Class 12 Maths NCERT Solutions Chapter 3 Matrices: Exercise PDFs, Key Topics, Weightage & Formulae

The Class 12 Chapter 3 Matrices typically have 6-8 marks weightage in the CBSE Class 12 Board Exams. Matrices are also a very important chapter for competitive exams for their direct weightage moreover its applications in solving linear equation systems in geometry, linear algebra, vector algebra, and more. Check the Key Topics discussed in Matrices Class 12 Solutions below;

Class 12 Matrices Key Topics

• Fundamentals of Matrices: Matrices are defined as an ordered rectangular array of numbers or functions. The numbers or functions are called the elements/ entries of the matrix. basic of Matrices are discussed.
• Types of Matrices: Row Matrix, Column Matrix, Square Matrix, Diagonal Matrix, Scalar Matrix, Identity Matrix, Zero/Null Matrix, Symmetric and Skew-Symmetric Matrices.
• Operations on Matrices: Addition and Subtraction, Scalar Multiplication, Matrix/ Vector Multiplication, Transpose of a Matrix.
• Other Important Concepts: Properties of Matrix Operations, Determinants and Inverses 
• Applications of Matrices: Solving linear equations system using matrices.

**Check out Dropped Topics of Class 12 Chapter 3 Matrices – 3.7 Elementary Operations (Transformation) of a Matrix, 3.8.1 Inverse of Matrices by Elementary Operations (Retain Ques. 18 of Exercise 3.4), Page 98 Example 26, Page 100-101 Ques. 1–3 and 12 (Miscellaneous Exercise), Page 102 Third Last Point of Summary.

Matrices Important Formulae for CBSE and Competitive Exams

Students can check important formulae and basic concepts of the Matrices chapter below for a better understanding of questions. Students can also use these formulae to solve exercises.

Matrix Basics

• Order of a Matrix:
  • If a matrix has $m$ rows and $n$ columns, its order is $m \times n$.
• Elements:
  • $A = [a_{ij}]$, where $a_{ij}$ represents the element in the $i^{\text{th}}$ row and $j^{\text{th}}$ column.

Matrix Operations

• Addition/Subtraction:

  $$(A \pm B)_{ij} = a_{ij} \pm b_{ij}$$

• Scalar Multiplication:

  $$(kA)_{ij} = k \cdot a_{ij}$$

• Matrix Multiplication:

  $$(AB)_{ij} = \sum_{k=1}^n a_{ik} \cdot b_{kj}$$
  • Condition: Number of columns in $A$ = Number of rows in $B$.

• Transpose of a Matrix:

  $$(A^T)_{ij} = a_{ji}$$

Properties of Matrix Operations

• Transpose Properties:

  $$(A^T)^T = A$$ $$(A + B)^T = A^T + B^T$$ $$(kA)^T = kA^T$$ $$(AB)^T = B^T A^T$$

• Symmetry:

  $$A^T = A \quad (\text{Symmetric})$$ $$A^T = -A \quad (\text{Skew-Symmetric})$$

• Identity Matrix Properties:

  $$AI = IA = A$$

Determinant and Inverse of Square Matrices

• Adjoint Formula for Inverse:

  $$A^{-1} = \frac{\text{Adj}(A)}{\det(A)}$$

• Determinant Property:

  $$\det(A^T) = \det(A)$$

Applications of Matrcies

• Solving Linear Equations: $$AX = B \implies X = A^{-1}B$$
• Consistency of Linear Equations:
  • If $\det(A) \neq 0$, the system has a unique solution.

We have uploaded the complete exercise-wise solution of Matrices Class 12 NCERT PDF in one place. This Class 12 Chapter 3 Matrices NCERT Solution PDF is beneficial for all students in both CBSE boards and competitive exams; JEE Main, CUSAT CAT, BITSAT, etc... Students can access for free the Matrices Class 12 solutions pdf download below;

Class 12 Math Chapter 3 Matrices Solution: Free PDF Download

The class 12 Matrices Chapter focuses on several important topics, such as the Type of Matrices, zero, symmetric, and asymmetric Matrices, the Transpose and Inverse of Matrices, and other Matrices operations. Students will encounter different problems based on these concepts in the exercises. Each exercise deals with different concepts, Class 12 Matrices Exercise 3.1 deals with the basics of Matrix, its types and fundamentals. Class 12 Matrices Exercise 3.2  deals with operations of Matrix, its multiplication both scaler and vector and more advanced concepts. Class 12 Matrices Exercise 3.3 deals with problems finding the transpose and inverse of Matrix. Students can check Exercise-wise Class 12 Chapter 3 Matrices NCERT Solutions below;

Matrices exercise 3.1 solutions focuses on the basic concepts of matrices. Definition, types of Matrices, and basic operations of matrices are several topics discussed in the solutions of Matrices Class 12 Exercise 3.1. Chapter 3 Matrices Exercise 3.1 includes 10 Questions (7 Short Answers, 3 MCQs). Students can find the solution of exercise 3.1 below;

Matrices Exercise 3.1 Solutions

Q2. If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

A.2. As, number of elements of matrix having order m × n = m.n.

(b) So, (possible) order of matrix with 24 elements are (1 × 24), (2 × 12), (3 × 8), (4 × 6), (6 × 4), (8 × 3), (12 × 2), 24 × 1).

Similarly, possible order of matrix with 13 elements are (1 × 13) and (13 × 1)

Q3.If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

A.3. As number of elements of matrix with order m × n

(E) Possible order of matrix with 18 elements are (1 × 18), (2 × 9), (3 × 6), (6 × 3), (9 × 2) and (18 × 1)

Similarly, possible order of matrix with 5 elements are (1 × 5) and (5 × 1)

Q4.Construct a 2 × 2 matrix,  $A=\left[a_{ij}\right]$ , whose elements are given by:

$a_{ij}=\frac{{\left(i+j\right)}^2}{2}$ (ii)  $a_{ij}=\frac{i}{j}$ (iii)  $a_{ij}=\frac{{\left(i+2j\right)}^2}{2}$

A.4. (E) (i) a^ij ${\frac{\left(i+j\right)}{2}}^2$ such that i = 1, 2 and j = 1 × 2 for 2 × 2 matrix

Therefore a₁₁ = $\frac{{\left(1+1\right)}^2}{2}=\frac{2^2}{2}=2$ A _2×2 = $\left[\begin{array}{cc}\hfill a_{11}\hfill & \hfill a_{12}\hfill \\ \hfill a_{21}\hfill & \hfill a_{22}\hfill \end{array}\right]$

a₁₂ = $\frac{{\left(1+2\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$

a₂₁ $\frac{{\left(2+1\right)}^2}{2}=\frac{3^2}{2}=\frac{9}{2}$  $\left[\begin{array}{cc}\hfill 2\hfill & \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill \\ \hfill \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\hfill & \hfill 8\hfill \end{array}\right]$

a₂₂ = ${\frac{\left(2+2\right)}{2}}^2=\frac{4^2}{2}=\frac{16}{2}=8$

Q5.Construct a 3 × 4 matrix, whose elements are given by:

(i)  $a_{ij}=\frac{1}{2}|-3i+j|$ (ii)  $a_{ij}=2i-j$

A.5. (E) (i) a^ij = $\frac{1}{2}$  $|-3i+j|$ such that i = 1, 2, 3 and j = 1, 2, 3, 4 for 3 × 4 matrix

So, a₁₁= $\frac{1}{2}$ .  $|-3.1+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-3+1|=\frac{1}{2}|-2|=\frac{2}{2}=1.$

a₁₂ = $\frac{1}{2}|-3.1+2|=\frac{1}{2}|-1|=\frac{1}{2}$

$a_{13}=\frac{1}{2}|-3.1+3|=\frac{1}{2}\times 0=0$

a₁₄ = $\frac{1}{2}|-3.1+4|=\frac{1}{2}|1|=\frac{1}{2}$

a₂₁ = $\frac{1}{2}|-3.2+1|=\frac{1}{2}|-6+1|=\frac{1}{2}|5|=\frac{5}{2}$

a₂₂ = $\frac{1}{2}|-3.2+2|=\frac{1}{2}|-6+2|=\frac{1}{2}|-4|=\frac{4}{2}=2$

a₂₃ = $\frac{1}{2}|-3.2+3|=\frac{1}{2}|-6+3|=\frac{1}{2}|-3|=\frac{3}{2}$

$a_{24}=\frac{1}{2}|-3\cdot 2+4|=\frac{1}{2}|-6+4|=\frac{1}{2}+4|-2|=\frac{2}{2}=1$

$a_{31}=\frac{1}{2}|-3\cdot 3+1|=\frac{1}{2}|-0+1|=\frac{1}{2}|-8|=\frac{8}{2}=4.$

$a_{32}=\frac{1}{2}|-3\cdot 2+2|=\frac{1}{2}|-9+2|=\mid -\frac{7}{2}=\frac{7}{2}$

$a_{33}=\frac{1}{2}|-3\cdot 3+3|=\frac{1}{2}|-9+3|=\frac{+6\mid }{2}=\frac{6}{2}=3$

$a_{34}=\frac{1}{2}|-3\cdot 3+4|=\frac{1}{2}|-9+4|=\mid \frac{\left|5\right|}{2}=\frac{5}{2}.$

Q6. Find the values of x, y and z from the following equations:

A.6. (i)  $\left[\begin{array}{ll}4\hfill & 3\hfill \\ x\hfill & 5\hfill \end{array}\right]=\left[\begin{array}{ll}y\hfill & z\hfill \\ 1\hfill & 5\hfill \end{array}\right]$

corresponding

By equating the elements of the matrices, we get,

x= 1

y= 4

z= 3.

(ii)  $\left[\begin{array}{cc}\hfill x+y\hfill & \hfill 2\hfill \\ \hfill 5+z\hfill & \hfill xy\hfill \end{array}\right]=\left[\begin{array}{ll}6\hfill & 2\hfill \\ 5\hfill & 8\hfill \end{array}\right]$

By equating the corresponding elements of the matrices we get,

x+ y = 6 (I)

5 + Z = 5  $\Rightarrow z=5-5\Rightarrow z=0$

xy = 8

$\Rightarrow$ x  $=\frac{8}{y}$  $\to (2)$

putting eq ${}^n(2)$ in (1) we get

$\frac{8}{y}$ + y = 6.

$\Rightarrow$ 8 + y² = 6y

$\Rightarrow$ y² 6y + 8 = 0.

$\Rightarrow$ y² - 4y - 2y + 8 = 0

$\Rightarrow$ y (y-4) -2 (y-4) = 0

$\Rightarrow$ (y-4) (y-2) = 0

$\Rightarrow$ y= 4 0r y = 2.

When y = 4,x= 6-y = 6-4 = and z = 0.

Wheny = 2,x = 6-y = 6-2 = 4 and z = 0.

By equating the corresponding elements of the matrices we get,

x+ y + z = 9 -------(i)

x + z = 7 --------(ii)

y + z = 7 -------(iii)

Subtracting eqⁿ (3) from (1) and (2) from (1) we get,

x + y + z -y - z = 9 - 7 and x + y + z - x - z = 9 - 5

$\Rightarrow$ x = 2 and y = 4 .

Putting x = 2 in eqⁿ (2)

2 + z = 5

$\Rightarrow$ z = 5 - 2 = 3.

So, x = 2,y = 4, z = 3.

Q7. Find the value of a, b, c and d from the equation:

$\left[\begin{array}{cc}\hfill a-b\hfill & \hfill 2a+c\hfill \\ \hfill 2a-b\hfill & \hfill 3c+d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 13\hfill \end{array}\right]$

A.7.  $\left[\begin{array}{cc}\hfill a-b\hfill & \hfill 2a+c\hfill \\ \hfill 2a-b\hfill & \hfill 3c+d\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill -1\hfill & \hfill 5\hfill \\ \hfill 0\hfill & \hfill 13\hfill \end{array}\right]$

(5) Equating the corresponding elements of the matrices we get,

a - b = -1 --------(i)

2a + c = 5 ------(2)

2a - b = 0 ----(3)

3c + d = 13 -----(4)

Subtracting eqⁿ(1) from (3) we get,

2a - b (a - b) = 0-(-1)

2a - b - a+ b = 0 + 1 = 1

$\Rightarrow$ [a= 1]

Put  $a=\mathrm{1\; eq}{}^n(2)$ we get,

2 × 1 + c = 5  $\Rightarrow$ c = 5 - 2 $\Rightarrow$ [c = 3]

Put  $c=3q^n$ (4) we get,

3 × 3 + d = 13  => d = 13 - 9 $\Rightarrow$ [d = 4.]

put  $a=1q^2(1)$ we get, 1 - b = -1 $\Rightarrow$ b = 1 + 1  $\Rightarrow$ [b= 2]

Q8.A=  ${\left[a_{ij}\right]}_{mxn\backslash }$ is a square matrix, if

(A) m n (C) m = n  (D) None of these

A.8.  $={\left[a_{ij}\right]}_{m\times n}$ is a square matrix if m = n

(E) Option C is correct.

Q9.Which of the given values of x and y make the following pair of matrices equal:

$\left[\begin{array}{cc}\hfill 3x+7\hfill & \hfill 5\hfill \\ \hfill y+1\hfill & \hfill 2-3x\hfill \end{array}\right],\left[\begin{array}{cc}\hfill 0\hfill & \hfill y-2\hfill \\ \hfill 8\hfill & \hfill 4\hfill \end{array}\right]$

$\mathrm{(A)\; x}=\frac{-1}{3},y=7$ (B) Not possible to find (C)  $y=7,x=\frac{-2}{3}$

(D)  $x=\frac{-1}{3},y=\frac{-2}{3}$

A.9. For the matrices to be equal, the corresponding elements

(B) heed to be equal so,

Fora₁₁, 3x+ 7 = 0

$\Rightarrow$ 3x = -7.

$\Rightarrow$ x =  $-\frac{7}{3}$

fora₁₂, 5 = y - 2 

$\Rightarrow$ y = + 5 + 2 = 7.

For a₂₁, y + 1 = 8

$$  $\Rightarrow$ y = 8 - 1 = 7.

fora₂₂, 2 - 3x = 4.

$\Rightarrow$ 3x = 2 - 4

$\Rightarrow x=-\frac{2}{3}$

As the variable x and y has more than one value which is not peacetable.

Option B is correct.

Q10.The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27 (B) 18 (C) 81 (D) 512

A.10.  A 3 × 3 order matrix will have 9 elements

(M) Since, the elements can be either 1 or  $\mathrm{0il},$ number of choices for each element is 2 .

The required no. of arrangement = 29(4,2 × 2 × 2 9 times )  $\Rightarrow$ 512

So, option D is correct.

Class 12 Matrices Exercise 3.2  deals with problems based on the operations performed on matrices such as addition, subtraction, scalar multiplication, and their properties. The Matrices exercise 3.2 includes 22 Questions 14 Long, 6 Short, and 2 MCQs. Students can check the complete exercise 3.2 solution below;

Matrix Exercise 3.2 Solutions

Q1 Let $A=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 4\hfill \\ \hfill 3\hfill & \hfill 2\hfill \end{array}\right],B=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right],C=\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

Find each of the following:

(i) A + B (ii) A – B (iii) 3A – C (iv) AB (v) BA

A.1. A $=\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right],$ B  $=\left[\begin{array}{rr}\hfill 1& \hfill 3\\ \hfill -2& \hfill 5\end{array}\right],$ C  $=\left[\begin{array}{rr}\hfill -2& \hfill 5\\ \hfill 3& \hfill 4\end{array}\right]$

(i) A + B =  $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2+1\hfill & \hfill 4+3\hfill \\ \hfill 3-2\hfill & \hfill 2+5\hfill \end{array}\right]=\left[\begin{array}{ll}3\hfill & 7\hfill \\ 1\hfill & 7\hfill \end{array}\right]$

(ii) A - B = $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-1\hfill & \hfill 4-3\hfill \\ \hfill 3-(-2)\hfill & \hfill 2-5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 1\hfill \\ \hfill 5\hfill & \hfill -3\hfill \end{array}\right].$

(iii) 3A - C = 3 $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{ll}3\times 2\hfill & 3\times 4\hfill \\ 3\times 3\hfill & 3\times 2\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill 12\hfill \\ \hfill 9\hfill & \hfill 6\hfill \end{array}\right]\left[\begin{array}{cc}\hfill -2\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]$

$$  $=\left[\begin{array}{cc}\hfill 6-(-2)\hfill & \hfill 12-5\hfill \\ \hfill 9-3\hfill & \hfill 6-4\hfill \end{array}\right]$ =  $\left[\begin{array}{cc}\hfill 8\hfill & \hfill 7\hfill \\ \hfill 6\hfill & \hfill 2\hfill \end{array}\right]$

(iv) AB =  $\left[\begin{array}{ll}2\hfill & 4\hfill \\ 3\hfill & 2\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill -2\hfill & \hfill 5\hfill \end{array}\right]=\left[\begin{array}{ll}2\times 1+4\times (-2)\hfill & 2\times 3+4\times 5\hfill \\ 3\times 1+2\times (-2)\hfill & 3\times 3+2\times 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2-8\hfill & \hfill 6+20\hfill \\ \hfill 3-4\hfill & \hfill 9+10\hfill \end{array}\right]$

=  $\left[\begin{array}{rr}\hfill -6& \hfill 26\\ \hfill -1& \hfill 19\end{array}\right]$

Q2.Compute the following:

(i)  $\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill a\hfill \end{array}\right]$ (ii)  $\left[\begin{array}{cc}\hfill a^2+b^2\hfill & \hfill b^2+c^2\hfill \\ \hfill a^2+c^2\hfill & \hfill a^2+b^2\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 2ab\hfill & \hfill 2bc\hfill \\ \hfill -2ac\hfill & \hfill -2ab\hfill \end{array}\right]$

(iii)  $\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 4\hfill & \hfill -6\hfill \\ \hfill 8\hfill & \hfill 5\hfill & \hfill 16\hfill \\ \hfill 2\hfill & \hfill 8\hfill & \hfill 5\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 12\hfill & \hfill 7\hfill & \hfill 6\hfill \\ \hfill 8\hfill & \hfill 0\hfill & \hfill 5\hfill \\ \hfill 3\hfill & \hfill 2\hfill & \hfill 4\hfill \end{array}\right]$ (iv)  $\left[\begin{array}{cc}\hfill co{s}^{2}x\hfill & \hfill si{n}^{2}x\hfill \\ \hfill si{n}^{2}x\hfill & \hfill co{s}^{2}x\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill si{n}^{2}x\hfill & \hfill co{s}^{2}x\hfill \\ \hfill co{s}^{2}x\hfill & \hfill si{n}^{2}x\hfill \end{array}\right]$

A.2. (I)  $\left[\begin{array}{cc}\hfill a\hfill & \hfill b\hfill \\ \hfill -b\hfill & \hfill a\hfill \end{array}\right]+\left[\begin{array}{ll}a\hfill & b\hfill \\ b\hfill & a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill a+a\hfill & \hfill b+b\hfill \\ \hfill -b+b\hfill & \hfill a+a\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 2a\hfill & \hfill 2b\hfill \\ \hfill 0\hfill & \hfill 2a\hfill \end{array}\right]$

Q4.if  $A=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right],B=\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$ and  $C=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$ then compute

(A+B) and (B – C). Also, verify that A + (B – C) = (A + B) – C.

A.4  A + B = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 4\hfill & \hfill 2\hfill & \hfill 5\hfill \\ \hfill 2\hfill & \hfill 0\hfill & \hfill 3\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 1+3\hfill & \hfill 2-1\hfill & \hfill -3+2\hfill \\ \hfill 5+4\hfill & \hfill 0+2\hfill & \hfill 2+5\hfill \\ \hfill 1+2\hfill & \hfill -1+0\hfill & \hfill 1+3\hfill \end{array}\right]$

$A+B=\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]$

$B-C=\left[\begin{array}{lll}3\hfill & -1\hfill & 2\hfill \\ 4\hfill & 2\hfill & 5\hfill \\ 2\hfill & 0\hfill & 3\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$  $=\left[\begin{array}{l}3-4-1-12-2\\ 4-02-35-2\\ 2-10-(2)33\end{array}\right]$

$B-C=\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$

Now

L.H.S. = a+ (b - c) = $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 5\hfill & \hfill 0\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -1\hfill & \hfill 1\hfill \end{array}\right]+\left[\begin{array}{ccc}\hfill -1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill 4\hfill & \hfill -1\hfill & \hfill 3\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill 0\hfill \end{array}\right]$  $=\left[\begin{array}{l}1-12-2-3+0\\ 5+40-12+3\\ 1+1-1+21+0\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$

R.H.S. = (a+ b) - c = $\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill -1\hfill \\ \hfill 9\hfill & \hfill 2\hfill & \hfill 7\hfill \\ \hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \end{array}\right]-\left[\begin{array}{ccc}\hfill 4\hfill & \hfill 1\hfill & \hfill 2\hfill \\ \hfill 0\hfill & \hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 3\hfill \end{array}\right]$  $=\left[\begin{array}{l}4-41-1-1-3\\ 9-02-37-2\\ 3-1-1-(2)4-3\end{array}\right]$

$=\left[\begin{array}{l}00-3\\ 9-15\\ 211\end{array}\right]$ = L.H.S.

∴A + (B - C) = (A + B) - C

Q5.If  $A=\left[\begin{array}{ccc}\hfill \frac{2}{3}\hfill & \hfill 1\hfill & \hfill \frac{5}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{2}{3}\hfill & \hfill \frac{4}{3}\hfill \\ \hfill \frac{7}{3}\hfill & \hfill 2\hfill & \hfill \frac{2}{3}\hfill \end{array}\right]$ and  $B=\left[\begin{array}{ccc}\hfill \frac{2}{5}\hfill & \hfill \frac{3}{5}\hfill & \hfill 1\hfill \\ \hfill \frac{1}{5}\hfill & \hfill \frac{2}{5}\hfill & \hfill \frac{4}{5}\hfill \\ \hfill \frac{7}{5}\hfill & \hfill \frac{6}{5}\hfill & \hfill \frac{2}{5}\hfill \end{array}\right]$ then compute 3A – 5B.

A.5. 3a - 5b = 3 × $\left[\begin{array}{ccc}\hfill \frac{2}{3}\hfill & \hfill 1\hfill & \hfill \frac{5}{3}\hfill \\ \hfill \frac{1}{3}\hfill & \hfill \frac{2}{3}\hfill & \hfill \frac{4}{3}\hfill \\ \hfill \frac{7}{3}\hfill & \hfill 2\hfill & \hfill \frac{2}{3}\hfill \end{array}\right]$ -  $5\times \left[\begin{array}{ccc}\hfill \frac{2}{5}\hfill & \hfill \frac{3}{5}\hfill & \hfill 1\hfill \\ \hfill \frac{1}{5}\hfill & \hfill \frac{2}{5}\hfill & \hfill \frac{4}{5}\hfill \\ \hfill \frac{7}{5}\hfill & \hfill \frac{6}{5}\hfill & \hfill \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{ccc}\hfill 3\times \frac{2}{3}\hfill & \hfill 3\times 1\hfill & \hfill 3\times \frac{5}{3}\hfill \\ \hfill 3\times \frac{1}{3}\hfill & \hfill 3\times \frac{2}{3}\hfill & \hfill 3\times \frac{4}{3}\hfill \\ \hfill 3\times \frac{7}{3}\hfill & \hfill 3\times 2\hfill & \hfill 3\times \frac{2}{3}\hfill \end{array}\right]-$  $\left[\begin{array}{ccc}\hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{3}{5}\hfill & \hfill 5\times 1\hfill \\ \hfill 5\times \frac{1}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill & \hfill 5\times \frac{4}{5}\hfill \\ \hfill 5\times \frac{7}{5}\hfill & \hfill 5\times \frac{6}{5}\hfill & \hfill 5\times \frac{2}{5}\hfill \end{array}\right]$

$=\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]-\left[\begin{array}{lll}2\hfill & 3\hfill & 5\hfill \\ 1\hfill & 2\hfill & 4\hfill \\ 7\hfill & 6\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ccc}\hfill 2-2\hfill & \hfill 3-3\hfill & \hfill 5-5\hfill \\ \hfill 1-1\hfill & \hfill 2-2\hfill & \hfill 4-4\hfill \\ \hfill 7-7\hfill & \hfill 6-6\hfill & \hfill 2-2\hfill \end{array}\right]$ =  $\left[\begin{array}{l}000\\ 600\\ 000\end{array}\right]$

Q6.Simplify  $cos\theta \left[\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right]+sin\theta \left[\begin{array}{cc}\hfill sin\theta \hfill & \hfill -cos\theta \hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill \end{array}\right]$

A.6..  $cos\theta \left[\begin{array}{cc}\hfill cos\theta \hfill & \hfill sin\theta \hfill \\ \hfill -sin\theta \hfill & \hfill cos\theta \hfill \end{array}\right]+sin\theta \left[\begin{array}{cc}\hfill sin\theta \hfill & \hfill -cos\theta \hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill \end{array}\right]$

(E)  $=\left[\begin{array}{l}co{r}^2\theta cot\theta sin\theta \\ -cos\theta sin\theta co{s}^2\theta \end{array}\right]$  $+\left[\begin{array}{l}sin2\theta -sin\theta cos\theta .\\ sin\theta cos\theta si{n}^2\theta \end{array}\right]$  $$

$=\left[\begin{array}{l}co{s}^2\theta +si{n}^2\theta cos\theta sin\theta -sin\theta cos\theta \\ -cos\theta sin\theta +sin\theta cos\theta co{s}^2\theta +si{n}^2\theta \end{array}\right]$

$=\left[\begin{array}{l}10\\ 01\end{array}\right]$

Q7.Find X and Y, if

(i)  $X+Y=\left[\begin{array}{cc}\hfill 7\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 5\hfill \end{array}\right]$ and  $X-Y=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 3\hfill \end{array}\right]$

(ii)  $2X+3Y=\left[\begin{array}{cc}\hfill 2\hfill & \hfill 3\hfill \\ \hfill 4\hfill & \hfill 0\hfill \end{array}\right]$ and  $3X+2Y=\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 5\hfill \end{array}\right]$

A.7. (i) x + y = $\left[\begin{array}{ll}7\hfill & 0\hfill \\ 2\hfill & 5\hfill \end{array}\right]$

x - y = $\left[\begin{array}{l}30\\ 03\end{array}\right]$

$\Rightarrow Y=\frac{1}{2}\left[\begin{array}{ll}4\hfill & 0\hfill \\ 2\hfill & 2\hfill \end{array}\right]=\left[\begin{array}{ll}\frac{1}{2}\times 4\hfill & \frac{1}{2}\times 0\hfill \\ \frac{1}{2}\times 2\hfill & \frac{1}{2}\times 2\hfill \end{array}\right]$  $=\left[\begin{array}{l}20\\ 11\end{array}\right]$

(ii)2x + 3y =  $\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]$ ___ (1)

2x + 2y =  $\left[\begin{array}{cc}\hfill 2\hfill & \hfill -2\hfill \\ \hfill -1\hfill & \hfill 5\hfill \end{array}\right]$ _____ (2)

Multiplying  eqⁿ by 2 and $q^n(u)$

2 × (2x+ 3y) = 2 ×  $\left[\begin{array}{l}23\\ 40\end{array}\right]$

$\Rightarrow$ 4x + 6y =  $\left[\begin{array}{ll}2\times 2\hfill & 2\times 3\hfill \\ 2\times 4\hfill & 2\times 0\hfill \end{array}\right]$

$\Rightarrow$ 4x + 6y =  $\left[\begin{array}{ll}4\hfill & 6\hfill \\ 8\hfill & 0\hfill \end{array}\right]$  $\to (iii).$

3 × (3x+ 2y) = 3 ×  $\left[\begin{array}{l}2-2\\ -15\end{array}\right]$

$\Rightarrow$ 9x + 6y =  $\left[\begin{array}{cc}\hfill 3\times (2)\hfill & \hfill 3\times (-2)\hfill \\ \hfill 3\times (-1)\hfill & \hfill 3\times 5\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 6\hfill & \hfill -6\hfill \\ \hfill -3\hfill & \hfill 15\hfill \end{array}\right]$  $\to (iv)$

Subtracting  eqⁿ  (iii) from (iv) we get,

4x+ 6y - (4x+ 6y) = $\left[\begin{array}{l}6-6\\ -315\end{array}\right]-$  $\left[\begin{array}{l}46\\ 80\end{array}\right]$

$\Rightarrow$ 5x =  $\left[\begin{array}{l}6-4-6-6\\ -3-815-0\end{array}\right]=$  $\left[\begin{array}{l}2-12\\ -1115\end{array}\right]$

$\Rightarrow x=\frac{1}{5}\left[\begin{array}{rr}\hfill 2& \hfill -12\\ \hfill -11& \hfill 15\end{array}\right]=\left[\begin{array}{rr}\hfill 2\times \frac{1}{5}& \hfill -12\times \frac{1}{5}\\ \hfill -11\times \frac{1}{5}& \hfill 15\times \frac{1}{5}\end{array}\right]=$  $\left[\begin{array}{l}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$-12$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$-11$}\!\left/ \!\raisebox{-1ex}{$53$}\right.\end{array}\right]$

From eqⁿ  (u);

$3y=\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]-2x=\left[\begin{array}{ll}2\hfill & 3\hfill \\ 4\hfill & 0\hfill \end{array}\right]-\left[\begin{array}{cc}\hfill 2\times \frac{2}{5}\hfill & \hfill 2\times \left(-\frac{12}{5}\right)\hfill \\ \hfill 2\times (-11/5)\hfill & \hfill 2\times 3\hfill \end{array}\right]$

$=\left[\begin{array}{l}23\\ 40\end{array}\right]-$  $\left[\begin{array}{l}\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$-24$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$-22$}\!\left/ \!\raisebox{-1ex}{$56$}\right.\end{array}\right]$

$=\left[\begin{array}{l}2\raisebox{1ex}{$-4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.3-\left(\raisebox{1ex}{$-24$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)\\ 4-\left(\raisebox{1ex}{$-22$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)0-6\end{array}\right]$

$=\left[\begin{array}{cc}\hfill \frac{5\times 2-4}{5}\hfill & \hfill \frac{3\times 5+24}{5}\hfill \\ \hfill \frac{4\times 5+22}{5}\hfill & \hfill -6\hfill \end{array}\right]$

$=\left[\begin{array}{cc}\hfill \frac{10-4}{5}\hfill & \hfill \frac{15+24}{5}\hfill \\ \hfill \frac{20+22}{5}\hfill & \hfill -6\hfill \end{array}\right]$

y  $=\frac{1}{3}\left[\begin{array}{cc}\hfill \raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill \raisebox{1ex}{$3q$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill \\ \hfill \raisebox{1ex}{$42$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\hfill & \hfill -6\hfill \end{array}\right]=$  $\left[\begin{array}{l}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{6}{5}\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{39}{5}\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \frac{42}{5}\frac{1}{3}\times \left(-6\right)\end{array}\right]$

$=\left[\begin{array}{l}\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\\ \raisebox{1ex}{$14$}\!\left/ \!\raisebox{-1ex}{$5-2$}\right.\end{array}\right]$

Q8.Find X, if  $Y=\left[\begin{array}{cc}\hfill 3\hfill & \hfill 2\hfill \\ \hfill 1\hfill & \hfill 4\hfill \end{array}\right]$ and  $2X+Y=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]$

A.8. Give, 2x + 4 = $\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]$

$\Rightarrow 2x=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-$ y  $=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -3\hfill & \hfill 2\hfill \end{array}\right]-\left[\begin{array}{ll}3\hfill & 2\hfill \\ 1\hfill & 4\hfill \end{array}\right]$  $=\left[\begin{array}{l}1-30-2\\ -3-12-4\end{array}\right]$

$\Rightarrow x=\frac{1}{2}\left[\begin{array}{ll}-2\hfill & -2\hfill \\ -4\hfill & -2\hfill \end{array}\right]=\left[\begin{array}{ll}-2\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \\ -4\times \frac{1}{2}\hfill & -2\times \frac{1}{2}\hfill \end{array}\right]=$  $\left[\begin{array}{l}-1-1\\ -2-1\end{array}\right]$

Q9.Find x and y, if  $2\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill x\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

A.9. Given, $2\left[\begin{array}{l}13\\ 0x\end{array}\right]+$  $\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]$  $=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2\times 1\hfill & \hfill 2\times 3\hfill \\ \hfill 2\times 0\hfill & \hfill 2\times x\hfill \end{array}\right]$  $+\left[\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\hfill 2+y\hfill & \hfill 6+0\hfill \\ \hfill \hfill & \hfill 2x+2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 5\hfill & \hfill 6\hfill \\ \hfill 1\hfill & \hfill 8\hfill \end{array}\right]$

Equating the corresponding elements of the matrices we get,

2 + y = 5

y = 5 - 2 = 3

And 2x + 2 = 8

2x = 8 - 2

$\Rightarrow x=\frac{6}{2}$  $$

x = 3

∅x = 3, y - 3.

Q10.Solve the equation for x, y, z and t, if  $2\left[\begin{array}{cc}\hfill x\hfill & \hfill z\hfill \\ \hfill y\hfill & \hfill t\hfill \end{array}\right]+3\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=3\left[\begin{array}{cc}\hfill 3\hfill & \hfill 5\hfill \\ \hfill 4\hfill & \hfill 6\hfill \end{array}\right]$

A.10.  $\begin{array}{l}\\ 2\left[\begin{array}{ll}x\hfill & z\hfill \\ y\hfill & t\hfill \end{array}\right]+3\left[\begin{array}{cc}\hfill 1\hfill & \hfill -1\hfill \\ \hfill 0\hfill & \hfill 2\hfill \end{array}\right]=3\left[\begin{array}{ll}3\hfill & 5\hfill \\ 4\hfill & 6\hfill \end{array}\right]\end{array}$

$\Rightarrow \left[\begin{array}{cc}\hfill 2\times x\hfill & \hfill 2\times 2\hfill \\ \hfill 2\times y\hfill & \hfill 2\times t\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 3\times 1\hfill & \hfill 3\times (-1)\hfill \\ \hfill 3\times 0\hfill & \hfill 3\times 2\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 3\times 3\hfill & \hfill 3\times 5\hfill \\ \hfill 3\times 4\hfill & \hfill 3\times 6\hfill \end{array}\right]$

$$  $\Rightarrow \left[\begin{array}{cc}\hfill 2x\hfill & \hfill 2z\hfill \\ \hfill 2y\hfill & \hfill 21\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 3\hfill & \hfill -3\hfill \\ \hfill 0\hfill & \hfill 6\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 15\hfill \\ \hfill 12\hfill & \hfill 18\hfill \end{array}\right].$

$\Rightarrow \left[\begin{array}{cc}\hfill 2x+3\hfill & \hfill 2z-3\hfill \\ \hfill 2y+0\hfill & \hfill 21+6\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill 9\hfill & \hfill 15\hfill \\ \hfill 12\hfill & \hfill 18\hfill \end{array}\right]$

Equating the corresponding elements of the matrices  $u$ get,

2x + 3 = 9

2x = 9 - 3

$\Rightarrow x=\frac{6}{2}$

x = 3

2z - 3 = 15

2z = 15 + 3

$\Rightarrow z=\frac{18}{2}$

z = 9

2y = 12

$\Rightarrow y=\frac{12}{2}$ => y = 6

2t + 6 = 18

2t = 18 - 6

2t = 12 => t = $\frac{12}{2}$ t = 6

3x + y = 5 _____ (ii)

Adding (i) and (i) we get,

2x - y + 3x + y = 10 + 5.

5x = 15

$\Rightarrow x=\frac{15}{5}$ => x = 3

5x = 15

$\Rightarrow x=\frac{15}{5}$ => x = 3

Putting x = 3 in (i) we gel,

2 × 3 - y = 10

6 - 10 = y

y = -4

Q12.Given  $3\left[\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill w\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill x\hfill & \hfill 6\hfill \\ \hfill -1\hfill & \hfill 2w\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 4\hfill & \hfill x+y\hfill \\ \hfill z+w\hfill & \hfill 3\hfill \end{array}\right]$ , find the values of x, y, z and w.

A.12. Given,  $x\left[\begin{array}{ll}x\hfill & y\hfill \\ z\hfill & w\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill x\hfill & \hfill 6\hfill \\ \hfill -1\hfill & \hfill 2w\hfill \end{array}\right]+\left[\begin{array}{cc}\hfill 4\hfill & \hfill x+y\hfill \\ \hfill z+w\hfill & \hfill 3\hfill \end{array}\right].$

$(B)\Rightarrow \left[\begin{array}{cc}\hfill 3x\hfill & \hfill 3y\hfill \\ \hfill 3z\hfill & \hfill 3w\hfill \end{array}\right]=\left[\begin{array}{cc}\hfill x+4\hfill & \hfill 6+x+y\hfill \\ \hfill -1+2+w\hfill & \hfill 2w+3\hfill \end{array}\right]$

Q13.If F  $\left(x\right)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$ , show that F(x) F(y) = F(x + y).

A.13. Given,  $f(x)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

So, F(y)  $=\left[\begin{array}{ccc}\hfill cosy\hfill & \hfill -siny\hfill & \hfill 0\hfill \\ \hfill siny\hfill & \hfill cosy\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

$\therefore f(x)f(y)=\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill cosx\hfill & \hfill -sinx\hfill & \hfill 0\hfill \\ \hfill sinx\hfill & \hfill cosx\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$

Q14.Show that

(i)  $\left[\begin{array}{cc}\hfill 5\hfill & \hfill -1\hfill \\ \hfill 6\hfill & \hfill 7\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\ne \left[\begin{array}{cc}\hfill 2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 5\hfill & \hfill -1\hfill \\ \hfill 6\hfill & \hfill 7\hfill \end{array}\right]$

(ii)  $\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\ne \left[\begin{array}{ccc}\hfill -1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill -1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 3\hfill & \hfill 4\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill 3\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 1\hfill & \hfill 0\hfill \end{array}\right]$