Given matrix, X = $\begin{bmatrix$}[r] 1 & 1 \ 8 & 3 \end{bmatrix} \text{ and } I = \begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix}, prove that X² = 4X + 5I.

Given,

X² = 4X + 5I

Solving for L.H.S.,

$X^2 = \begin{bmatrix$}[r] 1 & 1 \ 8 & 3 \end{bmatrix}\begin{bmatrix}[r] 1 & 1 \ 8 & 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 \times 1 + 1 \times 8 & 1 \times 1 + 1\times 3 \ 8 \times 1 + 3 \times 8 & 8 \times 1 + 3 \times 3 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 1 + 8 & 1 + 3 \ 8 + 24 & 8 + 9 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 & 4 \ 32 & 17 \end{bmatrix}.

Solving for R.H.S.,

$4X + 5I = 4\begin{bmatrix$}[r] 1 & 1 \ 8 & 3 \end{bmatrix} + 5\begin{bmatrix}[r] 1 & 0 \ 0 & 1 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 & 4 \ 32 & 12 \end{bmatrix} + \begin{bmatrix}[r] 5 & 0 \ 0 & 5 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 4 + 5 & 4 + 0 \ 32 + 0 & 12 + 5 \end{bmatrix} \\[1em] = \begin{bmatrix}[r] 9 & 4 \ 32 & 17 \end{bmatrix}.

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

Hence, proved that X² = 4X + 5I.