Note that
\[\mathbf{A}^2 = \begin{pmatrix} 2 & 7 \\ -1 & -3 \end{pmatrix}\]and
\[\mathbf{A}^3 = \mathbf{A} \mathbf{A}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -\mathbf{I}.\]Then
\begin{align*}
\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40} &= (\mathbf{A}^3)^9 + (\mathbf{A}^3)^{10} \mathbf{A} + (\mathbf{A}^3)^{13} \mathbf{A} \\
&= (-\mathbf{I})^9 + (-\mathbf{I})^{10} \mathbf{A} + (-\mathbf{I})^{13} \mathbf{A} \\
&= -\mathbf{I} + \mathbf{A} - \mathbf{A} \\
&= -\mathbf{I} = \boxed{\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}}.
\end{align*}
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Given
\[\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},\]compute $\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}.$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
Given
\[\mathbf{A} = \begin{pmatrix} 3 & 7 \\ -1 & -2 \end{pmatrix},\]compute $\mathbf{A}^{27} + \mathbf{A}^{31} + \mathbf{A}^{40}.$
test