Let \( A=\left[\begin{array}{cc}8 & 2 \\ 7 & -2\end{array}\right] \) and \( B=\left[\begin{array}{rr}-8 & 7 \\ -8 & 5\end{array}\right] \) Find \( -3 A+3 B \) a \( -3 A+3 B=\square \)

To find \( -3 A + 3 B \), first we need to compute \( -3 A \) and \( 3 B \). Starting with \( -3 A \): \[ -3 A = -3 \left[\begin{array}{cc}8 & 2 \\ 7 & -2\end{array}\right] = \left[\begin{array}{cc}-24 & -6 \\ -21 & 6\end{array}\right] \] Next, calculate \( 3 B \): \[ 3 B = 3 \left[\begin{array}{rr}-8 & 7 \\ -8 & 5\end{array}\right] = \left[\begin{array}{rr}-24 & 21 \\ -24 & 15\end{array}\right] \] Now, add the two results: \[ -3 A + 3 B = \left[\begin{array}{cc}-24 & -6 \\ -21 & 6\end{array}\right] + \left[\begin{array}{rr}-24 & 21 \\ -24 & 15\end{array}\right] = \left[\begin{array}{cc}-24-24 & -6+21 \\ -21-24 & 6+15\end{array}\right] \] Calculating the entries: \[ \left[\begin{array}{cc}-48 & 15 \\ -45 & 21\end{array}\right] \] Thus, we have: \[ -3 A + 3 B = \left[\begin{array}{cc}-48 & 15 \\ -45 & 21\end{array}\right] \]