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

Last updated date: 04th Aug 2024
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## NCERT Solutions for Miscellaneous Exercise Class 12 Chapter 3 Maths Matrices - Free PDF Download

Class 12 Maths NCERT Solutions for Chapter 3 Matrices includes solutions to all Miscellaneous Exercise problems. Matrices Miscellaneous Exercise Class 12 Chapter 3 Solutions are based on the concepts presented in Maths Chapter 3. This activity is crucial for both the CBSE Board examinations and competitive tests. Students can download the revised Class 12 Maths NCERT Solutions from our page which is prepared in a way so that you can understand it easily.

Table of Content
1. NCERT Solutions for Miscellaneous Exercise Class 12 Chapter 3 Maths Matrices - Free PDF Download
2. Access NCERT Solutions for Class 12 Maths Chapter 3 Matrices
2.1Miscellaneous Exercise
3. Class 12 Maths Chapter 3: Exercises Breakdown
4. CBSE Class 12 Maths Chapter 3 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 12 Maths
FAQs

The Class 12 Maths Chapter 3 Miscellaneous Exercise Solutions are aligned with the updated CBSE guidelines for Class 12, ensuring students are well-prepared for exams. Access the here.

## Access NCERT Solutions for Class 12 Maths Chapter 3 Matrices

### Miscellaneous Exercise

1. If $A$ and $B$ are symmetric matrices, prove that $AB-BA$ is a skew symmetric matrix.

Ans: $A$ and $B$ are symmetric matrix, therefore, we have:

$A'=A$ and $B'=B$                                                    …(1)

Here, $\left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)'$

$\Rightarrow \left( AB-BA \right)'=B'A'-A'B'$

From (1),

$\Rightarrow \left( AB-BA \right)'=BA-AB$

$\Rightarrow \left( AB-BA \right)'=-\left( AB-BA \right)$                           …(2)

From equation (2),

$AB-BA$ is a skew symmetric matrix.

2. Show that the matrix $B'AB$ is symmetric or skew symmetric accordingly as $A$ is symmetric or skew symmetric.

Ans: Let $A$ be a symmetric matrix, then $A'=A$               …(1)

$\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'$

$\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'$

$\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)$

From (1),

$\Rightarrow \left( B'AB \right)'=B'\left( AB \right)$

Thus, if $A$ is a symmetric matrix, then $B'AB$ is a symmetric matrix.

Let $A$ be a skew symmetric matrix, then $A'=-A$               …(2)

$\left( B'AB \right)'=\left\{ B'\left( AB \right) \right\}'$

$\Rightarrow \left( B'AB \right)'=\left( AB \right)'B'$

$\Rightarrow \left( B'AB \right)'=B'\left( A'B \right)$

From (2),

$\Rightarrow \left( B'AB \right)'=B'\left( -AB \right)$

$\Rightarrow \left( B'AB \right)'=-B'AB$

Thus, if $A$ is a skew-symmetric matrix, then $B'AB$ is a skew-symmetric matrix.

Therefore, the matrix $B'AB$ is symmetric or skew symmetric accordingly as $A$ is symmetric or skew symmetric.

3. Find the values of x, y, z if the matrix A =$\begin{pmatrix}0 & 2y& z \\x&y& -z \\ x& -y& z\\\end{pmatrix}$ satisfy the equation ${A}'A=I$

Ans: Given matrix A=$\begin{bmatrix} 0 & 2y& z \\ x & y & -z \\ x& -y& z \\ \end{bmatrix}$

Transpose of A = ${A}'=\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z& -z &z \\ \end{bmatrix}$

Given ${A}'A=I$

= $\begin{bmatrix} 0 & 2y &z \\ x& y&-z \\ x& -y &z \\ \end{bmatrix}\begin{bmatrix} 0 &x &x \\ 2y& y &-y \\ z &-z &z \\ \end{bmatrix}=\begin{bmatrix} 1 & 0 &0 \\ 0& 1& 0 \\ 0& 0 &1 \\ \end{bmatrix}$

${A}'$=$\begin{bmatrix} 0+x^{2}+x^{2} &0+xy-xy &0-xz+xz \\ 0+xy-xy & 4y^{2}+y^{2}+y^{2} &2yz-yz-yz \\ 0-zx+zx& 2yz-yz-yz &z^{2}+z^{2}+z^{2} \\ \end{bmatrix}=\begin{bmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{bmatrix}$

=$\begin{bmatrix} 2x^{2} &0 &0 \\ 0& 6y^{2}&0 \\ 0& 0 & 3z^{2} \\ \end{bmatrix}$ =$\begin{pmatrix} 1 &0 &0 \\ 0& 1& 0\\ 0& 0 &1 \\ \end{pmatrix}$

To find the values of x, y, z we have to equate the entries with corresponding matrix

$2x^{2}=1$

$\Rightarrow x^{2}$=$\dfrac{1}{2}$

$\Rightarrow$ x= $\pm \dfrac{1}{\sqrt{2}}$

$6y^{2}=1$

$\Rightarrow y^{2}$ = $\dfrac{1}{6}$

$\Rightarrow$ y = $\pm \dfrac{1}{\sqrt{6}}$

And

$3z^{2} = 1$

$\Rightarrow z^{2} = \dfrac{1}{3}$

$\Rightarrow$ z= $\pm \dfrac{1}{\sqrt{3}}$

4. For what values of $x$ , $\left[ \begin{matrix} 1 & 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$ ?

Ans: Given $\left[ \begin{matrix} 1 & 2 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 0 \\ 2 & 0 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} 1+4+1 & 2+0+0 & 0+2+2 \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} 6 & 2 & 4 \\\end{matrix} \right]\left[ \begin{matrix} 0 \\ 2 \\ x \\ \end{matrix} \right]=O$

$\Rightarrow \left[ 6\left( 0 \right)+2\left( 2 \right)+4\left( x \right) \right]=O$

$\Rightarrow \left[ 4+4x \right]=\left[ 0 \right]$

$\Rightarrow 4+4x=0$

$\therefore x=-1$

Thus, the required value of $x$ is $-1$ .

5. If $A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$ , show that ${{A}^{2}}-5A+7I=O$ .

Ans: Given $A=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$

$\therefore {{A}^{2}}=A\cdot A$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 3\left( 3 \right)+1\left( -1 \right) & 3\left( 1 \right)+1\left( 2 \right) \\ -1\left( 3 \right)+2\left( -1 \right) & -1\left( 1 \right)+2\left( 2 \right) \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 9-1 & 3+2 \\ -3-2 & -1+4 \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]$

LHS: ${{A}^{2}}-5A+7I$

$\therefore \left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]-5\left[ \begin{matrix} 3 & 1 \\ -1 & 2 \\ \end{matrix} \right]+7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 8 & 5 \\ -5 & 3 \\ \end{matrix} \right]-\left[ \begin{matrix} 15 & 5 \\ -5 & 10 \\ \end{matrix} \right]+\left[ \begin{matrix} 7 & 0 \\ 0 & 7 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} 0 & 0 \\ 0 & 0 \\ \end{matrix} \right]$

$\Rightarrow O$

RHS: $\Rightarrow O$

$LHS=RHS$

$\therefore {{A}^{2}}-5A+7I=O$ hence proved.

6. Find $X$ , if $\left[ \begin{matrix} x & -5 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$ .

Ans: Given $\left[ \begin{matrix} x & -5 & -1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$

$\Rightarrow \left[ \begin{matrix} x-2 & -10 & 2x-8 \\ \end{matrix} \right]\left[ \begin{matrix} x \\ 4 \\ 1 \\ \end{matrix} \right]=O$

$\Rightarrow \left[ x\left( x-2 \right)-40+2x-8 \right]=O$

$\Rightarrow \left[ {{x}^{2}}-48 \right]=\left[ 0 \right]$

$\Rightarrow {{x}^{2}}-48=0$

$\therefore x=\pm 4\sqrt{3}$

Thus, the required value of $x$ is $\pm 4\sqrt{3}$ .

7. A manufacture produces three products $X,Y,Z$ which he sells in two markets.

Annual sales are indicated below:

 Market Products $I$ $10000$ $2000$ $18000$ $II$ $6000$ $20000$ $8000$

(a) If unit sale prices of $X,Y,Z$ are Rs $2.50$ , Rs $1.50$ and Rs $1.00$ , respectively, find the total revenue in each market with the help of matrix algebra.

Ans: Here the total revenue in market $I$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 10000 & 2000 & 18000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.50 \\ 1.50 \\ 1.00 \\ \end{matrix} \right]$

$\Rightarrow 1000\times 2.50+2000\times 1.50+18000\times 1.00$

$\Rightarrow 46000$

And, the total revenue in market $II$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 6000 & 20000 & 8000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.50 \\ 1.50 \\ 1.00 \\ \end{matrix} \right]$

$\Rightarrow 6000\times 2.50+20000\times 1.50+8000\times 1.00$

$\Rightarrow 53000$

Thus, the total revenue in market $I$ is Rs $46000$ and the same in market $II$ is Rs $53000$.

(b) If the unit costs of the above three commodities are Rs $2.00$ , Rs $1.00$ and $50$ paise respectively. Find the gross profit.

Ans: Here, the total cost prices of all the products in the market $I$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 10000 & 2000 & 18000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.00 \\ 1.00 \\ 0.50 \\ \end{matrix} \right]$

$\Rightarrow 1000\times 2.00+2000\times 1.00+18000\times 0.50$

$\Rightarrow 31000$

As the total revenue in market $I$ is Rs $46000$ , the gross profit in this market is $Rs46000-Rs31000=Rs15000$ .

And, the total revenue in market $II$ can be represented in the form of a matrix as:

$\left[ \begin{matrix} 6000 & 20000 & 8000 \\ \end{matrix} \right]\left[ \begin{matrix} 2.00 \\ 1.00 \\ 0.50 \\ \end{matrix} \right]$

$\Rightarrow 6000\times 2.00+20000\times 1.00+8000\times 0.50$

$\Rightarrow 36000$

As the total revenue in market $II$ is Rs $53000$ , the gross profit in this market is $Rs53000-Rs36000=Rs17000$ .

8. Find the matrix $X$ so that $X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$.

Ans: Given $X\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

Here, $X$ has to be a $2\times 2$ matrix.

Now, let $X=\left[ \begin{matrix} a & c \\ b & d \\ \end{matrix} \right]$

Thus, we have:

$\left[ \begin{matrix} a & c \\ b & d \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

$\Rightarrow \left[ \begin{matrix} a+4c & 2a+5c & 3a+6c \\ b+4d & 2b+5d & 3b+6d \\ \end{matrix} \right]=\left[ \begin{matrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{matrix} \right]$

Comparing the corresponding elements of two matrices, we have:

$a+4c=-7$

$2a+5c=-8$

$3a+6c=-9$

$b+4d=2$

$2b+5d=4$

$3b+6d=6$

Now, solving the above equations we get,

$\therefore a=1,b=2,c=-2,d=0$

Thus, the required matrix $X$ is $\left[ \begin{matrix} 1 & -2 \\ 2 & 0 \\ \end{matrix} \right]$.

9. Choose the correct answer in the following questions:

If $A=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$ is such that ${{A}^{2}}=I$ then

1. $1+{{\alpha }^{2}}+\beta \gamma =0$

2. $1-{{\alpha }^{2}}+\beta \gamma =0$

3. $1-{{\alpha }^{2}}-\beta \gamma =0$

4. $1+{{\alpha }^{2}}-\beta \gamma =0$

Ans: Given $A=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$

$\therefore {{A}^{2}}=A\cdot A$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]\left[ \begin{matrix} \alpha & \beta \\ \gamma & -\alpha \\ \end{matrix} \right]$

$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} {{\alpha }^{2}}+\beta \gamma & \alpha \beta -\alpha \beta \\ \alpha \gamma -\alpha \gamma & \beta \gamma +{{\alpha }^{2}} \\ \end{matrix} \right]$

Now, ${{A}^{2}}=I$

$\Rightarrow \left[ \begin{matrix} {{\alpha }^{2}}+\gamma & 0 \\ 0 & \beta \gamma +{{\alpha }^{2}} \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]$

Equating the corresponding elements, we get:

$\beta \gamma +{{\alpha }^{2}}=1$

$\Rightarrow {{\alpha }^{2}}+\beta \gamma -1=0$

$\Rightarrow 1-{{\alpha }^{2}}-\beta \gamma =0$

10. If the matrix $A$ is both symmetric and skew symmetric, then

1. $A$ is a diagonal matrix

2. $A$ is a zero matrix

3. $A$ is a square matrix

4. None of these

Ans: If a matrix $A$ is both symmetric and skew symmetric, then

$A'=A$ and $A'=-A$

$\Rightarrow A=-A$

$\Rightarrow A+A=O$

$\Rightarrow A=O$

Thus, option (B) is correct.

11. If $A$ is a square matrix such that ${{A}^{2}}=A$ , then ${{\left( I+A \right)}^{3}}-7A$ is equal to

1. $A$

2. $I-A$

3. $I$

4. $3A$

Ans: ${{\left( I+A \right)}^{3}}-7A={{I}^{3}}+{{A}^{3}}+3A+3{{A}^{2}}-7A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{3}}+3A+3{{A}^{2}}-7A$

Given that ${{A}^{2}}=A$ ,

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+{{A}^{2}}\cdot A+3A+3{{A}^{2}}-7A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I+A\cdot A-A$

$\Rightarrow {{\left( I+A \right)}^{3}}-7A=I$

Thus, option (C) is correct.

## Conclusion

Matrices Class 12 Miscellaneous Exercise Solutions are important for understanding various concepts thoroughly. Class 12 Maths Chapter 3 Miscellaneous Exercise Solutions covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

## Class 12 Maths Chapter 3: Exercises Breakdown

 Exercise Number of Questions Exercise 3.1 10 Questions & Solutions Exercise 3.2 22 Questions & Solutions Exercise 3.3 12 Questions & Solutions Exercise 3.4 1 Questions & Solutions

## CBSE Class 12 Maths Chapter 3 Other Study Materials

 S.No. Important Links for Chapter 3 Matrices 1 Class 12 Matrices Revision Notes 2 Class 12 Matrices Important Questions 3 Class 12 Matrices Formula 4 Class 12 Matrices NCERT Exemplar Solution 5 Class 12 Matrices RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 3 - Matrices

1. What topics are covered in the Miscellaneous Exercise of Chapter 3?

The Miscellaneous Exercise in Chapter 3 covers a variety of problems related to matrices, including matrix operations, types of matrices, determinants, and applications of matrices in solving linear equations.

2. What are the main types of matrices discussed in Matrices Class 12 Miscellaneous Exercise Solutions?

The main types of matrices discussed are square matrices, diagonal matrices, scalar matrices, identity matrices, and zero matrices. Understanding their properties is essential.

3. How do I perform matrix operations in class 12 chapter 3?

Matrix operations include addition, subtraction, and multiplication of matrices. Each operation follows specific rules regarding the dimensions and elements of the matrices involved.

4. What is a determinant in class 12 chapter 3?

A determinant is a scalar value that can be computed from the elements of a square matrix. It provides important properties of the matrix and is used in solving linear equations and finding the inverse of a matrix.

5. How can matrices be used to solve linear equations?

Matrices can be used to solve linear equations by representing the system of equations as a matrix equation AX=B and then finding the solution X using methods such as matrix inversion or row reduction.

6. Are there specific formulas I need to remember to solve Matrix Miscellaneous Exercise Class 12?

Yes, remember the formulas for matrix operations, the method to calculate determinants, and the process to find the inverse of a matrix.

7. How can I solve the problems in the Miscellaneous Exercise?

To solve the problems, carefully read each question, understand the concepts involved, and apply the appropriate formulas and methods learned in the chapter. Practice regularly to enhance your skills.

8. Why are matrices important in math class 12 chapter 3?

Matrices are fundamental in various fields of mathematics and are used in physics, engineering, computer science, and economics. They help in modelling and solving complex problems.

9. How can the NCERT solutions of class 12 chapter 3 miscellaneous exercise help me?

NCERT solutions by Vedantu provided step-by-step explanations for each problem, helping you understand the methods and concepts required to solve them. They are a valuable resource for exam preparation.

10. Are there any tricky questions in Matrix Miscellaneous Exercise Class 12?

Yes, the Miscellaneous Exercise includes some challenging questions designed to test your understanding of the entire chapter. Practice them thoroughly to improve your problem-solving skills.