\( 1 \leftarrow \begin{array}{l}\text { Find } B-A \text {. } \\ B-A=\left[\begin{array}{ll}4 & 1 \\ 8 & 8\end{array}\right] B=\left[\begin{array}{rr}-6 & 9 \\ 6 & -5\end{array}\right] \\ B=\square\end{array} \)

To solve for \( B \), we start with the equation \( B - A = \begin{bmatrix} 4 & 1 \\ 8 & 8 \end{bmatrix} \). Rearranging gives us \( B = A + \begin{bmatrix} 4 & 1 \\ 8 & 8 \end{bmatrix} \). Given \( B = \begin{bmatrix} -6 & 9 \\ 6 & -5 \end{bmatrix} \), we can substitute \( B \) back into the equation: \[ \begin{bmatrix} -6 & 9 \\ 6 & -5 \end{bmatrix} = A + \begin{bmatrix} 4 & 1 \\ 8 & 8 \end{bmatrix} \] To find \( A \), we need to subtract \( \begin{bmatrix} 4 & 1 \\ 8 & 8 \end{bmatrix} \) from \( B \): \[ A = B - \begin{bmatrix} 4 & 1 \\ 8 & 8 \end{bmatrix} = \begin{bmatrix} -6-4 & 9-1 \\ 6-8 & -5-8 \end{bmatrix} = \begin{bmatrix} -10 & 8 \\ -2 & -13 \end{bmatrix} \] So, \( A = \begin{bmatrix} -10 & 8 \\ -2 & -13 \end{bmatrix} \) and \( B = \begin{bmatrix} -6 & 9 \\ 6 & -5 \end{bmatrix} \).