CAIE A-Level Further Math A2 2.2.5 Eigenvalues and Eigenvectors Questions

$\lambda=a:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} 0 -4-a 0 3 -1-a\end{array}\right|=\left(\begin{array}{c}(-4-a)(-1-a) 0 0\end{array}\right) \sim\left(\begin{array}{l}1 0 0\end{array}\right)$
Uses vector product (or equations) to find
corresponding eigenvectors.

$\lambda=-1: \quad\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} a+1 2 a+5 a+1 0 -3 0\end{array}\right|=\left(\begin{array}{c}3(a+1) 0 -3(a+1)\end{array}\right) \sim\left(\begin{array}{c}1 0 -1\end{array}\right)$$\lambda=-4:\left|\begin{array}{ccc}\mathbf{i} \mathbf{j} \mathbf{k} a+4 2 a+5 a+1 0 3 3\end{array}\right|=\left(\begin{array}{c}3 a+12 -(3 a+12) 3 a+12\end{array}\right) \sim\left(\begin{array}{c}1 -1 1\end{array}\right)$$\mathbf{P}=\left(\begin{array}{ccc}1 1 0 0 -1 0 -1 1\end{array}\right)$
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