NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4

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

NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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