NCERT Solutions for Class 12th Chapter 3 Maths Chapter 3 Matrices Ex 3.2

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Topics and Sub Topics in Class 11 Maths Chapter 3 Matrices:

Section Name	Topic Name
3	Matrices
3.1	Introduction
3.2	Matrix
3.3	Types of Matrices
3.4	Operations on Matrices
3.5	Transpose of a Matrix
3.6	Symmetric and Skew Symmetric Matrices
3.7	Elementary Operation (Transformation) of a Matrix
3.8	Invertible Matrices

NCERT Solutions for Class 12th Chapter 3 Maths Chapter 3 Matrices Ex 3.2

Ex 3.2 Class 12 Maths Question 1.
Let $A=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},B=\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},C=\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\qquad$
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
Solution:
Let $A=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},B=\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},C=\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\qquad$
(i) A + B
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 1

Ex 3.2 Class 12 Maths Question 2.
Compute the following:
$(i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}$
$(ii)\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & \quad { b }^{ 2 }+{ c }^{ 2 } \\ { a }^{ 2 }+{ c }^{ 2 } & \quad { a }^{ 2 }+{ b }^{ 2 } \end{bmatrix}+\begin{bmatrix} 2ab & \quad 2bc \\ -2ac & \quad -2ab \end{bmatrix}$
$(iii)\left[ \begin{matrix} \begin{matrix} -1 \\ 8 \\ 2 \end{matrix} & \begin{matrix} 4 \\ 5 \\ 8 \end{matrix} & \begin{matrix} -6 \\ 16 \\ 5 \end{matrix} \end{matrix} \right] +\left[ \begin{matrix} \begin{matrix} 12 \\ 8 \\ 3 \end{matrix} & \begin{matrix} 7 \\ 0 \\ 2 \end{matrix} & \begin{matrix} 6 \\ 5 \\ 4 \end{matrix} \end{matrix} \right]$
$(iv)\begin{bmatrix} { cos }^{ 2 }x & \quad { sin }^{ 2 }x \\ { sin }^{ 2 }x & { \quad cos }^{ 2 }x \end{bmatrix}+\begin{bmatrix} { sin }^{ 2 }x & \quad { cos }^{ 2 }x \\ { cos }^{ 2 }x & { \quad sin }^{ 2 }x \end{bmatrix}$
Solution:
$(i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}$
$=\begin{bmatrix} 2a & \quad 2b \\ 0 & \quad 2a \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 2

Ex 3.2 Class 12 Maths Question 3.
Compute the indicated products.
(i) $\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix}$
(ii) $\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \left[ \begin{matrix} 2 & 3 & 4 \end{matrix} \right]$
(iii) $\begin{bmatrix} 1 & -2 \\ 2 & \quad 3 \end{bmatrix}\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix} \right]$
(iv) $\left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix} \right] \left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix} \right]$
(v) $\left[ \begin{matrix} 2 \\ 3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \left[ \begin{matrix} \begin{matrix} 1 & 0 & 1 \end{matrix} \\ \begin{matrix} -1 & 2 & 1 \end{matrix} \end{matrix} \right]$
(vi) $\left[ \begin{matrix} \begin{matrix} 3 & -1 & 3 \end{matrix} \\ \begin{matrix} -1 & 0 & 2 \end{matrix} \end{matrix} \right] \left[ \begin{matrix} \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} & \begin{matrix} -3 \\ 0 \\ 1 \end{matrix} \end{matrix} \right]$
Solution:
(i) $\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix}$
= $\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 0 \\ 0 & { b }^{ 2 }+{ a }^{ 2 } \end{bmatrix}$
byjus class 12 maths Chapter 3 Matrices 3
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 3.1

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

Ex 3.2 Class 12 Maths Question 5.
If $A=\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] and\quad B=\left[ \begin{matrix} \frac { 2 }{ 5 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] ,$
then compute 3A – 5B.
Solution:
$3A-5B=3\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] -5\left[ \begin{matrix} \frac { 2 }{ 5 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] ,$
= $\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] -\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right]$

Ex 3.2 Class 12 Maths Question 6.
Simplify:
$cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$
Solution:
$cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$
byjus class 12 maths Chapter 3 Matrices 6

Ex 3.2 Class 12 Maths Question 7.
Find X and Y if
$(i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
$(ii)\quad 2X+3Y=\begin{bmatrix} 2 & 0 \\ 4 & 0 \end{bmatrix}and\quad 3X+2Y=\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
Solution:
$(i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 7

Ex 3.2 Class 12 Maths Question 8.
Find
$X\quad if\quad Y=\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}and\quad 2X+Y=\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$
Solution:
$Y=\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$
We are given that
byjus class 12 maths Chapter 3 Matrices 8
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 8.1

Ex 3.2 Class 12 Maths Question 9.
Find x and y, if $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Solution:
$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
=> $\begin{bmatrix} 2+y & \quad 6 \\ 1 & \quad 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
=> 2+y = 5 and 2x+2 = 8
=> y=3 and x=3
Hence x=3 and y=3

Ex 3.2 Class 12 Maths Question 10.
Solve the equation for x,y,z and t, if
$2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$
Solution:
$2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 10

Ex 3.2 Class 12 Maths Question 11.
If $x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$ then find the values of x and y
Solution:
$x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
=> $\left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
byjus class 12 maths Chapter 3 Matrices 11

Ex 3.2 Class 12 Maths 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.

Ex 3.2 Class 12 Maths Question 13.
If F(x) = $\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
then show that F(x).F(y) = F(x+y)
Solution:
F(x) = $\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
∴ F(y) = $\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 13

Ex 3.2 Class 12 Maths Question 14.
Show that
$(i)\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$
$(ii)\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{matrix} \right] \neq \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} \right]$
Solution:
$(i)L.H.S=\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$
$R.H.S=\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}=\begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$
L.H.S≠R.H.S
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 14

Ex 3.2 Class 12 Maths Question 15.
Find A² – 5A + 6I, if A = $\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right]$
Solution:
A² – 5A + 6I = $\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] -5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] +6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 15

Ex 3.2 Class 12 Maths Question 16.
If A = $\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right]$ Prove that A³-6A²+7A+2I = 0
Solution:
We have
A² = A x A
= $\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] =\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 16

Ex 3.2 Class 12 Maths Question 17.
If $A=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix},I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ then find k so that A²=kA-2I
Solution:
Given
$A=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix},I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Required: To find the value of k
Now A²=kA-2I
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 17

Ex 3.2 Class 12 Maths Question 18.
If $A=\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}$ and I is the identity matrix of order 2,show that
$I+A=I-A\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}$
Solution:
L.H.S= $I+A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 18

Ex 3.2 Class 12 Maths 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:
Let Rs 30,000 be divided into two parts and Rs x and Rs (30,000-x)
Let it be represented by 1 x 2 matrix [x (30,000-x)]
Rate of interest is 005 and 007 per rupee.
It is denoted by the matrix R of order 2 x 1.
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 19

Ex 3.2 Class 12 Maths Question 20.
The book-shop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books, 10 dozen Economics books. Their selling price are Rs 80, Rs 60 and Rs 40 each respectively. Find die total amount the book-shop will receive from selling all the books using matrix algebra.
Solution:
Number of Chemistry books = 10 dozen books
= 120 books
Number of Physics books = 8 dozen books = 96 books
Number of Economics books = 10 dozen books
= 120 books
NCERT Solutions for Class 12 Maths Chapter 3 Matrices 20

Assuming X, Y, Z, W and P are the 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 Question 21 and 22.

Ex 3.2 Class 12 Maths Question 21.
The restrictions 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) pis arbitrary, k = 3
(d) k = 2,p = 3
Solution:
Given : x_2xn, y_3xn, z_2xp, w_nx3,P_pxk
Now py +wy = P_pxk x y_3+k x w_nx3 x y_3xk
Clearly, k = 3 and p = n
Hence, option (a) is correct p x 2.

Ex 3.2 Class 12 Maths 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:
7X – 5Z = 7X_2xn – 5X_2xp
∴ We can add two matrices if their order is same n = P
∴ Order of 7X – 5Z is 2 x n.
Hence, option (b) is correct 2 x n.