Given matrix A = $\begin{bmatrix$}[r] 4 sin 30° & cos 0° \ cos 0° & 4 sin 30° \end{bmatrix}\text{ and } B = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix}. If AX = B,

(i) write the order of matrix X.

(ii) find the matrix 'X'.

(i) Let order of matrix X be a × b.

i.e. $\begin{bmatrix$}[r] 4 sin 30° & cos 0° \ cos 0° & 4 sin 30° \end{bmatrix}{2 \times 2} \times X{a \times b} = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix}_{2 \times 1}

Since product of matrix is possible, only when the number of columns in the first matrix is equal to no. of rows in second.

∴ a = 2.

Also the no. of columns of product (resulting matrix) is equal to no. of columns of second matrix.

∴ b = 1.

Hence, order of matrix X = 2 × 1.

(ii) Let matrix X = $\begin{bmatrix$}[r] x \ y \end{bmatrix}

Given,

$\Rightarrow AX = B \\[1em] \Rightarrow \begin{bmatrix$}[r] 4 sin 30° & cos 0° \ cos 0° & 4 sin 30° \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 \times \dfrac{1}{2} & 1 \ 1 & 4 \times \dfrac{1}{2} \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2 & 1 \ 1 & 2 \end{bmatrix}\begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 2x + y \ x + 2y \end{bmatrix} = \begin{bmatrix}[r] 4 \ 5 \end{bmatrix}

By definition of equality of matrices we get,

2x + y = 4
⇒ y = 4 - 2x …….(i)

x + 2y = 5

Substituting value of y from (i) in above equation we get,

⇒ x + 2(4 - 2x) = 5
⇒ x + 8 - 4x = 5
⇒ -3x = 5 - 8
⇒ -3x = -3
⇒ x = 1.

⇒ y = 4 - 2x = 4 - 2(1) = 2.

$\therefore X = \begin{bmatrix$}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 1 \ 2 \end{bmatrix}

Hence, X = $\begin{bmatrix$}[r] 1 \ 2 \end{bmatrix}.