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NCERT Solutions for Class 12 Maths Chapter 3: Matrices - Exercise 3.3

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NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Free PDF Download

Chapter 3 of Class 12 Maths in the NCERT curriculum focuses on Matrices, a fundamental concept in algebra. Exercise 3.3 delves into various operations on matrices, including addition, subtraction, and multiplication. Understanding these operations is crucial for solving complex problems in linear algebra and for applications in fields such as computer science, physics, and economics. Our NCERT solutions for Exercise 3.3 provide step-by-step explanations and detailed answers to help students grasp these concepts thoroughly. With these solutions, students can enhance their problem-solving skills and build a strong foundation in matrix theory.

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Table of Content
1. NCERT Solutions for Exercise 3.3 Class 12 Maths Chapter 3 - Free PDF Download
2. Glance on Maths Ex 3.3 Class 12 Chapter 3 Matrices 
3. Topics Covered in the NCERT Solutions for Class 12 Maths Chapter 3
4. Access NCERT Solutions for Class 12 Maths Chapter 3- Matrices Exercise 3.3
    4.1Exercise 3.3
5. Conclusion
6. Class 12 Maths Chapter 3: Exercises Breakdown
7. Other Study Materials for CBSE Class 12 Maths Chapter 3 Matrices
8. NCERT Solutions for Class 12 Maths | Chapter-wise List
9. Related Links for NCERT Class 12 Maths in Hindi
10. Important Related Links for NCERT Class 12 Maths
FAQs


Glance on Maths Ex 3.3 Class 12 Chapter 3 Matrices 

  • Revisiting square matrices, rectangular matrices, diagonal matrices, and identity matrices.

  • Encounter problems on addition, subtraction, scalar multiplication, and multiplication of matrices.

  • Understanding properties like associative and distributive properties for matrix addition, and properties of scalar multiplication.

  • Finding the transpose (swapping rows and columns) of a matrix.

  • Identifying matrices where elements are mirrored across the diagonal (symmetric) or negated across the diagonal (skew-symmetric).


Topics Covered in the NCERT Solutions for Class 12 Maths Chapter 3

  • Transpose of a Matrix

  • Properties of the transpose of the matrices

  • Symmetric and Skew Symmetric Matrices

Competitive Exams after 12th Science
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Access NCERT Solutions for Class 12 Maths Chapter 3- Matrices Exercise 3.3

Exercise 3.3

1. Find the transpose of each of the following matrices:

\[\begin{align} & \mathrm{(i)}\left[ \begin{array}{*{35}{l}} \mathrm{5} \\ \mathrm{1/2} \\ \mathrm{-1} \\ \end{array} \right] \\ & \mathrm{(ii)}\left[ \begin{matrix} \mathrm{1} & \mathrm{-1} \\ \mathrm{2} & \mathrm{3} \\ \end{matrix} \right] \\ & \mathrm{(iii)}\left[ \begin{matrix} \mathrm{-1} & \mathrm{5} & \mathrm{6} \\ \sqrt{\mathrm{3}} & \mathrm{5} & \mathrm{6} \\ \mathrm{2} & \mathrm{3} & \mathrm{-1} \\ \end{matrix} \right] \\ \end{align}\]

Ans:

Find transpose

\[(i)A=\left[ \begin{array}{*{35}{l}} 5 \\ 1/2 \\ -1 \\ \end{array} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} 5 & \frac{1}{2} & -1 \\ \end{matrix} \right]\] \[(ii)A=\left[ \begin{matrix} 1 & -1 \\ 2 & 3 \\ \end{matrix} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} 1 & 2 \\ -1 & 3 \\ \end{matrix} \right]\] \[(iii)A=\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt{3} & 5 & 6 \\ 2 & 3 & -1 \\ \end{matrix} \right]\Rightarrow {{A}^{T}}=\left[ \begin{matrix} -1 & \sqrt{3} & 2 \\ 5 & 5 & 3 \\ 6 & 6 & -1 \\ \end{matrix} \right]\]


2. If \[\mathrm{A=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{3} \\ \mathrm{5} & \mathrm{7} & \mathrm{9} \\ \mathrm{-2} & \mathrm{1} & \mathrm{1} \\ \end{matrix} \right]\] and \[\mathrm{B=}\left[ \begin{matrix} \mathrm{-4} & \mathrm{1} & \mathrm{-5} \\ \mathrm{1} & \mathrm{2} & \mathrm{0} \\ \mathrm{1} & \mathrm{3} & \mathrm{1} \\ \end{matrix} \right]\] , then verify that \[\begin{align} & \mathrm{(i)}\left( \mathrm{A+B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ +B }\!\!'\!\!\text{ } \\ & \mathrm{(ii)}\left( \mathrm{A-B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ -B }\!\!'\!\!\text{ } \\ \end{align}\]

Ans:

We have 

\[\begin{align} & A'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]B'=\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right] \\ & (i)A+B=\left[ \begin{matrix} -5 & 3 & -2 \\ 6 & 9 & 9 \\ -1 & 4 & 2 \\ \end{matrix} \right]\Rightarrow \left( A+B \right)'=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right] \\ & A'+B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -5 & 6 & -1 \\ 3 & 9 & 4 \\ -2 & 9 & 2 \\ \end{matrix} \right] \\ \end{align}\] Hence, we have verified that \[\left( A+B \right)'=A'+B'\] \[\begin{align} & (ii)A-B=\left[ \begin{matrix} 3 & 1 & 8 \\ 4 & 5 & 9 \\ -3 & -2 & 0 \\ \end{matrix} \right]\Rightarrow \left( A-B \right)'=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right] \\ & A'-B'=\left[ \begin{matrix} -1 & 5 & -2 \\ 2 & 7 & 1 \\ 3 & 9 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -4 & 1 & 1 \\ 1 & 2 & 3 \\ -5 & 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 3 & 4 & -3 \\ 1 & 5 & -2 \\ 8 & 9 & 0 \\ \end{matrix} \right] \\ \end{align}\]

Hence, we have verified that \[\left( A-B \right)'=A'-B'\]


3. If\[\mathrm{A }\!\!'\!\!\text{ =}\left[ \begin{matrix} \mathrm{3} & \mathrm{4} \\ \mathrm{-1} & \mathrm{2} \\ \mathrm{0} & \mathrm{1} \\ \end{matrix} \right]\] and \[\mathrm{B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \mathrm{1} & \mathrm{2} & \mathrm{3} \\ \end{matrix} \right]\] , then verify that \[\begin{align} & \mathrm{(i)}\left( \mathrm{A+B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ +B }\!\!'\!\!\text{ } \\ & \mathrm{(ii)}\left( \mathrm{A-B} \right)\mathrm{ }\!\!'\!\!\text{ =A }\!\!'\!\!\text{ -B }\!\!'\!\!\text{ } \\ \end{align}\]

Ans:

We have 

\[\begin{align} & A=\left[ \begin{matrix} 3 & -1 & 0 \\ 4 & 2 & 1 \\ \end{matrix} \right]B'=\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right] \\ & (i)A+B=\left[ \begin{matrix} 2 & 1 & 1 \\ 5 & 4 & 4 \\ \end{matrix} \right]\Rightarrow \left( A+B \right)'=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right] \\ & A'+B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]+\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 5 \\ 1 & 4 \\ 1 & 4 \\ \end{matrix} \right] \\ \end{align}\] Hence, we have verified that \[\left( A+B \right)'=A'+B'\] \[\begin{align} & (ii)A-B=\left[ \begin{matrix} 4 & -3 & -1 \\ 3 & 0 & -2 \\ \end{matrix} \right]\Rightarrow \left( A-B \right)'=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right] \\ & A'-B'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \\ \end{matrix} \right]-\left[ \begin{matrix} -1 & 1 \\ 2 & 2 \\ 1 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} 4 & 3 \\ -3 & 0 \\ -1 & -2 \\ \end{matrix} \right] \\ \end{align}\] Hence, we have verified that \[\left( A-B \right)'=A'-B'\]


4. If\[\mathrm{A }\!\!'\!\!\text{ =}\left[ \begin{matrix} \mathrm{-2} & \mathrm{3} \\ \mathrm{1} & \mathrm{2} \\ \end{matrix} \right]\mathrm{ }\!\!\And\!\!\text{ B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{0} \\ \mathrm{1} & \mathrm{2} \\ \end{matrix} \right]\] , then find \[\left( \mathrm{A+2B} \right)\mathrm{ }\!\!'\!\!\text{ }\]

Ans:

Solve for the condition\[\begin{align} & \therefore A=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right] \\ & \therefore A+2B=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+2\left[ \begin{matrix} -1 & 0 \\ 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} -2 & 1 \\ 3 & 2 \\ \end{matrix} \right]+\left[ \begin{matrix} -2 & 0 \\ 2 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} -4 & 1 \\ 5 & 6 \\ \end{matrix} \right] \\ & \therefore (A+2B)=\left[ \begin{matrix} -4 & 5 \\ 1 & 6 \\ \end{matrix} \right] \\ \end{align}\]


5. For the matrix \[\mathrm{A }\;\;\And\;\;\text{ B}\] , verify that \[\left( \mathrm{AB} \right)\mathrm{ }\!\!'\!\!\text{ =B }\!\!'\!\!\text{ A }\!\!'\!\!\text{ }\]where

\[\begin{align} & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{1} \\ \mathrm{-4} \\ \mathrm{3} \\ \end{matrix} \right]\mathrm{,B=}\left[ \begin{matrix} \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \end{matrix} \right] \\ & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{0} \\ \mathrm{1} \\ \mathrm{2} \\ \end{matrix} \right]\mathrm{,B=}\left[ \begin{matrix} \mathrm{1} & \mathrm{5} & \mathrm{7} \\ \end{matrix} \right] \\ \end{align}\] 

Ans:

\[\begin{align} & (i)~AB=\left[ \begin{matrix} 1 \\ -4 \\ 3 \\ \end{matrix} \right]\left[ \begin{matrix} -1 & 2 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\ -3 & 6 & 3 \\ \end{matrix} \right] \\ & \therefore (AB{)}'=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right] \\ & B'A'=\left[ \begin{matrix} -1 \\ 2 \\ 1 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & -4 & 3 \\ \end{matrix} \right]=\left[ \begin{matrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \\ \end{matrix} \right] \\ \end{align}\] Hence, we have verified \[\left( AB \right)'=B'A'\] \[\begin{align} & (ii)~AB=\left[ \begin{matrix} 0 \\ 1 \\ 2 \\ \end{matrix} \right]\left[ \begin{matrix} 1 & 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 0 & 0 \\ 1 & 5 & 7 \\ 2 & 10 & 14 \\ \end{matrix} \right] \\ & \therefore (AB{)}'=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right] \\ & B'A'=\left[ \begin{matrix} 1 \\ 5 \\ 7 \\ \end{matrix} \right]\left[ \begin{matrix} 0 & 1 & 2 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & 1 & 2 \\ 0 & 5 & 10 \\ 0 & 7 & 14 \\ \end{matrix} \right] \\ \end{align}\]

Hence, we have verified \[\left( AB \right)'=B'A'\]


6. If \[\begin{align} & \mathrm{(i)A=}\left[ \begin{matrix} \mathrm{cos }\;\;\alpha\;\;\text{ } & \mathrm{sin }\;\;\alpha\!\!\text{ } \\ \mathrm{-sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\;\;\text{ } \\ \end{matrix} \right]\mathrm{, then verify thatA }\;\;\;\;\text{ A=I} \\ & \mathrm{(ii)A=}\left[ \begin{matrix} \mathrm{sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\!\!\text{ } \\ \mathrm{-cos }\;\;\alpha\!\!\text{ } & \mathrm{sin }\;\;\alpha\!\!\text{ } \\ \end{matrix} \right]\mathrm{, then verify thatA }\;\;'\;\;\text{ A=I} \\ \end{align}\]

Ans:

Solve for the condition 

  1. \[\begin{align} & A=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & \therefore A'=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I \\ \end{align}\] \[\begin{align} & A=\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & \therefore A'=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} \sin \alpha & -\cos \alpha \\ \cos \alpha & \sin \alpha \\ \end{matrix} \right]\left[ \begin{matrix} \sin \alpha & \cos \alpha \\ -\cos \alpha & \sin \alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} {{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha & \sin \alpha \cos \alpha -\sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha -\sin \alpha \cos \alpha & {{\sin }^{2}}\alpha +{{\cos }^{2}}\alpha \\ \end{matrix} \right] \\ & A'A=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right]=I \\ \end{align}\]


7. (i) Show that the matrix \[\mathrm{A=}\left[ \begin{matrix} \mathrm{1} & \mathrm{-1} & \mathrm{5} \\ \mathrm{-1} & \mathrm{2} & \mathrm{1} \\ \mathrm{5} & \mathrm{1} & \mathrm{3} \\ \end{matrix} \right]\] is a symmetric matrix (ii) Show that the matrix \[\mathrm{A=}\left[ \begin{matrix} \mathrm{0} & \mathrm{1} & \mathrm{-1} \\ \mathrm{-1} & \mathrm{0} & \mathrm{1} \\ \mathrm{1} & \mathrm{-1} & \mathrm{0} \\ \end{matrix} \right]\] is a skew symmetric matrix

Ans:

(i) We have 

\[\begin{align} & A'=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \\ \end{matrix} \right]=A \\ & \therefore A'=A \\ \end{align}\] Hence, A is a symmetric matrix

(ii) We have \[\begin{align} & A'=\left[ \begin{matrix} 0 & -1 & 1 \\ 1 & 0 & -1 \\ -1 & 1 & 0 \\ \end{matrix} \right]=-\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \\ \end{matrix} \right]=-A \\ & \therefore A'=-A \\ \end{align}\]

Hence, A is a skew-symmetric matrix


8. For the matrix \[\mathrm{A=}\left[ \begin{matrix} \mathrm{1} & \mathrm{5} \\ \mathrm{6} & \mathrm{7} \\ \end{matrix} \right]\], verify that (i) \[\left( \mathrm{A+A }\!\!'\!\!\text{ } \right)\] is a symmetric matrix (ii) \[\left( \mathrm{A+A }\!\!'\!\!\text{ } \right)\] is a skew symmetric matrix 

Ans:

\[A=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right],A'=\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]\] \[\begin{align} & (i)A+A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]+\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right] \\ & \left( A+A' \right)'=\left[ \begin{matrix} 2 & 11 \\ 11 & 14 \\ \end{matrix} \right]=A+A' \\ \end{align}\] Hence, \[\left( A+A' \right)\]is a symmetric matrix \[\begin{align} & (ii)A-A'=\left[ \begin{matrix} 1 & 5 \\ 6 & 7 \\ \end{matrix} \right]-\left[ \begin{matrix} 1 & 6 \\ 5 & 7 \\ \end{matrix} \right]=\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right] \\ & \left( A-A' \right)'=\left[ \begin{matrix} 0 & 1 \\ -1 & 0 \\ \end{matrix} \right]=-\left[ \begin{matrix} 0 & -1 \\ 1 & 0 \\ \end{matrix} \right]=A+A' \\ \end{align}\]

Hence, \[\left( A-A' \right)\]is a skew-symmetric matrix


9. Find \[\frac{\mathrm{1}}{\mathrm{2}}\left( \mathrm{A+A }\;\;'\;\;\text{ } \right)\mathrm{ }\;\;\And\!\!\text{ }\frac{\mathrm{1}}{\mathrm{2}}\left( \mathrm{A-A }\;\;'\;\;\text{ } \right)\] , when \[\mathrm{A=}\left[ \begin{matrix} \mathrm{0} & \mathrm{a} & \mathrm{b} \\ \mathrm{-a} & \mathrm{0} & \mathrm{c} \\ \mathrm{-b} & \mathrm{-c} & \mathrm{0} \\ \end{matrix} \right]\]

Ans:

The given matrix is \[A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]\], then \[A'=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right]\] \[\begin{align} & \frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]+\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right] \\ \end{align}\] \[\begin{align} & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right]-\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \\ \end{matrix} \right] \\ \end{align}\]


10. Express the following matrices as the sum of a symmetric and a skew symmetric matrix:

\[\left[ \begin{matrix} \mathrm{3} & \mathrm{5} \\ \mathrm{1} & \mathrm{-1} \\ \end{matrix} \right]\]

Ans:

\[\begin{align} & (i)\frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]+\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 6 & 6 \\ 6 & -2 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A+A' \right)=\left[ \begin{matrix} 3 & 3 \\ 3 & -1 \\ \end{matrix} \right] \\ \end{align}\] Thus, \[\frac{1}{2}\left( A+A' \right)\] is a symmetric matrix. \[\begin{align} & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 3 & 5 \\ 1 & -1 \\ \end{matrix} \right]-\left[ \begin{matrix} 3 & 1 \\ 5 & -1 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\frac{1}{2}\left( \left[ \begin{matrix} 0 & 4 \\ -4 & 0 \\ \end{matrix} \right] \right) \\ & \frac{1}{2}\left( A-A' \right)=\left[ \begin{matrix} 0 & 2 \\ -2 & 0 \\ \end{matrix} \right] \\ \end{align}\]

Thus, \[\frac{1}{2}\left( A-A' \right)\] is a skew-symmetric matrix.


11. If A,B are symmetric matrix of same order, then \[\mathrm{AB-BA}\] is a 

  1. Skew symmetric matrix

  2. Symmetric matrix 

  3. Zero matrix

  4. Identity matrix

Ans:

The correct answer is A

\[A\And B\] are symmetric , therefore , we have 

\[A'=A\And B'=B\]

Consider 

\[\begin{align} & \left( AB-BA \right)'=\left( AB \right)'-\left( BA \right)' \\ & =B'A'-A'B' \\ & =BA-AB \\ & =-\left( AB-BA \right) \\ & \therefore \left( AB-BA \right)'=-\left( AB-BA \right) \\ \end{align}\] Thus, \[\left( AB-BA \right)'\] is a skew-symmetric matrix


12. If\[\mathrm{A=}\left[ \begin{matrix} \mathrm{cos }\;\;\alpha\;\;\text{ } & \mathrm{-sin }\;\;\alpha\!\!\text{ } \\ \mathrm{sin }\;\;\alpha\;\;\text{ } & \mathrm{cos }\;\;\alpha\!\!\text{ } \\ \end{matrix} \right]\] , then \[\mathrm{A+A }\;\;'\;\;\text{ =I}\] , if the value of \[\mathrm{ }\;\;\alpha\!\!\text{ }\] is \[\begin{align} & \mathrm{A}\mathrm{.}\frac{\mathrm{ }\;\;\pi\;\;\text{ }}{\mathrm{6}} \\ & \mathrm{B}\mathrm{.}\frac{\mathrm{ }\;\;\pi\;\;\text{ }}{\mathrm{3}} \\ & \mathrm{C}\mathrm{. }\;\;\pi\!\!\text{ } \\ & \mathrm{D}\mathrm{.}\frac{\mathrm{3 }\;\;\pi\!\!\text{ }}{\mathrm{2}} \\ \end{align}\]

Ans:

The correct answer is B

 \[\begin{align} & A=\left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A'=\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right] \\ & A+A'=I \\ & \therefore \left[ \begin{matrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \\ \end{matrix} \right]+\left[ \begin{matrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & \left[ \begin{matrix} 2\cos \alpha & 0 \\ 0 & 2\cos \alpha \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right] \\ & \alpha =\frac{\pi }{3} \\ \end{align}\]


Conclusion

In conclusion, class 12 ex 3.3 has thoroughly explored the fundamental concepts of matrices, including various operations, types, properties, and their applications. This exercise is essential for understanding how matrices are used in solving linear equations and performing geometric transformations. Mastery of these concepts is crucial for students aiming to excel in advanced mathematics and related fields.


Class 12 Maths Chapter 3: Exercises Breakdown

S.No.

Chapter 3 - Matrices Exercises in PDF Format

1

Class 12 Maths Chapter 3 Exercise 3.1 - 10 Questions & Solutions (5 Short Answers, 5 Long Answers)

2

Class 12 Maths Chapter 3 Exercise 3.2 - 22 Questions & Solutions (3 Short Answers, 19 Long Answers)

3

Class 12 Maths Chapter 3 Exercise 3.4 - 18 Questions & Solutions (18 Short Answers)

4

Class 12 Maths Chapter 3 Miscellaneous Exercise - 11 Questions & Solutions



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FAQs on NCERT Solutions for Class 12 Maths Chapter 3: Matrices - Exercise 3.3

1. What type of questions are covered in Ex 3.3 class 12?

Exercise 3.3 focuses on applying concepts of matrix transpose, symmetric and skew-symmetric matrices, and basic operations like addition, subtraction, and scalar multiplication on matrices.

2. How to approach solving questions in Exercise 3.3 Class 12 Maths?

  • Revise the concepts of transpose, symmetric and skew-symmetric matrices.

  • Understand the properties of these matrices mentioned in the NCERT chapter.

  • Practice with solved examples to grasp the problem-solving approach.

  • Attempt the exercise questions yourself. Refer to the solutions only if you get stuck.