(c) Let \[ D=\left[\begin{array}{lll}4 & -2 & 1 \\ 3 & -3 & 2\end{array}\right] \quad \text { and } B=\left[\begin{array}{lll}5 & -3 & 2 \\ 1 & -2 & 1\end{array}\right] \] Evaluate \( D B^{T} \), where \( B^{T} \) is the transpose of \( B \).

To find \( DB^{T} \), we first need to compute the transpose of matrix \( B \): \[ B^{T} = \left[\begin{array}{cc}5 & 1 \\ -3 & -2 \\ 2 & 1\end{array}\right] \] Now we perform the matrix multiplication \( D B^{T} \): \[ D = \left[\begin{array}{ccc}4 & -2 & 1 \\ 3 & -3 & 2\end{array}\right] \] and \[ B^{T} = \left[\begin{array}{cc}5 & 1 \\ -3 & -2 \\ 2 & 1\end{array}\right] \] We will multiply \( D \) by \( B^{T} \): 1. The entry in the first row and first column of the resultant matrix is calculated as: \[ 4 \cdot 5 + (-2) \cdot (-3) + 1 \cdot 2 = 20 + 6 + 2 = 28 \] 2. The entry in the first row and second column is calculated as: \[ 4 \cdot 1 + (-2) \cdot (-2) + 1 \cdot 1 = 4 + 4 + 1 = 9 \] 3. The entry in the second row and first column is: \[ 3 \cdot 5 + (-3) \cdot (-3) + 2 \cdot 2 = 15 + 9 + 4 = 28 \] 4. The entry in the second row and second column is: \[ 3 \cdot 1 + (-3) \cdot (-2) + 2 \cdot 1 = 3 + 6 + 2 = 11 \] Putting these together, we have: \[ DB^{T} = \left[\begin{array}{cc}28 & 9 \\ 28 & 11\end{array}\right] \] Thus, we find that: \[ DB^{T} = \left[\begin{array}{cc}28 & 9 \\ 28 & 11\end{array}\right] \]