If for two matrices M and N, N = $\begin{bmatrix$}[r] 3 & 2 \ 2 & -1 \end{bmatrix} and product M × N = [-1 4]; find matrix M.

Let order of matrix M be a × b.

$M${a \times b} \times \begin{bmatrix}[r] 3 & 2 \ 2 & -1 \end{bmatrix}{2 \times 2} = \begin{bmatrix}[r] -1 & 4 \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] 3 & 2 \ 2 & -1 \end{bmatrix} =\begin{bmatrix}[r] -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] x \times 3 + y \times 2 & x \times 2 + y \times -1 \end{bmatrix} = \begin{bmatrix}[r] -1 & 4 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 3x + 2y & 2x - y \end{bmatrix} = \begin{bmatrix}[r] -1 & 4 \end{bmatrix}

By definition of equality of matrices we get,

⇒ 3x + 2y = -1 and 2x - y = 4

From 2x - y = 4

⇒ y = 2x - 4

Substituting value of y in 3x + 2y = -1

⇒ 3x + 2(2x - 4) = -1

⇒ 3x + 4x - 8 = -1

⇒ 7x = -1 + 8

⇒ 7x = 7

⇒ x = 1.

⇒ y = 2x - 4 = 2(1) - 4 = 2 - 4 = -2.

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

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