(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 $$ B $$.

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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
$$
DB^T=\begin{bmatrix} 28 & 9 \ 28 & 11 \end{bmatrix}.
$$
$$ DB^{T} = \begin{bmatrix} 28 & 9 \ 28 & 11 \end{bmatrix} $$

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