(6) F \( M=\left|\begin{array}{cc}-1 & -5 \\ 3 & 2\end{array}\right| \) and \( m=\left(\left.\begin{array}{ll}0 & 4 \\ 6 & 3\end{array} \right\rvert\,\right. \) find \( \left|\begin{array}{ll}3 m-A\end{array}\right| \)

To find \( |3m - A| \), we will first compute \( 3m \) and then subtract matrix \( A \), where \( A = \left|\begin{array}{cc}-1 & -5 \\ 3 & 2\end{array}\right| \). First, let's calculate \( 3m \): \[ m = \left(\begin{array}{cc} 0 & 4 \\ 6 & 3 \end{array}\right) \implies 3m = 3 \cdot \left(\begin{array}{cc} 0 & 4 \\ 6 & 3 \end{array}\right) = \left(\begin{array}{cc} 0 & 12 \\ 18 & 9 \end{array}\right). \] Now, we can write the matrix \( A \): \[ A = \left(\begin{array}{cc} -1 & -5 \\ 3 & 2 \end{array}\right). \] Next, we need to calculate \( 3m - A \): \[ 3m - A = \left(\begin{array}{cc} 0 & 12 \\ 18 & 9 \end{array}\right) - \left(\begin{array}{cc} -1 & -5 \\ 3 & 2 \end{array}\right) = \left(\begin{array}{cc} 0 - (-1) & 12 - (-5) \\ 18 - 3 & 9 - 2 \end{array}\right) = \left(\begin{array}{cc} 1 & 17 \\ 15 & 7 \end{array}\right). \] Finally, we need to find the determinant of \( 3m - A \): \[ |3m - A| = \left| \begin{array}{cc} 1 & 17 \\ 15 & 7 \end{array} \right| = (1 \cdot 7) - (17 \cdot 15) = 7 - 255 = -248. \] Thus, the final answer is: \[ \left| 3m - A \right| = -248. \]