CBSE Class 12 Mathematics Matrices Notes Set 01

These Class 12 CBSE Mathematics notes for Chapter 3 Matrices cover types of matrices, equality, matrix operations, transpose, and symmetric and skew-symmetric matrices with all key properties. Nine solved examples with step marks, Do Yourself practice questions and their answers are also included.

Important Notes and Formulas

Definition

A rectangular arrangement of numbers in rows and columns is called a matrix. Such a rectangular arrangement of numbers is enclosed by small ( ) or big [ ] brackets. Generally a matrix is represented by a capital letter A, B, C, … etc. and its elements are represented by small letters a, b, c, x, y etc.
Following are some examples of a matrix:
\( A = \begin{bmatrix} p & q \\ r & s \end{bmatrix} \),   \( B = \begin{bmatrix} 2 & 4 & 6 \\ 1 & 0 & 3 \end{bmatrix} \),   \( C = \begin{bmatrix} 5 \\ 1 \end{bmatrix} \),   D = [2, 3, 4],   E = [7]

Order of Matrix

A matrix which has m rows and n columns is called a matrix of order m × n, and it is represented by \( A_{m \times n} \) or \( A = [a_{ij}]_{m \times n} \).
It is obvious that a matrix of order m × n contains mn elements. Every row of such a matrix contains n elements and every column contains m elements.

Types of Matrices

(1) Row matrix
If in a matrix there is only one row, then it is called a row matrix.
Thus \( A = [a_{ij}]_{m \times n} \) is a row matrix if m = 1.
e.g. [2  4  6] is a row matrix of order 1 × 3.

(2) Column matrix
A matrix which contains only one column is called a column matrix.
Thus \( A = [a_{ij}]_{m \times n} \) is a column matrix if n = 1.
e.g. \( \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix} \) is a column matrix of order 3 × 1.

(3) Square matrix
If the number of rows and the number of columns in a matrix are equal, then it is called a square matrix.
Thus \( A = [a_{ij}]_{m \times n} \) is a square matrix if m = n.

(4) Singleton matrix
A matrix having only one element is called a singleton matrix.
Thus \( A = [a_{ij}]_{m \times n} \) is a singleton matrix if m = n = 1.
e.g. [3], [4], [d], [−7] are singleton matrices.

(5) Null or zero matrix
If in a matrix all the elements are zero, then it is called a zero matrix and it is generally denoted by O.
Thus \( A = [a_{ij}]_{m \times n} \) is a zero matrix if \( a_{ij} = 0 \) for all i and j.

(6) Diagonal matrix
If all elements except the principal diagonal in a square matrix are zero, it is called a diagonal matrix.
Thus a square matrix \( A = [a_{ij}] \) is a diagonal matrix if \( a_{ij} = 0 \), when i ≠ j.
e.g. \( \begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{bmatrix} \) is a diagonal matrix of order 3 × 3, which can also be denoted by diag [2  3  4].

(7) Scalar matrix
If all the elements of the diagonal of a diagonal matrix are equal, it is called a scalar matrix.
Thus a square matrix \( A = [a_{ij}] \) is a scalar matrix if \( a_{ij} = \begin{cases} 0, & i \neq j \\ k, & i = j \end{cases} \) where k is a constant.   e.g. \( \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)

(8) Unit matrix (Identity matrix)
If all elements of the principal diagonal in a diagonal matrix are 1, then it is called a unit matrix. A unit matrix of order n is denoted by \( I_n \).
Thus a square matrix \( A = [a_{ij}] \) is a unit matrix if \( a_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases} \)   e.g. \( I_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \)
Note: Every unit matrix is a scalar matrix.

Equality of Matrices

Two matrices A and B are said to be equal if they are of the same order and their corresponding elements are equal.
e.g. if \( A = \begin{bmatrix} 5 & 6 & 7 \\ 2 & 4 & 8 \end{bmatrix} \) and \( B = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{bmatrix} \) are equal matrices, then
\( a_1 = 5,\ a_2 = 6,\ a_3 = 7,\ b_1 = 2,\ b_2 = 4,\ b_3 = 8 \)

Addition and Subtraction of Matrices

If \( A = [a_{ij}]_{m \times n} \) and \( B = [b_{ij}]_{m \times n} \) are two matrices of the same order, then their sum A + B is a matrix whose each element is the sum of the corresponding elements,
i.e. \( A + B = [a_{ij} + b_{ij}]_{m \times n} \)
e.g. If \( A = \begin{bmatrix} 5 & 2 \\ 1 & 3 \\ 4 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 5 \\ 2 & 2 \\ 3 & 3 \end{bmatrix} \), then \( A + B = \begin{bmatrix} 5+1 & 2+5 \\ 1+2 & 3+2 \\ 4+3 & 1+3 \end{bmatrix} = \begin{bmatrix} 6 & 7 \\ 3 & 5 \\ 7 & 4 \end{bmatrix} \)
Similarly their subtraction A − B is defined as \( A - B = [a_{ij} - b_{ij}]_{m \times n} \),
i.e. in the above example \( A - B = \begin{bmatrix} 5-1 & 2-5 \\ 1-2 & 3-2 \\ 4-3 & 1-3 \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ -1 & 1 \\ 1 & -2 \end{bmatrix} \)
Note: Matrix addition and subtraction are possible only when the matrices are of the same order.

Properties of matrix addition:
If A, B and C are matrices of the same order, then
(i) A + B = B + A (Commutative law)
(ii) (A + B) + C = A + (B + C) (Associative law)
(iii) A + O = O + A = A, where O is the zero matrix (additive identity of the matrix A).
(iv) A + (−A) = O = (−A) + A, where (−A) is obtained by changing the sign of every element of A. The matrix (−A) is known as the additive inverse of the matrix A.
(v) \( \left. \begin{matrix} A + B = A + C \\ B + A = C + A \end{matrix} \right\rbrace \Rightarrow B = C \) (Cancellation law)

Scalar Multiplication of Matrices

Let \( A = [a_{ij}]_{m \times n} \) be a matrix and k be a number. Then the matrix which is obtained by multiplying every element of A by k is called the scalar multiplication of A by k and it is denoted by kA.
Thus \( A = [a_{ij}]_{m \times n} \Rightarrow kA = [ka_{ij}]_{m \times n} \)
e.g. if \( A = \begin{bmatrix} 4 & 2 \\ 3 & 5 \\ 6 & 7 \end{bmatrix} \), then \( 5A = \begin{bmatrix} 20 & 10 \\ 15 & 25 \\ 30 & 35 \end{bmatrix} \)

Properties of scalar multiplication:
If A, B are matrices of the same order and m, n are any numbers, then the following results can be easily established.
(i) m(A + B) = mA + mB    (ii) (m + n)A = mA + nA    (iii) m(nA) = (mn)A = n(mA)

Multiplication of Matrices

If A and B are any two matrices, then their product AB is defined only when the number of columns in A is equal to the number of rows in B. If \( A = [a_{ij}]_{m \times n} \) and \( B = [b_{ij}]_{n \times p} \), then their product AB = C = \( [c_{ij}] \) is a matrix of order m × p, where
\( (AB)_{ij} = C_{ij} = \sum_{r=1}^{n} a_{ir} b_{rj} \)
e.g. If \( A = \begin{bmatrix} 1 & 4 & 2 \\ 2 & 3 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 \\ 2 & 2 \\ 1 & 3 \end{bmatrix} \), then \( AB = \begin{bmatrix} 1.1+4.2+2.1 & 1.2+4.2+2.3 \\ 2.1+3.2+1.1 & 2.2+3.2+1.3 \end{bmatrix} = \begin{bmatrix} 11 & 16 \\ 9 & 13 \end{bmatrix} \)

Properties of matrix multiplication:
If A, B and C are three matrices such that their products are defined, then
(i) AB ≠ BA (Generally not commutative)
(ii) (AB)C = A(BC) (Associative law)
(iii) IA = A = AI (I is the identity matrix for matrix multiplication)
(iv) A(B + C) = AB + AC (Distributive law)
(v) If AB = AC \( \not\Rightarrow \) B = C (Cancellation law is not applicable)

Transpose of a Matrix

If we interchange the rows and columns of a matrix A, then the matrix so obtained is called the transpose of A and it is denoted by \( A^T \) or \( A^t \) or A′.
From this definition it is obvious to note that
(i) Order of A is m × n ⇒ order of \( A^T \) is n × m
(ii) \( (A^T)_{ij} = (A)_{ji} \), ∀ i, j

Properties of transpose:
If A, B are matrices of suitable orders, then
(i) \( (A^T)^T = A \)    (ii) \( (A + B)^T = A^T + B^T \)
(iii) \( (A - B)^T = A^T - B^T \)    (iv) \( (kA)^T = kA^T \), k is any scalar
(v) \( (AB)^T = B^T A^T \)    (vi) \( (A_1 A_2 \ldots A_n)^T = A_n^T \ldots A_2^T A_1^T \)

Symmetric and Skew-Symmetric Matrices

(a) Symmetric matrix: A square matrix \( A = [a_{ij}] \) is called a symmetric matrix if \( a_{ij} = a_{ji} \) for all i, j or \( A^T = A \).
e.g. \( \begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 \end{bmatrix} \)
Note:
(i) Every unit matrix and square zero matrix are symmetric matrices.
(ii) Maximum number of different elements in a symmetric matrix is \( \frac{n(n+1)}{2} \).

(b) Skew-symmetric matrix: A square matrix \( A = [a_{ij}] \) is called a skew-symmetric matrix if \( a_{ij} = -a_{ji} \) for all i, j or \( A^T = -A \).
e.g. \( \begin{bmatrix} 0 & 2 & 3 \\ -2 & 0 & 1 \\ -3 & -1 & 0 \end{bmatrix} \)
Note: All principal diagonal elements of a skew-symmetric matrix are always zero because for any diagonal element \( a_{ii} = -a_{ii} \Rightarrow a_{ii} = 0 \).

Properties of symmetric and skew-symmetric matrices:
(i) If A is a square matrix, then \( A + A^T \), \( AA^T \), \( A^TA \) are symmetric matrices while \( A - A^T \) is a skew-symmetric matrix.
(ii) Every square matrix A can be uniquely expressed as the sum of a symmetric and a skew-symmetric matrix, i.e.
\( A = \underbrace{\left[\frac{1}{2}(A + A^T)\right]}_{\text{Symmetric}} + \underbrace{\left[\frac{1}{2}(A - A^T)\right]}_{\text{Skew-symmetric}} \)

Solved Examples

Illustration 1. Find x, y, z and w if \( \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
Answer:
Given \( \begin{bmatrix} x - y & 2x + z \\ 2x - y & 3z + w \end{bmatrix} = \begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix} \).
By definition of equality, we have
x − y = −1 ...(i)    2x − y = 0 ...(ii)
2x + z = 5 ...(iii)    3z + w = 13 ...(iv)
To find x and y, subtract (ii) from (i) to get
x − 2x = −1 ⇒ x = 1 and
then from (i), y = x + 1 = 1 + 1 = 2.
To find z and w, substitute x = 1 in (iii) to obtain z = 5 − 2x = 5 − 2 = 3 and
then from (iv), w = 13 − 3z = 13 − 3 × 3 = 4.
Hence, x = 1, y = 2, z = 3 and w = 4.

Illustration 2. If \( A = \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} \) and \( B = \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \), then compute (sin θ) A + (cos θ) B. [4 Marks]
Answer:
\( (\sin\theta) A + (\cos\theta) B = \sin\theta \begin{bmatrix} \sin\theta & -\cos\theta \\ \cos\theta & \sin\theta \end{bmatrix} + \cos\theta \begin{bmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{bmatrix} \)
\( = \begin{bmatrix} \sin^2\theta & -\sin\theta\cos\theta \\ \sin\theta\cos\theta & \sin^2\theta \end{bmatrix} + \begin{bmatrix} \cos^2\theta & \cos\theta\sin\theta \\ -\cos\theta\sin\theta & \cos^2\theta \end{bmatrix} \) [2]
\( = \begin{bmatrix} \sin^2\theta + \cos^2\theta & -\sin\theta\cos\theta + \sin\theta\cos\theta \\ \sin\theta\cos\theta - \cos\theta\sin\theta & \sin^2\theta + \cos^2\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) [2]

Illustration 3. (i) Find x, y if \( 3\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} - 2\begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} x & -4 \\ 3 & y \end{bmatrix} = O \). [1 Mark]
(ii) Find the matrix X such that 2A + B + X = O, where \( A = \begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} \). [4 Marks]
Answer:
(i) Given \( 3\begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} - 2\begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} + \begin{bmatrix} x & -4 \\ 3 & y \end{bmatrix} = O \)
⇒ \( \begin{bmatrix} 12 & 6 \\ 3 & 9 \end{bmatrix} - \begin{bmatrix} -4 & 2 \\ 6 & 4 \end{bmatrix} + \begin{bmatrix} x & -4 \\ 3 & y \end{bmatrix} = O \)
⇒ \( \begin{bmatrix} 12 - (-4) + x & 6 - 2 - 4 \\ 3 - 6 + 3 & 9 - 4 + y \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
⇒ 16 + x = 0 and 5 + y = 0 (By definition of equality)
∴ x = −16 and y = −5.
(ii) Given 2A + B + X = O
⇒ \( 2\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} + X = O \Rightarrow \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} + \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} + X = O \)
⇒ \( \begin{bmatrix} -2 + 3 & 4 - 2 \\ 6 + 1 & 8 + 5 \end{bmatrix} + X = O \Rightarrow \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} + X = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) [1]
⇒ \( \begin{bmatrix} 1 & 2 \\ 7 & 13 \end{bmatrix} + \begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \), where \( X = \begin{bmatrix} x & y \\ z & t \end{bmatrix} \) [1]
⇒ \( \begin{bmatrix} 1 + x & 2 + y \\ 7 + z & 13 + t \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
⇒ 1 + x = 0, 2 + y = 0, 7 + z = 0 and 13 + t = 0 [1]
⇒ x = −1, y = −2, z = −7 and t = −13.
Hence \( X = \begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix} \). [1]

Illustration 4. Find A and B, if \( 2A + 3B = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} \) and \( A - 2B = \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \). [4 Marks]
Answer:
Given \( 2A + 3B = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} \) ...(1)
and \( A - 2B = \begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \) ...(2)
Multiplying equation (2) by 2 and subtracting it from equation (1), we obtain
\( 2A + 3B - 2(A - 2B) = \begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} - 2\begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \)
⇒ \( 7B = \begin{bmatrix} 1 - 6 & -2 - 0 & 3 - 2 \\ 2 + 2 & 0 - 12 & -1 - 4 \end{bmatrix} \Rightarrow B = \frac{1}{7}\begin{bmatrix} -5 & -2 & 1 \\ 4 & -12 & -5 \end{bmatrix} \) [2]
Multiplying equation (1) by 2, equation (2) by 3 and adding, we get
\( 2(2A + 3B) + 3(A - 2B) = 2\begin{bmatrix} 1 & -2 & 3 \\ 2 & 0 & -1 \end{bmatrix} + 3\begin{bmatrix} 3 & 0 & 1 \\ -1 & 6 & 2 \end{bmatrix} \)
⇒ \( 7A = \begin{bmatrix} 2 + 9 & -4 + 0 & 6 + 3 \\ 4 - 3 & 0 + 18 & -2 + 6 \end{bmatrix} \Rightarrow A = \frac{1}{7}\begin{bmatrix} 11 & -4 & 9 \\ 1 & 18 & 4 \end{bmatrix} \) [2]

Illustration 5. If \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \), find x and y such that \( (xI + yA)^2 = A \). [4 Marks]
Answer:
We have, \( A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \)
∴ \( xI + yA = x\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + y\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \)
⇒ \( xI + yA = \begin{bmatrix} x & 0 \\ 0 & x \end{bmatrix} + \begin{bmatrix} 0 & y \\ -y & 0 \end{bmatrix} \)
⇒ \( xI + yA = \begin{bmatrix} x & y \\ -y & x \end{bmatrix} \) [1]
Now, \( (xI + yA)^2 = A \)
⇒ \( \begin{bmatrix} x & y \\ -y & x \end{bmatrix}\begin{bmatrix} x & y \\ -y & x \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \)
⇒ \( \begin{bmatrix} x^2 - y^2 & 2xy \\ -2xy & x^2 - y^2 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \) [1]
⇒ \( x^2 - y^2 = 0 \), 2xy = 1, −2xy = −1 and \( x^2 - y^2 = 0 \)
⇒ \( x^2 - y^2 = 0 \) and 2xy = 1
⇒ x = ±y and 2xy = 1
Now two cases arise.
Case I: When x = y and 2xy = 1
In this case, we have
x = y and 2xy = 1 ⇒ \( 2x^2 = 1 \Rightarrow x = \pm\frac{1}{\sqrt{2}} \)
∴ \( \left(x = \frac{1}{\sqrt{2}} \text{ and } y = \frac{1}{\sqrt{2}}\right) \) or \( \left(x = -\frac{1}{\sqrt{2}} \text{ and } y = -\frac{1}{\sqrt{2}}\right) \) [1]
Case II: When x = −y and 2xy = 1
In this case, we have
x = −y and 2xy = 1 ⇒ \( -2x^2 = 1 \Rightarrow x = \pm\frac{i}{\sqrt{2}} \), where i = iota = \( \sqrt{-1} \)
∴ \( \left(x = \frac{i}{\sqrt{2}} \text{ and } y = -\frac{i}{\sqrt{2}}\right) \) or \( \left(x = -\frac{i}{\sqrt{2}} \text{ and } y = \frac{i}{\sqrt{2}}\right) \) [1]

Illustration 6. Let \( A = \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} \) and I be the identity matrix of order 2. Show that \( I + A = (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \). [4 Marks]
Answer:
We have,
LHS: \( I + A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} = \begin{bmatrix} 1 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 1 \end{bmatrix} \) [½]
and \( I - A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 0 \end{bmatrix} = \begin{bmatrix} 1 & \tan\frac{\alpha}{2} \\ -\tan\frac{\alpha}{2} & 1 \end{bmatrix} \) [½]
RHS: \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & \tan\frac{\alpha}{2} \\ -\tan\frac{\alpha}{2} & 1 \end{bmatrix}\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} \)
⇒ \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & \tan\frac{\alpha}{2} \\ -\tan\frac{\alpha}{2} & 1 \end{bmatrix}\begin{bmatrix} \dfrac{1 - \tan^2\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}} & -\dfrac{2\tan\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}} \\ \dfrac{2\tan\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}} & \dfrac{1 - \tan^2\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}} \end{bmatrix} \) [1]
⇒ \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & t \\ -t & 1 \end{bmatrix}\begin{bmatrix} \dfrac{1 - t^2}{1 + t^2} & -\dfrac{2t}{1 + t^2} \\ \dfrac{2t}{1 + t^2} & \dfrac{1 - t^2}{1 + t^2} \end{bmatrix} \), where \( t = \tan\frac{\alpha}{2} \).
⇒ \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \dfrac{1 - t^2 + 2t^2}{1 + t^2} & \dfrac{-2t + t - t^3}{1 + t^2} \\ \dfrac{-t + t^3 + 2t}{1 + t^2} & \dfrac{2t^2 + 1 - t^2}{1 + t^2} \end{bmatrix} \) [1]
⇒ \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} \dfrac{1 + t^2}{1 + t^2} & \dfrac{-t(1 + t^2)}{1 + t^2} \\ \dfrac{t(1 + t^2)}{1 + t^2} & \dfrac{1 + t^2}{1 + t^2} \end{bmatrix} = \begin{bmatrix} 1 & -t \\ t & 1 \end{bmatrix} \)
⇒ \( (I - A)\begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix} = \begin{bmatrix} 1 & -\tan\frac{\alpha}{2} \\ \tan\frac{\alpha}{2} & 1 \end{bmatrix} = I + A \) [1]

Illustration 7. Let \( A = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} \) and \( f(x) = x^2 - 4x + 7 \). Show that f(A) = O. Use this result to find \( A^5 \). [4 Marks]
Answer:
We have, \( f(x) = x^2 - 4x + 7 \)
∴ \( f(A) = A^2 - 4A + 7I_2 \)
Now, \( A^2 = \begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix}\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 4 - 3 & 6 + 6 \\ -2 - 2 & -3 + 4 \end{bmatrix} = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} \) [½]
\( -4A = \begin{bmatrix} -8 & -12 \\ 4 & -8 \end{bmatrix} \) and \( 7I_2 = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \) [½]
∴ \( f(A) = A^2 - 4A + 7I_2 \)
⇒ \( f(A) = \begin{bmatrix} 1 & 12 \\ -4 & 1 \end{bmatrix} + \begin{bmatrix} -8 & -12 \\ 4 & -8 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)
⇒ \( f(A) = \begin{bmatrix} 1 - 8 + 7 & 12 - 12 + 0 \\ -4 + 4 + 0 & 1 - 8 + 7 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O \) [1]
Now, f(A) = O
⇒ \( A^2 - 4A + 7I_2 = O \)
⇒ \( A^2 = 4A - 7I_2 \)
⇒ \( A^3 = A^2 \cdot A = (4A - 7I_2)A = 4A^2 - 7I_2A \)
⇒ \( A^3 = 4(4A - 7I_2) - 7A \)
⇒ \( A^3 = 9A - 28I_2 \) [Using: \( A^2 = 4A - 7I_2 \)] [½]
⇒ \( A^4 = A^3A = (9A - 28I_2)A \)
⇒ \( A^4 = 9A^2 - 28A = 9(4A - 7I_2) - 28A \) [Using: \( A^2 = 4A - 7I_2 \)]
⇒ \( A^4 = 36A - 63I_2 - 28A = 8A - 63I_2 \)
⇒ \( A^5 = A^4A = (8A - 63I_2)A = 8A^2 - 63I_2A \) [½]
⇒ \( A^5 = 8(4A - 7I_2) - 63A = -31A - 56I_2 \) [Using: \( A^2 = 4A - 7I_2 \)]
⇒ \( A^5 = -31\begin{bmatrix} 2 & 3 \\ -1 & 2 \end{bmatrix} - 56\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
⇒ \( A^5 = \begin{bmatrix} -62 & -93 \\ 31 & -62 \end{bmatrix} + \begin{bmatrix} -56 & 0 \\ 0 & -56 \end{bmatrix} \) [½]
⇒ \( A^5 = \begin{bmatrix} -118 & -93 \\ 31 & -118 \end{bmatrix} \) [½]

Illustration 8. If \( A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \\ 5 & 6 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & -3 & 4 \\ 1 & 2 & 0 \end{bmatrix} \), prove that (AB)′ = B′A′. [4 Marks]
Answer:
Here, \( AB = \begin{bmatrix} 1 & -2 \\ 3 & 0 \\ 5 & 6 \end{bmatrix}\begin{bmatrix} 0 & -3 & 4 \\ 1 & 2 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 1 \times 0 + (-2) \times 1 & 1 \times (-3) + (-2) \times 2 & 1 \times 4 + (-2) \times 0 \\ 3 \times 0 + 0 \times 1 & 3 \times (-3) + 0 \times 2 & 3 \times 4 + 0 \times 0 \\ 5 \times 0 + 6 \times 1 & 5 \times (-3) + 6 \times 2 & 5 \times 4 + 6 \times 0 \end{bmatrix} = \begin{bmatrix} -2 & -7 & 4 \\ 0 & -9 & 12 \\ 6 & -3 & 20 \end{bmatrix} \)
⇒ \( (AB)' = \begin{bmatrix} -2 & 0 & 6 \\ -7 & -9 & -3 \\ 4 & 12 & 20 \end{bmatrix} \) ...(1) [2]
Now, \( B'A' = \begin{bmatrix} 0 & 1 \\ -3 & 2 \\ 4 & 0 \end{bmatrix}\begin{bmatrix} 1 & 3 & 5 \\ -2 & 0 & 6 \end{bmatrix} \)
\( = \begin{bmatrix} 0 \times 1 + 1 \times (-2) & 0 \times 3 + 1 \times 0 & 0 \times 5 + 1 \times 6 \\ (-3) \times 1 + 2 \times (-2) & -3 \times 3 + 2 \times 0 & (-3) \times 5 + 2 \times 6 \\ 4 \times 1 + 0 \times (-2) & 4 \times 3 + 0 \times 0 & 4 \times 5 + 0 \times 6 \end{bmatrix} \)
\( = \begin{bmatrix} -2 & 0 & 6 \\ -7 & -9 & -3 \\ 4 & 12 & 20 \end{bmatrix} \) ...(2) [2]
From (1) and (2), we have (AB)′ = B′A′.

Illustration 9. Express \( \begin{bmatrix} 2 & 4 & -1 \\ 3 & 5 & 8 \\ 1 & -2 & 1 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix. [4 Marks]
Answer:
Let \( A = \begin{bmatrix} 2 & 4 & -1 \\ 3 & 5 & 8 \\ 1 & -2 & 1 \end{bmatrix} \Rightarrow A^T = \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & -2 \\ -1 & 8 & 1 \end{bmatrix} \)
∴ \( A + A^T = \begin{bmatrix} 2 & 4 & -1 \\ 3 & 5 & 8 \\ 1 & -2 & 1 \end{bmatrix} + \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & -2 \\ -1 & 8 & 1 \end{bmatrix} = \begin{bmatrix} 4 & 7 & 0 \\ 7 & 10 & 6 \\ 0 & 6 & 2 \end{bmatrix} \)
⇒ \( \frac{1}{2}(A + A^T) = \frac{1}{2}\begin{bmatrix} 4 & 7 & 0 \\ 7 & 10 & 6 \\ 0 & 6 & 2 \end{bmatrix} = \begin{bmatrix} 2 & 7/2 & 0 \\ 7/2 & 5 & 3 \\ 0 & 3 & 1 \end{bmatrix} \) [1½]
Now, \( A - A^T = \begin{bmatrix} 2 & 4 & -1 \\ 3 & 5 & 8 \\ 1 & -2 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 3 & 1 \\ 4 & 5 & -2 \\ -1 & 8 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 10 \\ 2 & -10 & 0 \end{bmatrix} \)
⇒ \( \frac{1}{2}(A - A^T) = \frac{1}{2}\begin{bmatrix} 0 & 1 & -2 \\ -1 & 0 & 10 \\ 2 & -10 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1/2 & -1 \\ -1/2 & 0 & 5 \\ 1 & -5 & 0 \end{bmatrix} \), which is clearly skew-symmetric. [1½]
Now \( A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T) = \begin{bmatrix} 2 & 7/2 & 0 \\ 7/2 & 5 & 3 \\ 0 & 3 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1/2 & -1 \\ -1/2 & 0 & 5 \\ 1 & -5 & 0 \end{bmatrix} \). [1]

Do Yourself

Do Yourself 1
(i) Find x, y, a and b if \( \begin{bmatrix} 2x + 3y & a + b & 8 \\ 1 & 4x + y & 3a - 4b \end{bmatrix} = \begin{bmatrix} 7 & 1 & 8 \\ 1 & 9 & 10 \end{bmatrix} \).
(ii) If \( A - B = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \) and \( A + B = \begin{bmatrix} 3 & 5 & 7 \\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix} \), find A and B.

Do Yourself 2
(i) If \( A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \), show that \( (A - A^T) \) is a skew-symmetric matrix, where \( A^T \) is the transpose of matrix A.
(ii) Express the matrix \( \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix} \) as the sum of a symmetric and a skew-symmetric matrix.

Answers for Do Yourself

Do Yourself 1
(i) x = 2, y = 1, a = 2 and b = −1
(ii) \( A = \begin{bmatrix} 2 & 3 & 4 \\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 & 3 \\ -1 & 0 & 2 \\ 5 & 4 & 0 \end{bmatrix} \)

Do Yourself 2
(ii) \( \begin{bmatrix} 3 & -3/2 \\ -3/2 & -1 \end{bmatrix} + \begin{bmatrix} 0 & -5/2 \\ 5/2 & 0 \end{bmatrix} \)