Given that \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\), find the value ...

We have the equation \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\). We can solve for P by performing matrix multiplication and solving the resulting system of linear equations: \(\begin{pmatrix} 1 & -3 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} -6 \\ P \end{pmatrix} = \begin{pmatrix} -6 - 3P \\ P + 4P \end{pmatrix} = \begin{pmatrix} 3 \\ -26 \end{pmatrix}\) This gives us two equations: \(-6 - 3P = 3\) \(5P = -26\) Solving for P, we get: \(-6 - 3P = 3 \Rightarrow -3P = 9 \Rightarrow P = -3\) \(5P = -26 \Rightarrow P = \frac{-26}{5}\) Therefore, the value of P is -3. Note that the system of linear equations is consistent, meaning that there is a unique solution.