Find matrix B, if matrix A = $\begin{bmatrix$}[r] 1 & 5 \ 1 & 2 \end{bmatrix}, matrix C = $\begin{bmatrix$}[r] 2 \ 1 \end{bmatrix} and AB = 3C.

Let order of matrix B = m × n

Given,

AB = 3C

$\begin{bmatrix$}[r] 1 & 5 \ 1 & 2 \end{bmatrix}{2 \times 2}B{m \times n} = 3\begin{bmatrix}[r] 2 \ 1 \end{bmatrix}_{2 \times 1}

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.

∴ m = 2

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

∴ n = 1

Order of matrix B = m × n = 2 × 1.

Let matrix B = $\begin{bmatrix$}[r] a \ b \end{bmatrix}

Substituting matrix in AB = 3C we get,

$\Rightarrow \begin{bmatrix$}[r] 1 & 5 \ 1 & 2 \end{bmatrix}\begin{bmatrix}[r] a \ b \end{bmatrix} = 3\begin{bmatrix}[r] 2 \ 1 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 1 \times a + 5 \times b \ 1 \times a + 2 \times b \end{bmatrix} = \begin{bmatrix}[r] 6 \ 3 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] a + 5b \ a + 2b \end{bmatrix} = \begin{bmatrix}[r] 6 \ 3 \end{bmatrix}.

∴ a + 5b = 6 ……..(1)

a + 2b = 3 ……..(2)

Subtracting equation (2) from (1), we get :

⇒ a + 5b - (a + 2b) = 6 - 3

⇒ a - a + 5b - 2b = 3

⇒ 3b = 3

⇒ b = 1.

Substituting value of b in (1), we get :

⇒ a + 5(1) = 6

⇒ a + 5 = 6

⇒ a = 1.

B = $\begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 1 \ 1 \end{bmatrix}

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