NCERT Solutions for Class 12 Maths Chapter 3: Matrices - Exercise 3.4

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.

NCERT Solutions for Class 12 Maths PDF Download

• Chapter 1 - Relations and Functions

• Chapter 2 - Inverse Trigonometric Functions

• Chapter 3 - Matrices

• Chapter 4 - Determinants

• Chapter 5 - Continuity and Differentiability

• Chapter 6 - Application of Derivatives

• Chapter 7 - Integrals

• Chapter 8 - Application of Integrals

• Chapter 9 - Differential Equations

• Chapter 10 - Vector Algebra

• Chapter 11 - Three Dimensional Geometry

• Chapter 12 - Linear Programming

• Chapter 13 - Probability

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

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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.