Given matrix, X = $$\begin{bmatrix*}[r] 1 & 1 \\ 8 & 3 \end{bmatrix*}$$ and I = $$\begin{bmatrix*}[r] 1 & 0 \\ 0 & 1 \end{bmatrix*}$$,
Prove that X2 = 4X + 5I.
Show SolutionHide Solution↓
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.