NCERT Solutions | Class 12 Maths Chapter 3

NCERT Solutions | Class 12 Maths Chapter 3 | Matrix 

CBSE Solutions | Maths Class 12

Check the below NCERT Solutions for Class 12 Maths Chapter 3 Matrix Pdf free download. NCERT Solutions Class 12 Maths  were prepared based on the latest exam pattern. We have Provided Matrix Class 12 Maths NCERT Solutions to help students understand the concept very well.

NCERT | Class 12 Maths

NCERT Solutions Class 12 Maths

Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	12th
Subject:	Maths
Chapter:	3
Chapters Name:	Matrix
Medium:	English

Matrix | Class 12 Maths | NCERT Books Solutions

You can refer to MCQ Questions for Class 12 Maths Chapter 3 Matrix to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

NCERT Solutions for 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 = \(\\ \frac { 5 }{ 2 } \)

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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.
\(A={ \left[ { a }_{ ij } \right] }_{ 2\times 2 }=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \end{bmatrix}\)
\((i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 } \)

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.
\(A={ \left[ { a }_{ ij } \right] }_{ 3\times 4 }=\left[ \begin{matrix} { a }_{ 11 } \\ { a }_{ 21 } \\ { a }_{ 31 } \end{matrix}\begin{matrix} { a }_{ 12 } \\ { a }_{ 22 } \\ { a }_{ 32 } \end{matrix}\begin{matrix} { a }_{ 13 } \\ { a }_{ 23 } \\ { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right] \)
\((i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right| \)

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.
\(\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}\)

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) \(x=\frac { -1 }{ 3 } ,y=7\)

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.

 

NCERT Solutions for 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

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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.
\((i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}\)
\(=\begin{bmatrix} 2a & \quad 2b \\ 0 & \quad 2a \end{bmatrix}\)

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) \(\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix} \)
= \(\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 0 \\ 0 & { b }^{ 2 }+{ a }^{ 2 } \end{bmatrix}\)

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
\(A=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix} \right] ,C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{matrix} \right] \)

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.
\(cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix} \)

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.
\((i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

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

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.
\(2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}\)

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.
\(x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] \)
=> \(\left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] \)

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) = \(\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

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

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 = \(\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] -5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] +6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

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
= \(\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] =\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)

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

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=\(I+A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}\)

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.

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

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.

NCERT Solutions for 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

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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.
\((i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}\)
\(=\begin{bmatrix} 2a & \quad 2b \\ 0 & \quad 2a \end{bmatrix}\)

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) \(\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix} \)
= \(\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 0 \\ 0 & { b }^{ 2 }+{ a }^{ 2 } \end{bmatrix}\)

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
\(A=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix} \right] ,C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{matrix} \right] \)

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.
\(cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix} \)

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.
\((i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

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

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.
\(2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}\)

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.
\(x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] \)
=> \(\left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right] \)

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) = \(\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

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

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 = \(\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] -5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] +6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)

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
= \(\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] =\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)

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

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=\(I+A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}\)

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.

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

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.

NCERT Solutions for 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’ = \(\left[ \begin{matrix} 5 & \frac { 1 }{ 2 } & -1 \end{matrix} \right] \)

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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.
\(A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right] \)

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.
\(A’=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right] \)

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.
\(A’=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix},B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix} \)

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.
\((i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] \)
\(A’=\left[ \begin{matrix} 1 & -4 & 3 \end{matrix} \right] \)

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) \(A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix} \)
\(A’=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix} \)

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,
\(A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right] \)

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.
\(A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}\)
=> \(A’=\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}\)

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.
\(A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right] \)
\(A’=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right] \)

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 \(A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}\)
=> \(A’=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}\)

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

Thus option (b) is correct.

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4

Using Elementary transformation, find the inverse each of matrices, if it exists in ques 1 to 17.

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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

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

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

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

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

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

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

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

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

Ex 3.4 Class 12 Maths Ex 3.4 Class 12 Maths Question 10.
\(\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)
Solution.
Let \(A=\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}\)

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

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

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

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

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 \(A=\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right] \)

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

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

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

NCERT Class 12 Maths

Class 12 Maths Chapters | Maths Class 12 Chapter 3

Chapterwise NCERT Solutions for Class 12 Maths

NCERT Solutions For Class 12 Maths Chapter 1 Relations and Functions
NCERT Solutions For Class 12 Maths Chapter 2 Inverse Trigonometric Functions
NCERT Solutions For Class 12 Maths Chapter 3 Matrix
NCERT Solutions For Class 12 Maths Chapter 4 Determinants
NCERT Solutions For Class 12 Maths Chapter 5 Continuity and Differentiability
NCERT Solutions For Class 12 Maths Chapter 6 Application of Derivatives
NCERT Solutions For Class 12 Maths Chapter 7 Integrals
NCERT Solutions For Class 12 Maths Chapter 8 Application of Integrals
NCERT Solutions For Class 12 Maths Chapter 9 Differential Equations
NCERT Solutions For Class 12 Maths Chapter 10 Vector Algebra
NCERT Solutions For Class 12 Maths Chapter 11 Three-dimensional Geometry
NCERT Solutions For Class 12 Maths Chapter 12 Linear Programming
NCERT Solutions For Class 12 Maths Chapter 13 Probability

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