If A = $\begin{bmatrix$}[r] 0 & -1 \ 4 & -3 \end{bmatrix}, B = \begin{bmatrix}[r] -5 \ 6 \end{bmatrix} and 3A × M = 2B; find matrix M.

Let order of matrix M be a × b.

Given,

$\Rightarrow 3A \times M = 2B …….(i) \\[1em] \Rightarrow 3\begin{bmatrix$}[r] 0 & -1 \ 4 & -3 \end{bmatrix}{2 \times 2} \times M{a \times b} = 2\begin{bmatrix}[r] -5 \ 6 \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 M = 2 × 1.

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

Substituting value of A, M and B in (i) we get,

$\Rightarrow 3\begin{bmatrix$}[r] 0 & -1 \ 4 & -3 \end{bmatrix} \times \begin{bmatrix}[r] a \ b \end{bmatrix} = 2\begin{bmatrix}[r] -5 \ 6 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 & -3 \ 12 & -9 \end{bmatrix} \times \begin{bmatrix}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] -10 \ 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 0 \times a + (-3) \times b \ 12 \times a + (-9) \times b \end{bmatrix} = \begin{bmatrix}[r] -10 \ 12 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] -3b \ 12a - 9b \end{bmatrix} = \begin{bmatrix}[r] -10 \ 12 \end{bmatrix}

By definition of equality of matrices we get,

-3b = -10
⇒ b = $\dfrac{10}{3}$ ….(i)

12a - 9b = 12

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

$\Rightarrow 12a - 9 \times \dfrac{10}{3} = 12 \\[1em] \Rightarrow 12a - 30 = 12 \\[1em] \Rightarrow 12a = 42 \\[1em] \Rightarrow a = \dfrac{7}{2}.$

$\therefore \begin{bmatrix$}[r] a \ b \end{bmatrix} = \begin{bmatrix}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix}.

Hence, M = $\begin{bmatrix$}[r] \dfrac{7}{2} \\ \dfrac{10}{3} \end{bmatrix}.