(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 \( B \).

We are given \[ D=\begin{bmatrix} 4 & -2 & 1 \\ 3 & -3 & 2 \end{bmatrix} \quad \text{and} \quad B=\begin{bmatrix}5 & -3 & 2 \\ 1 & -2 & 1 \end{bmatrix}. \] The transpose of \( B \) is \[ B^T=\begin{bmatrix} 5 & 1 \\ -3 & -2 \\ 2 & 1 \end{bmatrix}. \] We want to evaluate \[ DB^{T}. \] Since \( D \) is a \( 2 \times 3 \) matrix and \( B^T \) is a \( 3 \times 2 \) matrix, the resulting product will be a \( 2 \times 2 \) matrix. First, compute the entry in the first row and first column of \( DB^T \) by taking the dot product of the first row of \( D \) and the first column of \( B^T \): \[ (4)(5) + (-2)(-3) + (1)(2) = 20 + 6 + 2 = 28. \] Next, compute the entry in the first row and second column of \( DB^T \) by taking the dot product of the first row of \( D \) and the second column of \( B^T \): \[ (4)(1) + (-2)(-2) + (1)(1) = 4 + 4 + 1 = 9. \] Now, compute the entry in the second row and first column of \( DB^T \) by taking the dot product of the second row of \( D \) and the first column of \( B^T \): \[ (3)(5) + (-3)(-3) + (2)(2) = 15 + 9 + 4 = 28. \] Finally, compute the entry in the second row and second column of \( DB^T \) by taking the dot product of the second row of \( D \) and the second column of \( B^T \): \[ (3)(1) + (-3)(-2) + (2)(1) = 3 + 6 + 2 = 11. \] Thus, the product \( DB^T \) is \[ DB^T=\begin{bmatrix} 28 & 9 \\ 28 & 11 \end{bmatrix}. \]