Let \( P=\left[\begin{array}{cc}-1 & 6 \\ 7 & 1\end{array}\right] \) and \( A=\left[\begin{array}{cc}0 & 9 \\ -3 & 8\end{array}\right] \) Calculate \( (P A)^{T} \) without calculating \( P A \)

To find \( (P A)^T \) without directly calculating \( P A \), we can use the property of transposes that states: \[ (AB)^T = B^T A^T \] for any matrices \( A \) and \( B \). In our case, we have \( P \) and \( A \), so we can apply this property: \[ (P A)^T = A^T P^T \] Now, we need to calculate \( A^T \) and \( P^T \). 1. **Calculate \( A^T \)**: \[ A = \begin{bmatrix} 0 & 9 \\ -3 & 8 \end{bmatrix} \] The transpose \( A^T \) is obtained by swapping rows and columns: \[ A^T = \begin{bmatrix} 0 & -3 \\ 9 & 8 \end{bmatrix} \] 2. **Calculate \( P^T \)**: \[ P = \begin{bmatrix} -1 & 6 \\ 7 & 1 \end{bmatrix} \] The transpose \( P^T \) is: \[ P^T = \begin{bmatrix} -1 & 7 \\ 6 & 1 \end{bmatrix} \] Now we can compute \( (P A)^T \) using the transposes we found: \[ (P A)^T = A^T P^T = \begin{bmatrix} 0 & -3 \\ 9 & 8 \end{bmatrix} \begin{bmatrix} -1 & 7 \\ 6 & 1 \end{bmatrix} \] Next, we perform the matrix multiplication: \[ A^T P^T = \begin{bmatrix} 0 & -3 \\ 9 & 8 \end{bmatrix} \begin{bmatrix} -1 & 7 \\ 6 & 1 \end{bmatrix} \] Calculating the elements of the resulting matrix: - First row, first column: \[ 0 \cdot (-1) + (-3) \cdot 6 = 0 - 18 = -18 \] - First row, second column: \[ 0 \cdot 7 + (-3) \cdot 1 = 0 - 3 = -3 \] - Second row, first column: \[ 9 \cdot (-1) + 8 \cdot 6 = -9 + 48 = 39 \] - Second row, second column: \[ 9 \cdot 7 + 8 \cdot 1 = 63 + 8 = 71 \] Putting it all together, we have: \[ (P A)^T = \begin{bmatrix} -18 & -3 \\ 39 & 71 \end{bmatrix} \] Thus, the final result is: \[ \boxed{\begin{bmatrix} -18 & -3 \\ 39 & 71 \end{bmatrix}} \]