Given the matrices \( A \) and \( B \) shown below, find \( -\frac{1}{2} B-3 A \). \[ A=\left[\begin{array}{cc} 4 & 6 \\ -4 & 3 \end{array}\right] \quad B=\left[\begin{array}{cr} 12 & -6 \\ 24 & -8 \end{array}\right] \]

Step 1 ：\(-\frac{1}{2} B = \left[\begin{array}{cc}-6 & 3\\-12 & 4\end{array}\right]\)

Step 2 ：\(-3 A = \left[\begin{array}{cc} -12 & -18\\12 & -9\end{array}\right]\)

Step 3 ：\(-\frac{1}{2} B - 3 A = \left[\begin{array}{cc} (-6) + (-12) & 3+(-18)\\(-12)+12 & 4+(-9)\end{array}\right] = \left[\begin{array}{cc} -18 & -15\\0 & -5\end{array}\right]\)