RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4

Rajasthan Board RBSE Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Textbook Exercise Questions and Answers.

RBSE Class 12 Maths Solutions Chapter 3 Matrices Ex 3.4

Question 1.
\(\left[\begin{array}{rr} 1 & -1 \\ 2 & 3 \end{array}\right]\)
Answer:

Question 2.
\(\left[\begin{array}{ll} 2 & 1 \\ 1 & 1 \end{array}\right]\)
Answer:

Question 3.
\(\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 1 & 3 \\ 2 & 7 \end{array}\right]\)
For elementary operations
A = IA

Question 4.
\(\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 2 & 3 \\ 5 & 7 \end{array}\right]\)
For elementary operations
A = IA

Question 5.
\(\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right]\)
For elementary operations
A = IA

Question 6.
\(\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 2 & 5 \\ 1 & 3 \end{array}\right]\)
For elementary operations
A = IA

Question 7.
\(\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 3 & 1 \\ 5 & 2 \end{array}\right]\)
For elementary operations
A = IA

Question 8.
\(\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 4 & 5 \\ 3 & 4 \end{array}\right]\)
For elementary operations
A = IA

Question 9.
\(\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{cc} 3 & 10 \\ 2 & 7 \end{array}\right]\)
For elementary operations
A = IA

Question 10.
\(\left[\begin{array}{rr} 3 & -1 \\ -4 & 2 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{rr} 3 & -1 \\ -4 & 2 \end{array}\right]\)
For elementary operations
A = IA

Question 11.
\(\left[\begin{array}{ll} 2 & -6 \\ 1 & -2 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 2 & -6 \\ 1 & -2 \end{array}\right]\)
For elementary operations
A = IA

Question 12.
\(\left[\begin{array}{rr} 6 & -3 \\ -2 & 1 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{rr} 6 & -3 \\ -2 & 1 \end{array}\right]\)
For elementary operations
A = IA

All the elements of second row are zero.
Thus, A^-1 does not exists.

Question 13.
\(\left[\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{rr} 2 & -3 \\ -1 & 2 \end{array}\right]\)
For elementary operations

Question 14.
\(\left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right]\)
Answer:
Let A = \(\left[\begin{array}{ll} 2 & 1 \\ 4 & 2 \end{array}\right]\)
For elementary operations
A = IA

All the elements of first row of left matrix are zero. Thus A^-1 does not exist i.e., inverse of A does not exists.

Question 15.
\(\left[\begin{array}{rrr} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{array}\right]\)
Answer:

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

Question 17.
\(\left[\begin{array}{rrr} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right]\)
Answer:

Question 18.
Matrices A and B will be inverse of each other if
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0, BA = I
(D) AB = BA = I
Answer:
AB = BA = I, then A and B will be inversible to each other.
Hence, option (D) is correct.