NCERT Solutions for Exercise 3.4 Class 12 Maths Chapter 3

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.

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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

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

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

Answer:

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.

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

Answer:

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

$A=IA$

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

$R_1\rightarrow R_1-R_2$

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

$R_2\rightarrow R_2-R_1$

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

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

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.

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

Answer:

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

$A=IA$

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

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.

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

Answer:

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

$A=IA$

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

$R_2\rightarrow R_2-2R_1$

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

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

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

$R_2\rightarrow R_2+R_1$

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

$R_1\rightarrow -R_1$

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

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

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

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

Answer:

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

$A=IA$

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

$R_2\rightarrow R_2-3R_1$

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

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

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

$R_2\rightarrow R_2-R_1$

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

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

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.

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

Answer:

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

$A=IA$

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

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.

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

Answer:

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

$A=IA$

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

$R_2\rightarrow R_2-R_1$

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

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

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

$R_2\rightarrow R_2-2R_1$

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

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

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.

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

Answer:

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

$A=IA$

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

$R_1\rightarrow R_1-R_2$

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

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

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

$R_1\rightarrow R_1-R_2$

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

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.

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

Answer:

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

$A=IA$

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

$R_1\rightarrow R_1-R_2$

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

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

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

$R_1\rightarrow R_1-3R_2$

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

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.

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

Answer:

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

$A=IA$

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

$R_2\rightarrow R_2+R_1$

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

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

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

$R_2\rightarrow R_2+R_1$

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

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

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

$R_1\rightarrow R_1-R_2$

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

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

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.

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

Answer:

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

$A=IA$

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

$R_1\rightarrow R_1-3R_1$

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

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

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

$R_1\rightarrow -R_1$

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

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

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

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.

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

Answer:

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

$A=IA$

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

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

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

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

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

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.

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

Answer:

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

$A=IA$

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

$R_2\rightarrow 2R_2+R_1$

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

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

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

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

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

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.

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

Answer:

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

$A=IA$

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

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

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

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

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

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.

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

Answer:

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

$A=IA$

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

$R_1\rightarrow R_1-R_3$

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

$R_1\rightarrow -R_1$

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

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

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

$R_3\rightarrow R_3-3R_1$

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

$R_2\leftrightarrow R_3$

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

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

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

$R_1\rightarrow R_1-R_2$

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

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

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

$R_2\rightarrow R_2+R_3$

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

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.

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

Answer:

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

$A=IA$

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

$R_2\rightarrow R_2+3R_1$ 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.

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

Answer:

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

$A=IA$

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

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

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

$R_2\rightarrow R_2-5R_1$

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

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

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

$R_3\rightarrow 2R_3$

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

$R_1\rightarrow R_1+\frac{R_3}{2}$ 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) $AB = BA = 0$

(D) $AB = BA = I$

Answer:

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.

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.
Inclusive Approach: Solutions for ex 3.4 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
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.

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Frequently Asked Question (FAQs)

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 ?

i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

Read Complete Answer

Divyan 06 August,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

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.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

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.

Read Complete Answer

Yuvan 01 September,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

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.

Read Complete Answer

Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

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.

Read Complete Answer

kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

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.

Good luck with your studies!

Read Complete Answer

kumardurgesh1802 10 July,2024

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

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

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.

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

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

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

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

NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.4

VMC VIQ Scholarship Test

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Assess NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

Matrices Exercise: 3.4

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

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

Key Features Of NCERT Solutions for Exercise 3.4 Class 12 Maths Chapter 3

JEE Main Important Mathematics Formulas

NCERT solutions of class 12 subject wise

Frequently Asked Question (FAQs)

Also Read

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Upcoming School Exams

Certifications By Top Providers

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