(\frac { 1 }{ 2 }\), since sin^-1(1 – \(\frac { 1 }{ 2 }\)) ≠ \(\frac { π }{ 2 }\) + 2 sin^-1\(\frac { 1 }{ 2 }\)
∴ x = 0.

Question 17.
tan^-1 \(\frac { x }{ y }\) – tan^-1 \(\frac{x-y}{x+y}\)
Solution:
a. \(\frac { π }{ 2 }\)
b. \(\frac { π }{ 3 }\)
c. \(\frac { π }{ 4 }\)
d. \(\frac { -3π }{ 4 }\)
Solution:
c. \(\frac { π }{ 4 }\)

These NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-2-ex-2-2/

NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Exercise 2.2

Ex 2.2 Class 12 NCERT Solutions Question 1.
3sin^-1x = sin^-1(3x – 4x³). x ∈ [-\(\frac { 1 }{ 2 }\), \(\frac { 1 }{ 2 }\)]
Solution:
Let θ = sin^-1 ⇒ x = sin θ
∴ 3x – 4x³ = 3sinθ – 4sin³θ = sin3 θ
∴ 3θ = sin^-1(3x – 4x³)
⇒ 3sin^-1x = sin^-1(3x – 4x³)

Class 12th Maths Chapter 2 Exercise 2.2 Question 2.
3cos^-1x = cos^-1(4x³ – 3x). x ∈ [\(\frac { 1 }{ 2 }\), 1]
Solution:
Let θ cos^-1 ⇒ x = cos θ
4x³ – 3x = 4cos³θ – 3cosθ = cos 3 θ
⇒ 3θ = cos^-1(4x³ – 3x)
⇒ 3cos^-1x = cos^-1(4x³ – 3x)

Maths Ex 2.2 Class 12 NCERT Solutions Question 3.
tan^-1\(\frac { 2 }{ 11 }\) + tan^-1\(\frac { 7 }{ 24 }\) = tan^-1\(\frac { 1 }{ 2 }\)
Solution:

Ncert Ex 2.2 Class 12  Question 4.
2tan^-1\(\frac { 1 }{ 2 }\) + tan^-1\(\frac { 1 }{ 7 }\) = tan^-1\(\frac { 31 }{ 17 }\)
Solution:

Exercise 2.2 Maths Class 12 Solutions Question 5.
tan^-1\(\frac{\sqrt{1+x^{2}}-1}{x}\), x ≠ 0
Solution:

Question 6.
tan^-1\(\frac{1}{\sqrt{x^{2}-1}}\), |x| > 1
Solution:

Question 7.
tan^-1\(\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)\), x < π
Solution:

Question 8.
tan^-1\(\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right)\), x < π
Solution:

Question 9.
tan^-1\(\frac{x}{\sqrt{a^{2}-x^{2}}}\), |x| < a
Solution:

Question 10.
tan^-1\(\left(\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}\right), a>0 ; \frac{-a}{\sqrt{3}} \leq x \leq \frac{a}{\sqrt{3}}\).
Solution:

Question 11.
tan^-1[2 cos(2 sin^-1\(\frac{1}{2}\))]
Solution:

Question 12.
cot(tan^-1 a + cot^-1 a)
Solution:
Since tan^-1 a + cot^-1 a = \(\frac{π}{2}\),
cot(tan^-1 a + cot^-1 a = cot(\(\frac{π}{2}\)) = 0

Question 13.
tan\(\frac{1}{2}\)[\(\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]\))], |x| < 1, y > 0 and xy < 1
Solution:

Question 14.
If sin(sin^-1\(\frac{1}{5}\) + cos^-1x) = 1, then find the value of x.
Solution:

Question 15.
If tan^-1\(\frac{x-1}{x-2}\) + tan^-1\(\frac{x+1}{x+2}\), then find the value of x.
Solution:

Question 16.
sin^-1\(\left(\sin \frac{2 \pi}{3}\right)\)
Solution:

Question 17.
tan^-1\(\left(\tan \frac{3 \pi}{4}\right)\)
Solution:

Question 18.
tan\(\left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)
Solution:

Question 19.
cos^-1\(\left(\cos \frac{7 \pi}{6}\right)\) is equal to
a. \(\frac{7π}{6}\)
b. \(\frac{5π}{6}\)
c. \(\frac{π}{3}\)
d. \(\frac{π}{6}\)
Solution:
b. \(\frac{5π}{6}\)

Question 20.
sin\(\left(\frac{\pi}{3}-\sin ^{-1}\left(\frac{-1}{2}\right)\right)\) is equal to
a. \(\frac{1}{2}\)
b. \(\frac{1}{3}\)
c. \(\frac{1}{4}\)
d. 1
Solution:
d. 1

Question 21.
tan^-1\(\sqrt{3}\) – cot^-1\(\sqrt{3}\) is equal to
a. π
b. – \(\frac{π}{2}\)
c. 0
d. 2\(\sqrt{3}\)
Solution:
b. – \(\frac{π}{2}\)
tan^-1\(\sqrt{3}\) – cot^-1(-\(\sqrt{3}\))
= \(\sqrt{3}\) – (π – cot^-1\(\sqrt{3}\))
= (tan^-1\(\sqrt{3}\) + cot^-1\(\sqrt{3}\)) – π
= \(\frac{π}{2}\) – π = – \(\frac{π}{2}\)