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

In the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4, you will get questions related to elementary operation on the matrix. Elementary operations or transformations are a set of operations that are allowed to perform on the matrix to get transformed matrix. These transformations are useful in the finding inverse of a matrix if exists. If the inverse of a matrix exists then it is called an invertible matrix. There are some questions in exercise 3.4 Class 12 Maths where you need to find the inverse of a matrix using elementary operation. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 is very important as generally one question is directly asked from this exercise in the CBSE board final exam. Class 12 maths ch 3 ex 3.4 required more practice and the chances of silly mistakes are more here, so you must practice all the questions in order to get them correct in the board exam. You can check NCERT solutions here.

Also, see

• Matrices Exercise 3.1
• Matrices Exercise 3.2
• Matrices Exercise 3.3
• Matrices Miscellaneous Exercise

## Matrices Exercise: 3.4

Question 1 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$

Use the elementary transformation we can find the inverse as follows

$A=\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix}1&-1\\2&3 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix}1&-1\\0&5 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{5}$

$\Rightarrow$ $\begin{bmatrix}1&-1\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A$

$R_1\rightarrow R_1+R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}A$

$\therefore A^{-1}= \begin{bmatrix}\frac{3}{5}&\frac{1}{5}\\\frac{-2}{5}&\frac{1}{5} \end{bmatrix}$

Question 2 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2&1\\1&1\end{bmatrix}$

$A=\begin{bmatrix} 2&1\\1&1\end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 2&1\\1&1\end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\1&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}A$

$A^{-1}= \begin{bmatrix}1&-1\\-1& 2 \end{bmatrix}$

Thus we have obtained the inverse of the given matrix through elementary transformation

Question 3 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$

$A=\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 1 &3 \\ 2 & 7 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

Using elementary transformations

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix}1&3\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-3R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}A$

$\therefore A^{-1}=$$\begin{bmatrix}7&-3\\-2&1 \end{bmatrix}$.

Question 4 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$

$A=\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 &3 \\ 5 & 7 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$$\begin{bmatrix} 2 &3 \\ 1 & 1 \end{bmatrix}$$= \begin{bmatrix}1&0\\-2&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-3R_2$

$\Rightarrow$$\begin{bmatrix} -1 &0 \\ 1 & 1 \end{bmatrix}$$= \begin{bmatrix}7&-3\\-2&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\Rightarrow$$\begin{bmatrix} -1 &0 \\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}7&-3\\5&-2\end{bmatrix}A$

$R_1\rightarrow -R_1$

$\Rightarrow$$\begin{bmatrix} 1 &0 \\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}-7&3\\5&-2\end{bmatrix}A$

$\therefore A^{-1}=$$\begin{bmatrix}-7&3\\5&-2\end{bmatrix}$

Question 5 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$

$A =\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & 1\\ 7 & 4 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-3R_1$

$\Rightarrow$$\begin{bmatrix} 2 & 1\\ 1 & 1 \end{bmatrix}$$= \begin{bmatrix}1&0\\-3&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}$$= \begin{bmatrix}4&-1\\-3&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}4&-1\\-7&2 \end{bmatrix}A$

$\therefore A^{-1}=$$\begin{bmatrix}4&-1\\-7&2 \end{bmatrix}$.

Thus the inverse of matrix A is obtained.

Question 6 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$

$A=\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & 5\\ 1 &3 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

Use the elementary transformation

$R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 2\\ 1 &3 \end{bmatrix}$$= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 2\\ 0 &1 \end{bmatrix}$$= \begin{bmatrix}1&-1\\-1&2 \end{bmatrix}A$

$R_1\rightarrow R_1-2R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$$= \begin{bmatrix}3&-5\\-1&2 \end{bmatrix}A$

$\therefore A^{-1}=$$\begin{bmatrix}3&-5\\-1&2 \end{bmatrix}$.

Question 7 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$

$A=\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 3 & 1\\ 5 & 2 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-R_1$

$\Rightarrow$ $\begin{bmatrix} 3 & 1\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\-1&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 2 & 1 \end{bmatrix}$ $= \begin{bmatrix}2&-1\\-1&1 \end{bmatrix}A$

$R_2\rightarrow R_2-2R_1$

$\Rightarrow$ $\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$ $= \begin{bmatrix}2&-1\\-5&3 \end{bmatrix}A$

$\therefore A^{-1}=$ $\begin{bmatrix}2&-1\\-5&3 \end{bmatrix}$.

Thus the inverse of matrix A is obtained using elementary transformation.

Question 8 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$

$A=\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 4 &5 \\ 3 & 4 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 1 &1 \\ 3 & 4 \end{bmatrix}$$= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-3R_1$

$\Rightarrow$ $\begin{bmatrix}1&1\\0&1 \end{bmatrix}$ $= \begin{bmatrix}1&-1\\-3&4 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$ $\begin{bmatrix}1&0\\0&1 \end{bmatrix}$ $= \begin{bmatrix}4&-5\\-3&4 \end{bmatrix}A$

Thus using elementary transformation inverse of A is obtained as

$\therefore A^{-1}=$$\begin{bmatrix}4&-5\\-3&4 \end{bmatrix}$.

Question 9 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$

$A=\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 3 & 10\\ 2 & 7 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 1& 3\\ 2 & 7 \end{bmatrix}$$= \begin{bmatrix}1&-1\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-2R_1$

$\Rightarrow$$\begin{bmatrix} 1& 3\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}1&-1\\-2&3 \end{bmatrix}A$

$R_1\rightarrow R_1-3R_2$

$\Rightarrow$$\begin{bmatrix} 1& 0\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}7&-10\\-2&3 \end{bmatrix}A$

Thus using elementary transformation the inverse of A is obtained as

$\therefore A^{-1}=$$\begin{bmatrix}7&-10\\-2&3 \end{bmatrix}$.

Question 10 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$

$A=\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 3 & -1\\ -4 & 2 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\Rightarrow$$\begin{bmatrix} 3 & -1\\ -1 & 1 \end{bmatrix}$$= \begin{bmatrix}1&0\\1&1 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1+2R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 1\\ -1 & 1 \end{bmatrix}$$= \begin{bmatrix}3&2\\1&1 \end{bmatrix}A$

$R_2\rightarrow R_2+R_1$

$\begin{bmatrix} 1 & 1\\ 0 & 2 \end{bmatrix}$$= \begin{bmatrix}3&2\\4&3 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{2}$

$\begin{bmatrix} 1 & 1\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}3&2\\2&\frac{3}{2} \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}A$

$\therefore A^{-1}=$$\begin{bmatrix}1&\frac{1}{2}\\2&\frac{3}{2} \end{bmatrix}$.

Thus the inverse of A is obtained using elementary transformation.

Question 11 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

$A=\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & -6\\ 1 & -2 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-3R_1$

$\Rightarrow$$\begin{bmatrix} -1 & 0\\ 1 & -2 \end{bmatrix}$$= \begin{bmatrix}1&-3\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2+R_1$

$\Rightarrow$$\begin{bmatrix} -1 & 0\\ 0 & -2 \end{bmatrix}$$= \begin{bmatrix}1&-3\\1&-2 \end{bmatrix}A$

$R_1\rightarrow -R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 0 & -2 \end{bmatrix}$$= \begin{bmatrix}-1&3\\1&-2 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{-2}$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}$$= \begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix}A$

thus the inverse of matrix A is

$\therefore A^{-1}=$$\begin{bmatrix}-1&3\\\frac{1}{-2}&1 \end{bmatrix}$.

Question 12 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$

$A=\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$

$A=IA$

$\Rightarrow$ $\begin{bmatrix} 6 & -3\\ -2 & 1 \end{bmatrix}$ $= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{6}$

$\Rightarrow$ $\begin{bmatrix} 1& -\frac{1}{2}\\ -2 & 1 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{6}&0\\0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2+2R_1$

$\Rightarrow$ $\begin{bmatrix} 1& -\frac{1}{2}\\ 0 & 0 \end{bmatrix}$ $= \begin{bmatrix}\frac{1}{6}&0\\\frac{1}{3}&1 \end{bmatrix}A$

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

Question 13 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$

$A=\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & -3\\ -1 & 2 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow 2R_2+R_1$

$\Rightarrow$$\begin{bmatrix} 2 & -3\\ 0 &1 \end{bmatrix}$$= \begin{bmatrix}1&0\\1&2 \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1+3R_2$

$\Rightarrow$$\begin{bmatrix} 2 & 0\\ 0 &1 \end{bmatrix}$$= \begin{bmatrix}4&6\\1&2 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{2}$

$\Rightarrow$$\begin{bmatrix} 1 & 0\\ 0 &1 \end{bmatrix}$$= \begin{bmatrix}2&3\\1&2 \end{bmatrix}A$

so the inverse of matrix A is

$\therefore A^{-1}=$$\begin{bmatrix}2&3\\1&2 \end{bmatrix}$.

Question 14 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$

$A=\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & 1\\ 4 & 2 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&1 \end{bmatrix}A$

$R_2\rightarrow \frac{R_2}{2}$

$\Rightarrow$$\begin{bmatrix} 2 & 1\\ 2 & 1 \end{bmatrix}$$= \begin{bmatrix}1&0\\0&\frac{1}{2} \end{bmatrix}A$

$\Rightarrow$ $R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 0 & 0\\ 2 & 1 \end{bmatrix}$$= \begin{bmatrix}1&\frac{-1}{2}\\0&\frac{1}{2} \end{bmatrix}A$

Hence, we can see all upper values of matirix are zeros in L.H.S so $A^{-1}$ does not exists.

Question 15 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$

$A=\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & -3 & 3\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$$= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow R_1-R_3$

$\Rightarrow$$\begin{bmatrix} -1 & -1 & 1\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$$= \begin{bmatrix}1&0&-1\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow -R_1$

$\begin{bmatrix} 1 & 1 & -1\\ 2& 2 & 3\\ 3 & -2 & 2 \end{bmatrix}$$= \begin{bmatrix}-1&0&1\\0&1&0 \\0&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_2\rightarrow R_2-2R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 1 & -1\\ 0& 0 & 5\\ 3 & -2 & 2 \end{bmatrix}$$= \begin{bmatrix}-1&0&1\\2&1&-2 \\0&0&1 \end{bmatrix}A$

$R_3\rightarrow R_3-3R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 1 & -1\\ 0& 0 & 5\\ 0 & -5 & 5 \end{bmatrix}$$= \begin{bmatrix}-1&0&1\\2&1&-2 \\3&0&-2 \end{bmatrix}A$

$R_2\leftrightarrow R_3$

$\Rightarrow$$\begin{bmatrix} 1 & 1 & -1\\ 0& -5 & 5\\ 0 & 0 & 5 \end{bmatrix}$$= \begin{bmatrix}-1&0&1\\3&0&-2 \\2&1&-2 \end{bmatrix}A$

$R_2\rightarrow \frac{-R_2}{5}$

$\Rightarrow$$\begin{bmatrix} 1 & 1 & -1\\ 0& 1 & -1\\ 0 & 0 & 5 \end{bmatrix}$$= \begin{bmatrix}-1&0&1\\\frac{-3}{5}&0&\frac{2}{5} \\2&1&-2 \end{bmatrix}A$

$R_1\rightarrow R_1-R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & -1\\ 0 & 0 & 5 \end{bmatrix}$$= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-3}{5}&0&\frac{2}{5} \\2&1&-2 \end{bmatrix}A$

$R_3\rightarrow \frac{R_3}{5}$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & -1\\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-3}{5}&0&\frac{2}{5} \\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}A$

$R_2\rightarrow R_2+R_3$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & 0\\ 0& 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-1}{5}&\frac{1}{5}&0\\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}A$

Thos the Inverse of A is

$\therefore A^{-1}=$.$\begin{bmatrix}-\frac{2}{5}&0&\frac{3}{5}\\\frac{-1}{5}&\frac{1}{5}&0\\\frac{2}{5}&\frac{1}{5}&-\frac{2}{5} \end{bmatrix}$.

Question 16 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$

$A=\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 1 &3 & -2\\ -3& 0 &-5 \\ 2 & 5 & 0 \end{bmatrix}$$= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_2\rightarrow R_2+3R_1$ and $R_3\rightarrow R_3-2R_1$

$\Rightarrow$$\begin{bmatrix} 1 &3 & -2\\ 0& 9 &-11 \\ 0 & -1 & 4 \end{bmatrix}$$= \begin{bmatrix}1&0&0\\3&1&0 \\-2&0&1 \end{bmatrix}A$

$R_1\rightarrow R_1+3R_3$ and $R_2\rightarrow R_2+8R_3$

$\Rightarrow$$\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & -1 & 4 \end{bmatrix}$$= \begin{bmatrix}-5&0&3\\-13&1&8 \\-2&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_3\rightarrow R_3+R_2$

$\Rightarrow$$\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 25 \end{bmatrix}$$= \begin{bmatrix}-5&0&3\\-13&1&8 \\-15&1&9 \end{bmatrix}A$

$R_3\rightarrow \frac{R_3}{25}$

$\Rightarrow$$\begin{bmatrix} 1 &0 & 10\\ 0& 1 &21 \\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix}-5&0&3\\-13&1&8 \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A$

$R_1\rightarrow R_1-10R_3$ and $R_2\rightarrow R_2-21R_3$

$\Rightarrow$$\begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}A$

Thus the inverse of three by three matrix A is

$\therefore A^{-1}=$.$\begin{bmatrix}1&\frac{-2}{5}&\frac{-3}{5}\\\frac{-2}{5}&\frac{4}{25}&\frac{11}{25} \\-\frac{3}{5}&\frac{1}{25}&\frac{9}{25} \end{bmatrix}$.

Question 17 Using elementary transformations, find the inverse of each of the matrices, if it exists
in Exercises 1 to 17.

$\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$

$A=\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$

$A=IA$

$\Rightarrow$$\begin{bmatrix} 2 & 0 & -1\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$$= \begin{bmatrix}1&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_1\rightarrow \frac{R_1}{2}$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 5 & 1 & 0\\ 0 & 1 & 3 \end{bmatrix}$$= \begin{bmatrix}\frac{1}{2}&0&0\\0&1&0 \\0&0&1 \end{bmatrix}A$

$R_2\rightarrow R_2-5R_1$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 1 & 3 \end{bmatrix}$$= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\0&0&1 \end{bmatrix}A$

$\Rightarrow$ $R_3\rightarrow R_3-R_2$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & \frac{1}{2} \end{bmatrix}$$= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\\frac{5}{2}&-1&1 \end{bmatrix}A$

$R_3\rightarrow 2R_3$

$\Rightarrow$$\begin{bmatrix} 1 & 0 & -\frac{1}{2}\\ 0 & 1 & \frac{5}{2}\\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix}\frac{1}{2}&0&0\\-\frac{5}{2}&1&0 \\ 5&-2&2 \end{bmatrix}A$

$R_1\rightarrow R_1+\frac{R_3}{2}$ and $R_2\rightarrow R_2-\frac{5}{2}R_3$

$\Rightarrow$$\begin{bmatrix} 1 &0 & 0\\ 0& 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$$= \begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}A$

Thus the inverse of A is obtained as

$\therefore A^{-1}=$.$\begin{bmatrix} 3&-1&1\\-15&6&-5 \\ 5&-2&2 \end{bmatrix}$.

Question:18 Matrices A and B will be inverse of each other only if

(A) $AB = BA$

(B) $AB = BA = 0$

(C) $AB = 0, BA = I$

(D) $AB = BA = I$

We know that if A is a square matrix of order n and there is another matrix B of same order n, such that $AB=BA=I$, then B is inverse of matrix A.

In this case, it is clear that A is inverse of B.

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

Option D is correct.

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

In Class 12th Maths chapter 3 exercise 3.4, there are 18 questions out of which 17 questions are related to finding the inverse of a matrix using elementary operations if exists. One multiple choice type question is related to the condition of an invertible matrix. There are 3 solved examples and some theorems given before the NCERT text book exercise 3.4 Class 12 Maths. Examples related to finding the inverse of a matrix are also given in Class 12 Maths ch 3 ex 3.4. Theorems given in the NCERT syllabus are important to get conceptual clarity.

Also Read| Matrices Class 12 Maths Chapter Notes

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