Let matrix A = $\begin{bmatrix$}[r] 4 & -1 \ 2 & 0 \end{bmatrix} \text{ and matrix B} = 2 \times \begin{bmatrix}[r] 3 \ 2 \end{bmatrix}, find matrix M so that AM = B.

Given,

Matrix A = $\begin{bmatrix$}[r] 4 & -1 \ 2 & 0 \end{bmatrix} \text{ and Matrix B} = 2 \times \begin{bmatrix}[r] 3 \ 2 \end{bmatrix} = \begin{bmatrix}[r] 6 \ 4 \end{bmatrix}

Order of matrix A = 2 × 2

Order of matrix B = 2 × 1

We know that,

Two matrix can be multiplied if no. of columns of 1st matrix and no. of rows in 2nd matrix are equal.

Also, the resultant matrix has no. of rows equal to no. of rows of 1st matrix and no. of columns equal to no. of columns of 2nd matrix.

Given,

⇒ AM = B

∴ Order of matrix M = 2 × 1.

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

$\Rightarrow AM = B \\[1em] \Rightarrow \begin{bmatrix$}[r] 4 & -1 \ 2 & 0 \end{bmatrix}\begin{bmatrix}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 6 \ 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4 \times a + (-1) \times b \ 2 \times a + 0 \times b \end{bmatrix} = \begin{bmatrix}[r] 6 \ 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 4a - b \ 2a \end{bmatrix} = \begin{bmatrix}[r] 6 \ 4 \end{bmatrix} \\[1em] \Rightarrow 2a = 4 \text{ and } 4a - b = 6\\[1em] \Rightarrow a = \dfrac{4}{2} = 2.

Substituting value of a in equation 4a - b = 6, we get :

⇒ 4(2) - b = 6

⇒ 8 - b = 6

⇒ b = 8 - 6 = 2.

M = $\begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] 2 \ 2 \end{bmatrix}.

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