Given \(P = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix}, \quad Q = \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix}\) we need to find \(P + 2Q\). First, calculate \(2Q\): \(2Q = 2 \begin{pmatrix} 3 & -7 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 \cdot 3 & 2 \cdot -7 \\ 2 \cdot 1 & 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix}\) Now add \(P\) and \(2Q\): \(P + 2Q = \begin{pmatrix} 1 & 3 \\ 2 & -5 \end{pmatrix} + \begin{pmatrix} 6 & -14 \\ 2 & 4 \end{pmatrix} = \begin{pmatrix} 1 + 6 & 3 + (-14) \\ 2 + 2 & -5 + 4 \end{pmatrix}\) Calculating each entry, we find: \(= \begin{pmatrix} 7 & -11 \\ 4 & -1 \end{pmatrix}\) Thus, the result of \(P + 2Q\) is \(\begin{pmatrix} 7 & -11 \\ 4 & -1 \end{pmatrix}\)