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 part-1

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.


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.

Why Vedantu?

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 (Frequently Asked Questions)

Q1. Where can I find NCERT Solutions for Class 12 Maths Chapter 3 Matrices Exercise 3.4?

Ans: 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.

Q2. How are NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 helpful?

Ans: 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.

Q3. What are the elementary operations of a Matrix?

Ans: 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:

  1. The interchange of any two rows or two columns

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

  3. 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. 

Q4. Why does a rectangular matrix not possess an inverse matrix?

Ans: 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.

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