Let \( A=\left[\begin{array}{rr}3 & 2 \\ -1 & 1\end{array}\right] \) and \( B=\left[\begin{array}{rr}2 & 4 \\ -2 & k\end{array}\right] \). What value(s) of \( k \), if any, will make \( A B=B A \) ? Select the correct choice below and, if necessary, fill in the answer box within your choice. A. \( k=\square \) (Use a comma to separate answers as needed.) B. No value of \( k \) will make \( A B=B A \)

To determine the value(s) of \( k \) such that \( AB = BA \), we first calculate \( AB \) and \( BA \). Calculating \( AB \): \[ AB = \left[\begin{array}{rr}3 & 2 \\ -1 & 1\end{array}\right] \left[\begin{array}{rr}2 & 4 \\ -2 & k\end{array}\right] = \left[\begin{array}{rr}3 \cdot 2 + 2 \cdot (-2) & 3 \cdot 4 + 2 \cdot k \\ -1 \cdot 2 + 1 \cdot (-2) & -1 \cdot 4 + 1 \cdot k\end{array}\right] = \left[\begin{array}{rr}6 - 4 & 12 + 2k \\ -2 - 2 & -4 + k\end{array}\right] = \left[\begin{array}{rr}2 & 12 + 2k \\ -4 & -4 + k\end{array}\right] \] Calculating \( BA \): \[ BA = \left[\begin{array}{rr}2 & 4 \\ -2 & k\end{array}\right] \left[\begin{array}{rr}3 & 2 \\ -1 & 1\end{array}\right] = \left[\begin{array}{rr}2 \cdot 3 + 4 \cdot (-1) & 2 \cdot 2 + 4 \cdot 1 \\ -2 \cdot 3 + k \cdot (-1) & -2 \cdot 2 + k \cdot 1\end{array}\right] = \left[\begin{array}{rr}6 - 4 & 4 + 4 \\ -6 - k & -4 + k\end{array}\right] = \left[\begin{array}{rr}2 & 4 + 4 \\ -6 - k & -4 + k\end{array}\right] \] Setting \( AB = BA \): \[ \left[\begin{array}{rr}2 & 12 + 2k \\ -4 & -4 + k\end{array}\right] = \left[\begin{array}{rr}2 & 4 + 4 \\ -6 - k & -4 + k\end{array}\right] \] From the first column (first row): \[ 2 = 2 \quad (\text{Always true}) \] From the first column (second row): \[ -4 = -6 - k \implies k = -2 \] From the second column (first row): \[ 12 + 2k = 4 + 4 \implies 12 + 2k = 8 \implies 2k = 8 - 12 \implies 2k = -4 \implies k = -2 \] From the second column (second row): \[ -4 + k = -4 + k \quad (\text{Always true}) \] Therefore, the only value of \( k \) that satisfies \( AB = BA \) is \( k = -2 \). So the answer is: A. \( k=-2 \)