Category: CBSE

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3

Ex 4.3 Class 12 Maths Question 1.
Find the area of the triangle with vertices at the point given in each of the following:
(i) (1,0), (6,0) (4,3)
(ii) (2,7), (1,1), (10,8)
(iii) (-2,-3), (3,2), (-1,-8)
Solution:
(i) Area of triangle = \(\frac { 1 }{ 2 } \left| \begin{matrix} 1\quad & 0 & \quad 1 \\ 6\quad & 0 & \quad 1 \\ 4\quad & 3 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [1(0-3)+1(18-0)]
= 7.5 sq units

Ex 4.3 Class 12 Maths Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:
The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)

Ex 4.3 Class 12 Maths Question 3.
Find the value of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4,0), (0,2)
(ii) (-2,0), (0,4), (0, k).
Solution:
(i) Area of ∆ = 4 (Given)
\(\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \\ 4\quad & 0 & \quad 1 \\ 0\quad & 2 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ==> k = 0
Case(b): -k + 4 = -4 ==> k = 8
Hence, k = 0,8
(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)

Ex 4.3 Class 12 Maths Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Given: Points (1,2), (3,6)
Equation of the line is

Ex 4.3 Class 12 Maths Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ = \(\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \\ 5\quad & 4 & \quad 1 \\ k\quad & 4 & \quad 1 \end{matrix} \right| \)
= \(\\ \frac { 1 }{ 2 } \) [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.3

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2

Get Free NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Determinants Exercise 4.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 4 Class 12 Determinants Ex 4.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Determinants chapter are the following:

Section Name	Topic Name
4	Determinants
4.1	Introduction
4.2	Determinant
4.3	Properties of Determinants
4.4	Area of a Triangle
4.5	Adjoint and Inverse of a Matrix
4.6	Applications of Determinants and Matrices
4.7	Summary

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2

Using the property of determinants and without expanding in Q 1 to 5, prove that

Ex 4.2 Class 12 Maths Question 1.
\(\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| =0\)
Solution:
L.H.S = \(\left| \begin{matrix} x & a & x \\ y & b & y \\ z & c & z \end{matrix} \right| +\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right| \)
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Ex 4.2 Class 12 Maths Question 2.
\(\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0\)
Solution:
L.H.S = \(\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0\)

Ex 4.2 Class 12 Maths Question 3.
\(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0\)
Solution:
\(\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{matrix} \right| \)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }-{ 9C }_{ 2 }=0\)

Ex 4.2 Class 12 Maths Question 4.
\(\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| =0\)
Solution:
L.H.S = \(\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 5.
\(\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| =2\left| \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} \right| \)
Solution:
L.H.S = ∆ = \(\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| \)

By using properties of determinants in Q 6 to 14, show that

Ex 4.2 Class 12 Maths Question 6.
\(\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0\)
Solution:
L.H.S = ∆ = \(\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| \) …(i)

Ex 4.2 Class 12 Maths Question 7.
\(\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\)
Solution:
L.H.S = \(\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 8.
(a) \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)\)
(b) \(\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 3 } & { b }^{ 3 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(a+b+c)\)
Solution:
(a) L.H.S = \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 9.
\(\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| =(x-y)(y-z)(z-x)(xy+yz+zx)\)
Solution:
Let ∆ = \(\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right|\)
Applying R1–>R1 – R2, R2–>R2 – R3

Ex 4.2 Class 12 Maths Question 10.
(a) \(\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right| =(5x+4){ (4-x) }^{ 2 }\)
(b) \(\left| \begin{matrix} y+x & y & y \\ y & y+k & y \\ y & y & y+k \end{matrix} \right| ={ k }^{ 2 }(3y+k) \)
Solution:
(a) L.H.S = \(\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right|
\)

Ex 4.2 Class 12 Maths Question 11.
(a) \(\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| ={ (a+b+c) }^{ 3 } \)
(b) \(\left| \begin{matrix} x+y+2z & \quad z & \quad z \\ x & \quad y+z+2x & \quad x \\ y & y & \quad z+x+2y \end{matrix} \right| ={ 2(x+y+z) }^{ 3 } \)
Solution:
(a) L.H.S = \(\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| \)
= \(\left( a+b+c \right) \left| \begin{matrix} 1 & \quad 1 & \quad 1 \\ 2b & \quad b-c-a & \quad 2b \\ 2c & \quad 2c & \quad c-a-b \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 12.
\(\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| ={ { (1-x }^{ 3 }) }^{ 2 } \)
Solution:
L.H.S = \(\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 13.
\(\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| ={ (1+{ a }^{ 2 }+{ b }^{ 2 }) }^{ 3 } \)
Solution:
L.H.S = \(\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| \)

Ex 4.2 Class 12 Maths Question 14.
\(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } \)
Solution:
Let ∆ = \(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| \)
\(\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab+0 & \quad ac+0 \\ ab+0\quad & \quad b^{ 2 }+1 & \quad bc+0 \\ ca+0\quad & \quad cb+0 & \quad { c }^{ 2 }+1 \end{matrix} \right| \)
This may be expressed as the sum of 8 determinants

Ex 4.2 Class 12 Maths Question 15.
If A be a square matrix of order 3×3, then | kA | is equal to
(a) k|A|
(b) k² |A|
(c) k³ |A|
(d) 3k|A|
Solution:
Option (c) is correct.

Ex 4.2 Class 12 Maths Question 16.
Which of the following is correct:
(a) Determinant is a square matrix
(b) Determinant is a number associated to a matrix
(c) Determinant is a number associated to a square matrix
(d) None of these
Solution:
Option (c) is correct

NCERT Solutions for Class 12 Maths Chapter 4 Determinants Hindi Medium Ex 4.2

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English

NCERT Solutions for Class 12th Chapter 3 Maths Chapter 3 Matrices Ex 3.4

Get Free NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Matrices Exercise 3.4 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 3 Class 12 Matrices Ex 3.4 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

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 3.4

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

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 Solutions for Class 12 Maths Chapter 3 Matrices (आव्यूह) Hindi Medium Ex 3.4

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English

NCERT Solutions for Class 12th Chapter 3 Maths Chapter 3 Matrices Ex 3.3

Get Free NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Matrices Exercise 3.3 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 3 Class 12 Matrices Ex 3.3 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

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 3.3

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

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 (आव्यूह) Hindi Medium Ex 3.3

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English

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

Get Free NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Matrices Exercise 3.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 3 Class 12 Matrices Ex 3.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

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

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 (आव्यूह) Hindi Medium Ex 3.2

More Resources for NCERT Solutions Class 12:

• Maths NCERT Solutions
• RD Sharma Class 12 Solutions
• NCERT Solutions Class 12 Physics
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions Class 12 Biology
• NCERT Solutions Class 12 Hindi
• NCERT Solutions Class 12 Vistas English
• NCERT Solutions Class 12th Flamingo English