Problem Statement
Consider the matrix $\mathbf{A} = \begin{pmatrix} 5 & -3 \\ 6 & -4 \end{pmatrix}$.
**(a)** [4 marks] Find an invertible matrix $\mathbf{P}$ and a diagonal matrix $\mathbf{D}$ such that $\mathbf{A} = \mathbf{P}\mathbf{D}\mathbf{P}^{-1}$.
**(b)** [4 marks] Hence, find an exact expression for $\mathbf{A}^n$ in terms of $n$, where $n \in \mathbb{Z}^+$.
Verified Solution & Marking Scheme
Part (a) [M1]
Eigenvalues: $\det(\mathbf{A} - \lambda\mathbf{I}) = (5 - \lambda)(-4 - \lambda) - (-18) = \lambda^2 - \lambda - 20 + 18 = \lambda^2 - \lambda - 2 = (\lambda - 2)(\lambda + 1) = 0$
Part (a) [A1]
Eigenvalues: $\lambda_1 = 2, \lambda_2 = -1 \implies \mathbf{D} = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$
Part (a) [M1]
For $\lambda = 2$: $3x - 3y = 0 \implies \mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. For $\lambda = -1$: $6x - 3y = 0 \implies y = 2x \implies \mathbf{v}_2 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$
Part (a) [A1]
$\mathbf{P} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$
Part (b) [M1]
$\mathbf{P}^{-1} = \frac{1}{2 - 1}\begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$
Part (b) [M1]
$\mathbf{A}^n = \mathbf{P}\mathbf{D}^n\mathbf{P}^{-1} = \begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 2^n & 0 \\ 0 & (-1)^n \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$
Part (b) [A1]
$= \begin{pmatrix} 2^n & (-1)^n \\ 2^{n+1} & 2(-1)^n \end{pmatrix} \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$
Part (b) [A1]
$\mathbf{A}^n = \begin{pmatrix} 2^{n+1} - (-1)^n & -2^n + (-1)^n \\ 2^{n+2} - 2(-1)^n & -2^{n+1} + 2(-1)^n \end{pmatrix}$