Problem Statement
Given the matrix $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$ and $I$ is the $2 \times 2$ identity matrix.
(i) Find $A^2$.
(ii) Hence, find the scalar matrix $M$ such that $A^2 - 6A + 11I = M$.
Verified Solution & Marking Scheme
Compute A²
$A^2 = A \cdot A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$
$A^2 = \begin{pmatrix} 2(2) + (-1)(3) & 2(-1) + (-1)(4) \\ 3(2) + 4(3) & 3(-1) + 4(4) \end{pmatrix} = \begin{pmatrix} 4 - 3 & -2 - 4 \\ 6 + 12 & -3 + 16 \end{pmatrix} = \begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix}$
Compute -6A and 11I
$-6A = -6 \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} = \begin{pmatrix} -12 & 6 \\ -18 & -24 \end{pmatrix}$
$11I = 11 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$
Evaluate A² - 6A + 11I
$M = \begin{pmatrix} 1 & -6 \\ 18 & 13 \end{pmatrix} + \begin{pmatrix} -12 & 6 \\ -18 & -24 \end{pmatrix} + \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$
$M = \begin{pmatrix} 1 - 12 + 11 & -6 + 6 + 0 \\ 18 - 18 + 0 & 13 - 24 + 11 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = \mathbf{O} \; (\text{Null Matrix})$