Problem Statement
If $A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$, show that $A^2 - 4A + I = O$. Hence, find $A^{-1}$.
Verified Solution & Marking Scheme
Compute A²
$$A^2 = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 4 + 3 & 6 + 6 \\ 2 + 2 & 3 + 4 \end{pmatrix} = \begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix}$$
Evaluate A² - 4A + I
$$\begin{pmatrix} 7 & 12 \\ 4 & 7 \end{pmatrix} - 4\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 7 - 8 + 1 & 12 - 12 + 0 \\ 4 - 4 + 0 & 7 - 8 + 1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = O$$
Deduce A⁻¹ by Multiplying by A⁻¹
$$A^2 - 4A + I = O \implies A^{-1}(A^2 - 4A + I) = A^{-1}O$$
$$A - 4I + A^{-1} = O \implies A^{-1} = 4I - A$$
$$A^{-1} = 4\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 4 - 2 & 0 - 3 \\ 0 - 1 & 4 - 2 \end{pmatrix} = \begin{pmatrix} 2 & -3 \\ -1 & 2 \end{pmatrix}$$