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

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

Edited By Ramraj Saini | Updated on Dec 03, 2023 03:51 PM IST | #CBSE Class 12th

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

NCERT Solutions for Exercise 3.4 Class 12 Maths Chapter 3 Matrices are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. 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.

12th class Maths exercise 3.4 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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## Matrices Exercise: 3.4

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

Use the elementary transformation we can find the inverse as follows

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

$A=IA$

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

$R_2\rightarrow R_2-2R_1$

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

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

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

$R_1\rightarrow R_1+R_2$

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

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

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

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

$A=IA$

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

$R_1\rightarrow R_1-R_2$

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

$R_2\rightarrow R_2-R_1$

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

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

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

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

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

$A=IA$

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

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}$.

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

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

$A=IA$

$\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}$

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

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

$A=IA$

$\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.

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

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

$A=IA$

$\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}$.

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

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

$A=IA$

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

$R_2\rightarrow R_2-R_1$

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

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

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

$R_2\rightarrow R_2-2R_1$

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

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

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

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

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

$A=IA$

$\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}$.

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

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

$A=IA$

$\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}$.

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

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

$A=IA$

$\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.

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

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

$A=IA$

$\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}$.

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

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

$A=IA$

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

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

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

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

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

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

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

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

$A=IA$

$\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}$.

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

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

$A=IA$

$\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.

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

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

$A=IA$

$\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

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

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

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

$A=IA$

$\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

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

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

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

$A=IA$

$\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

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

(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

## Benefits of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4:-

• Class 12 Maths chapter 3 exercise 3.4 solutions are helpful for the students in the board exams as well as in competitive exams.
• NCERT problems are solved by experts who have good knowledge and experience so you can rely upon them.
• NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 are beneficial for the students who are not able to solve some difficult problems.
• Students are advised to solve all the NCERT problems including examples as most of the questions in the board exams are directly asked from the NCERT textbook.
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## Key Features Of NCERT Solutions for Exercise 3.4 Class 12 Maths Chapter 3

• Comprehensive Coverage: The solutions encompass all the topics covered in ex 3.4 class 12, ensuring a thorough understanding of the concepts.
• Step-by-Step Solutions: In this class 12 maths ex 3.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
• Accuracy and Clarity: Solutions for class 12 ex 3.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
• Conceptual Clarity: In this 12th class maths exercise 3.4 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
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• Relevance to Curriculum: The solutions for class 12 maths ex 3.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.
##### JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

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## NCERT solutions of class 12 subject wise

Subject wise NCERT Exampler solutions

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1. If a matrix A = [ 1 2 3] then order of matrix A ?

The order of matrix A is 1x3.

2. What of the definition of elements in a matrix ?

The numbers or functions that exist in the rows and columns of a particular matrix are called elements of the matrix. Definitions, examples and questions are given in the NCERT book. For more practice problems NCERT exemplar for class 12 Maths can be used. Practice class 12 ex 3.4 to command the concepts.

3. write some types of matrix ?

A square matrix, zero or null matrix, scalar matrix, diagonal matrix,  row matrix, and column matrix are some types of matrices.

4. Two matrices A and B are inverse of each other only if ?

If AB = BA = I, then A and B are inverses of each other.

5. What is the condition for addition of two matrices ?

The order of matrices has to be the same in order to add them.

6. Can we add matrix A of order 2x3 and matrix B of 3x2 ?

No, as the order of both the matrices are not the same.

7. Can we multiply matrix A of order 2x3 and matrix B of 3x2 ?

Yes, matrices A and B can be multiplied

8. What is order of the matrix we get when we multiply matrix A of order 2x3 and matrix B of 3x2 ?

The order of the solution matrix will be 2x2.

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### Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

Yuvan 01 September,2024

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

• No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
• You have to appear for the 2025 12th board exams.
• Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
• Aim to register before late October to avoid extra fees.
• Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

 Option 1) Option 2) Option 3) Option 4)

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

 Option 1) 2.45×10−3 kg Option 2)  6.45×10−3 kg Option 3)  9.89×10−3 kg Option 4) 12.89×10−3 kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

 Option 1) Option 2) Option 3) Option 4)

A particle is projected at 600   to the horizontal with a kinetic energy . The kinetic energy at the highest point

 Option 1) Option 2) Option 3) Option 4)

In the reaction,

 Option 1)   at STP  is produced for every mole   consumed Option 2)   is consumed for ever      produced Option 3) is produced regardless of temperature and pressure for every mole Al that reacts Option 4) at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, will contain 0.25 mole of oxygen atoms?

 Option 1) 0.02 Option 2) 3.125 × 10-2 Option 3) 1.25 × 10-2 Option 4) 2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

 Option 1) decrease twice Option 2) increase two fold Option 3) remain unchanged Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

 Option 1) Molality Option 2) Weight fraction of solute Option 3) Fraction of solute present in water Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

 Option 1) twice that in 60 g carbon Option 2) 6.023 × 1022 Option 3) half that in 8 g He Option 4) 558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

 Option 1) less than 3 Option 2) more than 3 but less than 6 Option 3) more than 6 but less than 9 Option 4) more than 9