+68 Given a square matrix $\mathbf{A}$, the characteristic polynomial of $\mathbf{A}$ is defined as \[P_{\mathbf A}(t) = \det (t \mathbf{I} - \mathbf{A}),\]
where $\mathbf{I}$ is the identity matrix. For example, if $\mathbf{A} = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix},$ then the characteristic polynomial of$\mathbf{A}$ is \[P_{\mathbf A}(t) = \det (t \mathbf{I} - \mathbf{A}) = \det \left(t \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 1 & 2 \\3 & 4 \end{pmatrix} \right) ,\] and rewriting the difference as a single matrix, this becomes \[\det \begin{pmatrix} t - 1 & -2 \\ -3 & t - 4 \end{pmatrix} = (t-1)(t-4) - (-2)(-3) = t^2 - 5t -2.\]
(a) Compute the characteristic polynomial of the matrix \[\mathbf{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}.\]
Express your answer in the form $p_2 t^2 + p_1 t + p_0$.
(b) For the polynomial in part (a), find \[p_2 \mathbf{A}^2 + p_1 \mathbf{A} + p_0 \mathbf{I}.\]
