Class 12th Chapter -3 Matrices | NCERT Maths Solution | NCERT Solution

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

Table of Contents

NCERT Solutions for Class 12 Maths Chapter : 3 Matrices

Class 12 Maths Chapter 3 Matrices Ex 3.1

Ex 3.1 Class 12 Maths Question 1.
In the matrix $A=\left[ \begin{matrix} 2 \\ 35 \\ \sqrt { 3 } \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix}\begin{matrix} 19 \\ 5/2 \\ -5 \end{matrix}\begin{matrix} -7 \\ 12 \\ 17 \end{matrix} \right]$

(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃
Solution:
(i) The matrix A has three rows and 4 columns.
The order of the matrix is 3 x 4.
(ii) There are 3 x 4 = 12 elements in the matrix A
(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = – 5, a₂₄ = 12, a₂₃ = $\\ \frac { 5 }{ 2 }$

Ex 3.1 Class 12 Maths Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
(i) 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6
Thus there are 8 matrices having 24 elements their order are (1 x 24), (24 x 1), (2 x 12), (12 x 2),(3 x 8), (8 x 3), (4 x 6), (6 x 4).
(ii) 13 = 1 x 13,
There are 2 matrices of 13 elements of order (1 x 13) and (13 x 1).

Ex 3.1 Class 12 Maths Question 3.
If a matrix has 18 elements, what are the possible orders it can have ? What, if it has 5 elements.
Solution:
We know that if a matrix is of order m × n, it has mn elements.
=> 18 = 1 x 18 = 2 x 9 = 3 x 6
Thus, all possible ordered pairs of the matrix
having 18 elements are:
(1,18), (18,1), (2,9), (9,2), (3,6), (6,3)
If it has 5 elements, then possible order are: (1,5), (5,1)

Ex 3.1 Class 12 Maths Question 4.
Construct a 2 x 2 matrix, A= [a_ij] whose elements are given by:
$(i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 }$
$(ii)\quad { a }_{ ij }=\frac { i }{ j }$
$(iii)\quad { a }_{ ij }=\frac { { (i+2j) }^{ 2 } }{ 2 }$
Solution:
$A={ \left[ { a }_{ ij } \right] }_{ 2\times 2 }=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \end{bmatrix}$
$(i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q4.1

Ex 3.1 Class 12 Maths Question 5.
Construct a 3 x 4 matrix , whose elements are given by:
$(i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right|$
$(ii){ a }_{ ij }=2i-j$
Solution:
$A={ \left[ { a }_{ ij } \right] }_{ 3\times 4 }=\left[ \begin{matrix} { a }_{ 11 } \\ { a }_{ 21 } \\ { a }_{ 31 } \end{matrix}\begin{matrix} { a }_{ 12 } \\ { a }_{ 22 } \\ { a }_{ 32 } \end{matrix}\begin{matrix} { a }_{ 13 } \\ { a }_{ 23 } \\ { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right]$
$(i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right|$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q5.1

Ex 3.1 Class 12 Maths Question 6.
Find the values of x, y, z from the following equations:
$(i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
$(ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
$(iii)\left[ \begin{matrix} \begin{matrix} x+ & y+ & z \end{matrix} \\ \begin{matrix} x & +y \end{matrix} \\ \begin{matrix} y & +z \end{matrix} \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 5 \\ 7 \end{matrix} \right]$
Solution:
$(i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
Clearly x = 1,y = 4,z = 3
$(ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
Now 5 + z = 5 => z = 0
Now x + y = 6 and xy = 8
∴ y = 6 – x and x(6 – x) = 8
6x – x² = 8
x² – 6x + 8 = 0
(x – 4)(x – 2) = 0
=>x = 2,4
When x = 2, y = 6 – 2 = 4
and when x = 4,y = 6 – 4 = 2
Hence x = 2,y = 4,z = 0 or x = 4,y = 2,z = 0.
(iii) Equating the corresponding elements.
=> x+y+z=9 …..(i)
x+z = 5 …(ii)
y+ z = 7 …(iii)
Adding eqs. (ii) & (iii)
x + y + 2z = 12
=> (x+y+z) + z = 12,
9+z = 12 (from equ (i))
z = 3
x + z = 5
=>x + 3 = 5 => x = 2
and y+z = 7
=>y+3 = 7
=> y = 4
=> x = 2, y = 4 and z = 3

Ex 3.1 Class 12 Maths Question 7.
Find the values of a,b,c and d from the equation:
$\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
Solution:
$\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q7.1

Ex 3.1 Class 12 Maths Question 8.
A = [a_ij]_m×n is a square matrix, if
(a) m < n (b) n > n
(c) m = n
(d) none of these
Solution:
For a square matrix m=n.
Thus option (c) m = n, is correct.

Ex 3.1 Class 12 Maths Question 9.
Which of the given values of x and y make the following pairs of matrices equal:
$\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$
(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:
$\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$
(a) $x=\frac { -1 }{ 3 } ,y=7$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q9.1

Ex 3.1 Class 12 Maths Question 10.
The number of all possible matrices of order 3×3 with each entry 0 or 1 is
(a) 27
(b) 18
(c) 81
(d) 512
Solution:
There are 3 x 3 matrix or 9 entries in matrix each place can be filled with 0 or 1
∴ 9 Places can be filled in 2⁹ = 512 ways
Number of such matrices = 512
Option (d) is correct.

Class 12 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 Ex 3.2 Q1.1

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 Ex 3.2 Q2.1

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) $\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix}$
= $\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 0 \\ 0 & { b }^{ 2 }+{ a }^{ 2 } \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q3.1

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 Ex 3.2 Q4.1

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:
$cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q6.1

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 Ex 3.2 Q7.1

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
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q8.1

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 Ex 3.2 Q10.1

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:
$x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
=> $\left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q11.1

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 Ex 3.2 Q13.1

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 Ex 3.2 Q14.1

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 Ex 3.2 Q15.1

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 Ex 3.2 Q16.1

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 Ex 3.2 Q17.1

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 Ex 3.2 Q18.1

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 Ex 3.2 Q19.1

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 Ex 3.2 Q20.1

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.

Class 12 Maths Chapter 3 Matrices Ex 3.3

Ex 3.3 Class 12 Maths 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:
(i) let A = $\left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right]$
∴ transpose of A = A’ = $\left[ \begin{matrix} 5 & \frac { 1 }{ 2 } & -1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q1.1

Ex 3.3 Class 12 Maths 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:
$A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q2.1

Ex 3.3 Class 12 Maths 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:
$A'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q3.1

Ex 3.3 Class 12 Maths 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:
$A'=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix},B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q4.1

Ex 3.3 Class 12 Maths 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:
$(i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right]$
$A'=\left[ \begin{matrix} 1 & -4 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q5.1

Ex 3.3 Class 12 Maths 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:
(i) $A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix}$
$A'=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q6.1

Ex 3.3 Class 12 Maths 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) For a symmetric matrix a_ij = a_ji
Now,
$A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q7.1

Ex 3.3 Class 12 Maths 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:
$A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$
=> $A'=\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q8.1

Ex 3.3 Class 12 Maths 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:
$A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right]$
$A'=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q9.1

Ex 3.3 Class 12 Maths 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:
(i) let $A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$
=> $A'=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q10.1

Ex 3.3 Class 12 Maths 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.

Ex 3.3 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q12.1
Thus option (b) is correct.

Class 12 Maths Chapter 3 Matrices Ex 3.4

Ex 3.4 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q1.1

Ex 3.4 Class 12 Maths Question 2.
$\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q2.1

Ex 3.4 Class 12 Maths Question 3.
$\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q3.1

Ex 3.4 Class 12 Maths Question 4.
$\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q4.1

Ex 3.4 Class 12 Maths Question 5.
$\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q5.1

Ex 3.4 Class 12 Maths Question 6.
$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q6.1

Ex 3.4 Class 12 Maths Question 7.
$\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q7.1

Ex 3.4 Class 12 Maths Question 8.
$\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q8.1

Ex 3.4 Class 12 Maths Question 9.
$\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q9.1

Ex 3.4 Class 12 Maths Ex 3.4 Class 12 Maths Question 10.
$\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q10.1

Ex 3.4 Class 12 Maths Question 11.
$\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q11.1

Ex 3.4 Class 12 Maths Question 12.
$\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q12.1

Ex 3.4 Class 12 Maths Question 13.
$\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q13.1

Ex 3.4 Class 12 Maths Question 14.
$\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q14.1

Ex 3.4 Class 12 Maths Question 15.
$\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q15.1

Ex 3.4 Class 12 Maths Question 16.
$\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right]$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q16.1

Ex 3.4 Class 12 Maths Question 17.
$\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q17.1

Ex 3.4 Class 12 Maths 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

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NCERT Solutions for Class 12 Maths Chapter : 3 Matrices

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Class 12th Chapter -3 Matrices | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 3 Matrices

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