# NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 (Ex 3.4)

## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 (Ex 3.4)

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 supplied by Vedantu to help the Class 12 students for understanding Math easy. The Class 12 Maths NCERT Solutions Chapter 3 Exercise 3.4 contains the matrices, which is hard to understand for many students. But the easy and step by step explanation of the Math Class 12 Chapter 3 Exercise 3.4 problems, which is prepared by our expert teachers, will help them to score more mark in the examination.

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## Access NCERT Solutions for Class 12 Maths Chapter 3 – Matrices

Exercise 3.4

1. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 1 & -1 \\ 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -2 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to \dfrac{1}{5}{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ \dfrac{-2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{-2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} \dfrac{3}{5} & \dfrac{1}{5} \\ \dfrac{-2}{5} & \dfrac{1}{5} \\ \end{matrix} \right]$

2. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 5 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\ \end{matrix} \right]$

3. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 1 & 3 \\ 2 & 7 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 1 & 3 \\ 2 & 7 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 1 & 3 \\ 2 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 3 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -2 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-3{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 7 & -3 \\ -2 & 1 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 7 & -3 \\ -2 & 1 \\ \end{matrix} \right]$

4. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 3 \\ 5 & 7 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 3 \\ 5 & 7 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 3 \\ 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 2 & 3 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -2 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\leftrightarrow {{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 1 \\ 2 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 \\ 1 & 0 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\leftrightarrow {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 \\ 5 & -2 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & 3 \\ 5 & -2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 7 & 3 \\ 5 & -2 \\ \end{matrix} \right]$

5. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 1 \\ 7 & 4 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 1 \\ 7 & 4 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 1 \\ 7 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-3{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 2 & 1 \\ 1 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ -3 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & -1 \\ -7 & 2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 4 & -1 \\ -7 & 2 \\ \end{matrix} \right]$

6. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 5 \\ 1 & 3 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 5 \\ 1 & 3 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 5 \\ 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to \dfrac{1}{2}{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & \dfrac{5}{2} \\ 0 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & \dfrac{5}{2} \\ 0 & \dfrac{1}{2} \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 \\ \dfrac{-1}{2} & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-5{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & \dfrac{1}{2} \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & -5 \\ \dfrac{-1}{2} & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to 2{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & -5 \\ -1 & 2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 3 & -5 \\ -1 & 2 \\ \end{matrix} \right]$

7. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 3 & 1 \\ 5 & 2 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 3 & 1 \\ 5 & 2 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 3 & 1 \\ 5 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to 2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 6 & 2 \\ 5 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 5 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & -1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-5{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & -1 \\ -10 & 6 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to \dfrac{1}{2}{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & -1 \\ -5 & 3 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 2 & -1 \\ -5 & 3 \\ \end{matrix} \right]$

8. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 4 & 5 \\ 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 1 \\ 3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-3{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \\ \end{matrix} \right]$

9. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 3 & 10 \\ 2 & 7 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 3 & 10 \\ 2 & 7 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 3 & 10 \\ 2 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 3 \\ 2 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 3 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -1 \\ -2 & 3 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-3{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 7 & -10 \\ -2 & 3 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 7 & -10 \\ -2 & 3 \\ \end{matrix} \right]$

10. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 3 & -1 \\ -4 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ -4 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to -{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}+4{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -1 \\ -4 & -3 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to -\dfrac{1}{2}{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & -1 \\ 2 & \dfrac{3}{2} \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & \dfrac{1}{2} \\ 2 & \dfrac{3}{2} \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 1 & \dfrac{1}{2} \\ 2 & \dfrac{3}{2} \\ \end{matrix} \right]$

11. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & -6 \\ 1 & -2 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & -6 \\ 1 & -2 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & -6 \\ 1 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\leftrightarrow {{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 2 & -6 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 0 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ 1 & -2 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to -\dfrac{1}{2}{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -2 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 \\ \dfrac{-1}{2} & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+2{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 3 \\ \dfrac{-1}{2} & 1 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} -1 & 3 \\ \dfrac{-1}{2} & 1 \\ \end{matrix} \right]$

12. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 6 & -3 \\ -2 & 1 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 6 & -3 \\ -2 & 1 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 6 & -3 \\ -2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to \dfrac{1}{6}{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & \dfrac{-1}{2} \\ -2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{6} & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}+2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -\dfrac{1}{2} \\ 0 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{6} & 3 \\ \dfrac{1}{3} & 1 \\ \end{matrix} \right]A$

Since, we can see all the zeros in the second row of the matrix on the L.H.S, $\therefore {{A}^{-1}}$ does not exist.

13. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & -3 \\ -1 & 2 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & -3 \\ -1 & 2 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & -3 \\ -1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ -1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}+{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & -1 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 1 \\ 1 & 2 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 2 & 3 \\ 1 & 2 \\ \end{matrix} \right]$

14. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 1 \\ 4 & 2 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 1 \\ 4 & 2 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 1 \\ 4 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-\dfrac{1}{2}{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 0 & 0 \\ 4 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & \dfrac{-1}{2} \\ 0 & 1 \\ \end{matrix} \right]A$

Since, we can see all the zeros in the first row of the matrix on the L.H.S, $\therefore {{A}^{-1}}$ does not exist.

15.  Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 0 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}+3{{R}_{1}}$ and ${{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+3{{R}_{3}}$ and ${{R}_{2}}\to {{R}_{2}}+8{{R}_{3}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & -1 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} -5 & 0 & 3 \\ -13 & 1 & 8 \\ -2 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{3}}\to {{R}_{3}}+{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & 0 & 25 \\ \end{matrix} \right]=\left[ \begin{matrix} -5 & 0 & 3 \\ -13 & 1 & 8 \\ -15 & 1 & 9 \\ \end{matrix} \right]A$

Applying ${{R}_{3}}\to \dfrac{1}{25}{{R}_{3}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & 10 \\ 0 & 1 & 21 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -5 & 0 & 3 \\ -13 & 1 & 8 \\ -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}-10{{R}_{3}}$ and ${{R}_{2}}\to {{R}_{2}}-21{{R}_{3}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\ -\dfrac{2}{5} & \dfrac{4}{25} & \dfrac{11}{25} \\ -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 1 & -\dfrac{2}{5} & -\dfrac{3}{5} \\ -\dfrac{2}{5} & \dfrac{4}{25} & \dfrac{11}{25} \\ -\dfrac{3}{5} & \dfrac{1}{25} & \dfrac{9}{25} \\ \end{matrix} \right]$

16. Find the inverse of each of the matrices, if it exists.

$\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \\ \end{matrix} \right]$

Ans: Let $A=\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \\ \end{matrix} \right]$

Since $A=IA$

$\therefore \left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to \dfrac{1}{2}{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & -\dfrac{1}{2} \\ 5 & 1 & 0 \\ 0 & 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{2}}\to {{R}_{2}}-5{{R}_{1}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & -\dfrac{1}{2} \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 & 0 \\ -\dfrac{5}{2} & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{3}}\to {{R}_{3}}-{{R}_{2}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & -\dfrac{1}{2} \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 0 & \dfrac{1}{2} \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 & 0 \\ -\dfrac{5}{2} & 1 & 0 \\ \dfrac{5}{2} & -1 & 1 \\ \end{matrix} \right]A$

Applying ${{R}_{3}}\to 2{{R}_{3}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & -\dfrac{1}{2} \\ 0 & 1 & \dfrac{5}{2} \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} \dfrac{1}{2} & 0 & 0 \\ -\dfrac{5}{2} & 1 & 0 \\ 5 & -2 & 2 \\ \end{matrix} \right]A$

Applying ${{R}_{1}}\to {{R}_{1}}+\dfrac{1}{2}{{R}_{3}}$ and ${{R}_{2}}\to {{R}_{2}}-\dfrac{5}{2}{{R}_{3}}$

$\Rightarrow \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \\ \end{matrix} \right]A$

$\therefore {{A}^{-1}}=\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 &-2 & 2 \\ \end{matrix} \right]$

17. Matrices $A$ and $B$ will be inverse of each other only if

1. $AB=BA$

2. $AB=0,BA=I$

3. $AB=BA=0$

4. $AB=BA=I$

Ans: Since, if $A$ is a square matrix of order $m$ , and if there exists another square matrix $B$ of the same order $m$ , such that $AB=BA=I$ , then $B$ is said to be the inverse of $A$. In such a case, it is clear that $A$ is the inverse of $B$.

Thus, matrices $A$ and $B$ will be inverse of each other only if $AB=BA=I$.

Thus, option (D) is correct.

### Key Learnings from NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

Chapter 3 - Matrices is a very important chapter in Class 12 Maths. Exercise 3.4 Class 12 Maths NCERT Solutions is mainly based on basic concepts of elementary operations of a matrix and the inverse of a matrix. Below are some key learnings from this chapter.

• Elementary operation (transformation) of a matrix: Generally, there are six operations (or transformations) on a matrix, three of these operations are due to rows and the other three are due to columns. Interchanging the elements of two columns or rows, multiplying the elements in a column (or row) by a non-zero number and multiplying a column (or row) by a non-zero number, and adding the result to another column (or row) are the six transformations.

• Invertible matrices: If X is a square matrix of order a and if there is another square matrix Y of the same order so that XY = YX = I, then Y is called the inverse matrix of X and is denoted by X⁻¹ and matrix X is said to be invertible.

• If matrix Y is the inverse of matrix X, then matrix X is also the inverse of matrix Y.

• The inverse of a matrix by elementary operations: The inverse of a matrix X is a matrix that results in the identity (when multiplied by A).

This exercise consists of questions such as finding the inverse of a matrix by using these transformations.

### Benefits of NCERT Solutions Class 12 Maths Chapter 3

Some of the prime benefits of solving Class 12 Maths NCERT Solutions Chapter 3 are listed below:

1. The solutions are provided in a very simple way by our expert teachers who have years of experience.

2. By following these solutions, students will have a better understanding of the fundamental concepts of the matrix.

3. The solutions are solved by keeping in mind the CBSE format so that students can score good marks by practising these questions.

## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

All the solutions of our Math’s problem have been explained in an easy method with showing each step to make you understand the principle and reasons completely. Many of the students lose interest just in the name of Math. The Class 12 Maths includes all of those chapters you have learnt previously. So it may make you feel overloaded sometimes. Matrices are one of those chapters, which is very easy once if you understand the properties clearly.

Our expert teachers have worked for hand in hand with us to deliver you the best solutions for Class 12 Maths NCERT Solutions Chapter 3 Exercise 3.4. The solutions give you an overview of all the processes followed by giving solved answers to the chapter. By going through this Math solution, students can achieve their goals in the board examination.

Class 12 Maths Ex 3.4 Solutions are created by our expert teachers. Matrices are a very interesting chapter, to begin with. But the appearance of the chapter and the large format of the questions frighten the students. This is also the chapter which will bring definite marks in the exam. So it is advisable to refer to our Matrices NCERT Solutions Exercise 3.4 to score better in your board exam.

Scoring a good number in your board exam will open up the entrance to your Dream College or institute. And Math is the basic subject one may need in college entrance exams if you are from science or commerce group.  Our NCERT Solutions for Class 12 Chapter 3 Exercise 3.4 will guide you properly to lessen up your exam tension.

Our Class 12 Maths Matrices Exercise 3.4 has followed the guidelines of NCERT and elaborately explained all the formulas and procedure in easy and simple languages. The basic knowledge has been given of all those used terms and terminologies. Each formula has been explained in both language and examples. The examples are broken to the basic forms, so that you can understand the need to do that step and why it needed elaborately. After explaining the methods, our Class 12 Maths NCERT Solutions Chapter 3 Exercise 3.4 comes to the question section. Here, all of the problems from NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.4 have been explained from first to last without skipping any steps. Our expert team members have worked very hard to provide you with all the easiest means of solutions.  With this as your guide, it ensures you that you will get 100% benefit in your board exam.

### Class 12 Maths Chapter 3 Exercises

 Chapter 3 - Matrix Exercises in PDF Format Exercise 3.1 10 Questions & Solutions (5 Short Answers, 5 Long Answers) Exercise 3.2 22 Questions & Solutions (3 Short Answers, 19 Long Answers) Exercise 3.3 12 Questions & Solutions (4 Short Answers, 8 Long Answers) Exercise 3.4 18 Questions & Solutions (18 Short Answers)

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Vedantu has always worked hard to provide you with quality studies and solutions. Math is a tricky subject, only if you are studying it for passing your board exam. But once you have gotten the taste of its magic powers, there will be no turning back for you. Our Class 12 Maths Chapter 3 Exercise 3.4 will help you to increase interest in learning Math.

Along with our easy and simple solutions, there is also something else for you. Vedantu has an online learning app to study all of the subjects. Not just for Class 12, we have also solutions and study materials for all of the classes. On top of that, you can learn from home just by downloading the Vedantu mobile4 learning app with stable internet connections. The solutions for Exercise 3.4 Maths Class 12 can be easily available to you just by enrolling with us.

You can also download the free PDF versions of the solutions and sample solution papers from our mobile learning app.  These Exercise 3.4 Class 12 Maths NCERT Solutions has the complete syllabus covered, which will help you to complete your syllabus in time and gives you enough time for revisions.

## FAQs on NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 (Ex 3.4)

1. Where can I find NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4?

Vedantu, India’s top e-learning site, provides NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4. It is available in a free to download PDF format. Exercise-wise NCERT Solutions for Chapter 3 Matrices are curated by subject matter experts at Vedantu. The material aids in effective exam preparation as well as clearing the fundamentals of the chapter. NCERT Solutions for Class 12 Maths Chapter 3 Ex 3.4 are based on the latest NCERT guidelines and exam pattern. Hence, students can avail them to score well in the exams. The exercise’s solutions help students to get a thorough knowledge of the chapter and simple explanations to the problems.

2. How are NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 helpful?

The online solutions for NCERT Class 12 Maths Chapter 3 Matrices is prepared for doubt resolution and exam preparation. It is of great importance to learn the exercise well and provide confidence to solve the problems. These are prepared by experts who have immense knowledge and experience. Hence students can get a detailed explanation of the problems in an easy to understand manner. NCERT Solutions for CBSE Class 12 Maths Chapter 3 Exercise 3.4 can be utilized by students to solve their doubts regarding the exercise. NCERT Solutions for Class 12 Chapter 3 Matrices for Exercise 3.4 improves your chances of scoring well in the exam.

3. What are the elementary operations of a Matrix?

There are six elementary operations of a Matrix (3 due to rows and 3 due to columns). The elementary operations of a Matrix are also called the Transformation of a Matrix. Following are the elementary operations or transformations of a matrix:

• The interchange of any two rows or two columns

• The multiplication of the elements of any row or column by a non-zero number

• The addition to the elements of any row or column, the corresponding elements of any other row or column multiplied by any non zero number.

4. Why does a rectangular matrix not possess an inverse matrix?

A rectangular matrix does not possess an inverse matrix since for the products BA and AB to be defined equal, matrices A and B must be square matrices of the same order.

5. How important is Exercise 3.4 of Chapter 3 of Class 12 Maths?

Exercise 3.4 of Chapter 3 of Class 12 Maths is an important exercise from an exam point of view. It is not only important for board exams but also for competitive exams. Students of Class 12 should understand the basic concepts given in the exercise to solve the questions. NCERT Solutions of Chapter 3 of Class 12 Maths can help students get an in-depth understanding of the basic concepts and ideas for solving the questions given in the exercise easily. They can practice questions from the textbook and score good marks in Class 12 Maths.

6. What are the important points given in Exercise 3.4 of Chapter 3 of Class 12 Maths?

In Exercise 3.4 of Chapter 3 of Class 12 Maths, students will study the following important points:

• Six elementary operations on matrices out of which three are due to rows and three due to columns.

• What are invertible matrices? It is an important concept of matrices that students will be able to understand by solving different types of questions given in Exercise 3.4

• Why a rectangular matrix is not an inverse matrix

• Using elementary operations to find the inverse of a matrix.

7. Which are the difficult questions to solve in Exercise 3.4 of Chapter 3 of Class 12 Maths?

Students may face difficulty in solving question number 15, 16, and 17 of Exercise 3.4 of Chapter 3 of Class 12 Maths. To solve these questions students need to understand the concept of transformations of matrices. They should practice NCERT solutions for Class 12 Maths Exercise 3.4 given on Vedantu. All solutions for Exercise 3.4 are given by experts using simple methods that are easy to understand and students will not face any difficulty in solving the questions.

8. How much time will I need to prepare Exercise 3.4 of Chapter 3 of Class 12 Maths for my exams?

Students have to focus on the main concepts given in Exercise 3.4 of Chapter 3 of Class 12 Maths. They will need at least a week to prepare Exercise 3.4 of Class 12 Maths. They have to spend 3-4 hours daily to solve the questions given in the exercise. The time may vary from student to student depending on different factors such as capability, efficiency, and speed of the students. Students can also take help from the NCERT Solutions for Exercise 3.4 of Chapter 3 of Class 12 Maths given on Vedantu.

9. How many theorems are related to the inverse of matrix Exercise 3.4 of Chapter 3 of Class 12 Maths?

Two theorems are related to Exercise 3.4 based on the inverse of a matrix. Theorem three states uniqueness of inverse which means that the inverse of a square matrix is unique if it exists and theorem four which states that if A and B are invertible matrices having the same order, then  (AB)⁻¹ = B⁻¹ A⁻¹. Students can understand the theorems by solving the textbook questions given in Exercise 3.4. Students can get solutions for all textbook questions for Exercise 3.4 on Vedantu. SHARE TWEET SHARE SUBSCRIBE