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

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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.

Also see-

NCERT solutions of class 12 subject wise

Subject wise NCERT Exampler solutions

Happy learning!!!

Frequently Asked Questions (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.

Also Read

NCERT Class 12 Maths Notes - Download Chapter wise Free PDFs

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NCERT Class 12 Physics Chapter 7 Notes Alternating Current - Download PDF

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

Can I change my board from CBSE to CHSE Odisha in Class 12th?

Hello,

Yes, you can switch from CBSE to CHSE Odisha in Class 12th, but there are a few conditions:

Eligibility : Check if CHSE Odisha allows lateral entry into Class 12th. Generally, students can transfer from CBSE to state boards with prior approval.
Document Requirements : You may need to provide transcripts, mark sheets, and a transfer certificate.
Subject Alignment : Ensure that the subjects you studied in CBSE align with the CHSE curriculum.
Application Process : Contact the respective CHSE Odisha authority or school for specific transfer procedures and deadlines.

Hope it helps !

Read Complete Answer

Prachi Kumari 30 December,2024

quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

Jefferson 25 September,2024

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.

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

Read Complete Answer

Kanishka kaushikii 17 September,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,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

Read Complete Answer

Yuvan 30 August,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

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Central Board of Secondary Education Class 12th Examination

JEE Test Series

Applications Open

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

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

NCERT Solutions For Class 12 Maths Chapter 3 Exercise 3.4

Pearson | PTE

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

NCERT solutions of class 12 subject wise

Frequently Asked Questions (FAQs)

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