In each case given below, find :

(a) the order of matrix M.

(b) the matrix M.

(i) $M \times \begin{bmatrix$}[r] 1 & 1 \ 0 & 2 \end{bmatrix} =\begin{bmatrix}[r] 1 & 2 \end{bmatrix}

(ii) $\begin{bmatrix$}[r] 1 & 4 \ 2 & 1 \end{bmatrix} \times M = \begin{bmatrix}[r] 13 \ 5 \end{bmatrix}

(i) Let order of matrix M be a × b.

$M${a \times b} \times \begin{bmatrix}[r] 1 & 1 \ 0 & 2 \end{bmatrix}{2 \times 2} = \begin{bmatrix}[r] 1 & 2 \end{bmatrix}_{1 \times 2}

Since, the product of matrices is possible, only when the number of columns in the first matrix is equal to the number of rows in the second.

∴ b = 2

Also, the no. of rows of product (resulting) matrix is equal to no. of rows of first matrix.

∴ a = 1

Order of matrix M = a × b = 1 × 2.

Let M = $\begin{bmatrix$}[r] x & y \end{bmatrix}.

$\Rightarrow \begin{bmatrix$}[r] x & y \end{bmatrix} \times \begin{bmatrix}[r] 1 & 1 \ 0 & 2 \end{bmatrix} =\begin{bmatrix}[r] 1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x \times 1 + y \times 0 & x \times 1 + y \times 2 \end{bmatrix} = \begin{bmatrix}[r] 1 & 2 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x & x + 2y \end{bmatrix} = \begin{bmatrix}[r] 1 & 2 \end{bmatrix}

By definition of equality of matrices we get,

x = 1

x + 2y = 2
⇒ 1 + 2y = 2
⇒ 2y = 1
⇒ y = $\dfrac{1}{2}$.

∴ M = $\begin{bmatrix$}[r] x & y \end{bmatrix} = \begin{bmatrix}[r] 1 & \dfrac{1}{2} \end{bmatrix}.

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

(ii) Let order of matrix M be a × b.

i.e. $\begin{bmatrix$}[r] 1 & 4 \ 2 & 1 \end{bmatrix}{2 \times 2} \times M{a \times b} = \begin{bmatrix}[r] 13 \ 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 = 2 × 1.

Let M = $\begin{bmatrix$}[r] x \ y \end{bmatrix}

$\Rightarrow \begin{bmatrix$}[r] 1 & 4 \ 2 & 1 \end{bmatrix} \times \begin{bmatrix}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 13 \ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 1 \times x + 4 \times y \ 2 \times x + 1 \times y \end{bmatrix} = \begin{bmatrix}[r] 13 \ 5 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x + 4y \ 2x + y \end{bmatrix} = \begin{bmatrix}[r] 13 \ 5 \end{bmatrix}

By definition of equality of matrices we get,

x + 4y = 13
⇒ x = 13 - 4y ……(i)

2x + y = 5
⇒ 2(13 - 4y) + y = 5
⇒ 26 - 8y + y = 5
⇒ -7y = -21
⇒ y = 3.

⇒ x = 13 - 4y = 13 - 4(3) = 13 - 12 = 1.

$\therefore M = \begin{bmatrix$}[r] x \ y \end{bmatrix} = \begin{bmatrix}[r] 1 \ 3 \end{bmatrix}.

Hence, $M = \begin{bmatrix$}[r] 1 \ 3 \end{bmatrix}.