If X = $\begin{bmatrix$}[r] 4 & 1 \ -1 & 2 \end{bmatrix}, show that 6X - X² = 9I where I is the unit matrix.

We have to prove 6X - X² = 9I,

$\text{L.H.S. } = 6X - X^2 \\[1em] 6X - X^2 = 6\begin{bmatrix$}[r] 4 & 1 \ -1 & 2 \end{bmatrix} - \begin{bmatrix}[r] 4 & 1 \ -1 & 2 \end{bmatrix}\begin{bmatrix}[r] 4 & 1 \ -1 & 2 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 24 & 6 \ -6 & 12 \end{bmatrix} - \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] 24 & 6 \ -6 & 12 \end{bmatrix} - \begin{bmatrix}[r] 16 - 1 & 4 + 2 \ -4 - 2 & -1 + 4 \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}

$\text{R.H.S. } = 9I \\[1em] 9I = 9\begin{bmatrix$}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 & 0 \ 0 & 9 \end{bmatrix}.

Since, L.H.S. = $\begin{bmatrix$}[r] 9 & 0 \ 0 & 9 \end{bmatrix} = R.H.S. Hence, proved that 6X - X² = 9I.