If M = $\begin{bmatrix$}[r] 4 & 1 \ -1 & 2 \end{bmatrix}, show that : 6M - M² = 9I; where I is a 2 × 2 unit matrix.

$M^2 = \begin{bmatrix$}[r] 4 & 1 \ -1 & 2 \end{bmatrix}\begin{bmatrix}[r] 4 & 1 \ -1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 \times 4 + 1 \times (-1) & 4 \times 1 + 1 \times 2 \ -1 \times 4 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 16 - 1 & 4 + 2 \ -4 + (-2) & -1 + 4 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 15 & 6 \ -6 & 3 \end{bmatrix}.

Substituting value of M² in L.H.S. of 6M - M² = 9I,

$\phantom{=} 6\begin{bmatrix$}[r] 4 & 1 \ -1 & 2 \end{bmatrix} - \begin{bmatrix}[r] 15 & 6 \ -6 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 24 & 6 \ -6 & 12 \end{bmatrix} - \begin{bmatrix}[r] 15 & 6 \ -6 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 24 - 15 & 6 - 6 \ -6 - (-6) & 12 - 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 & 0 \ 0 & 9 \end{bmatrix} \\[1em] = 9\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = 9\text{I}

Since, L.H.S. = R.H.S.

Hence, proved that 6M - M² = 9I.