Let's solve each part of the problem step-by-step using the given matrices.

The matrices provided are:

$A = \begin{bmatrix} \frac{2}{5} & \frac{2}{5} & -\frac{4}{5} \\ \frac{2}{5} & \frac{2}{5} & \frac{2}{10} \\ \frac{2}{5} & -\frac{8}{5} & \frac{2}{10} \end{bmatrix}, \quad B = \begin{bmatrix} -3 \\ 5 \\ 8 \end{bmatrix}, \quad C = \begin{bmatrix} -1 & 3 & 4 \\ 2 & 4 & 2 \end{bmatrix}, \quad D = \begin{bmatrix} \frac{1}{2} \\ -1 \end{bmatrix}$

(a) Calculate $B + 2(C^T D^T)$

1. Transpose of $C$: $C^T = \begin{bmatrix} -1 & 2 \\ 3 & 4 \\ 4 & 2 \end{bmatrix}$

2. Transpose of $D$: $D^T = \begin{bmatrix} \frac{1}{2} & -1 \end{bmatrix}$

3. Calculate $C^T D^T$: $C^T D^T = \begin{bmatrix} -1 & 2 \\ 3 & 4 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{2} \\ -1 \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} - 2 \\ \frac{3}{2} - 4 \\ 2 - 2 \end{bmatrix} = \begin{bmatrix} -\frac{5}{2} \\ -\frac{5}{2} \\ 0 \end{bmatrix}$

4. Calculate $2(C^T D^T)$: $2(C^T D^T) = 2 \begin{bmatrix} -\frac{5}{2} \\ -\frac{5}{2} \\ 0 \end{bmatrix} = \begin{bmatrix} -5 \\ -5 \\ 0 \end{bmatrix}$

5. Add $B$ and $2(C^T D^T)$: $B + 2(C^T D^T) = \begin{bmatrix} -3 \\ 5 \\ 8 \end{bmatrix} + \begin{bmatrix} -5 \\ -5 \\ 0 \end{bmatrix} = \begin{bmatrix} -8 \\ 0 \\ 8 \end{bmatrix}$

So, the answer for (a) is: $\begin{bmatrix} -8 \\ 0 \\ 8 \end{bmatrix}$

(b) Calculate $20A + BDC$

1. Calculate $BD$: $BD = \begin{bmatrix} -3 \\ 5 \\ 8 \end{bmatrix} \begin{bmatrix} \frac{1}{2} & -1 \end{bmatrix} = \begin{bmatrix} -\frac{3}{2} & 3 \\ \frac{5}{2} & -5 \\ 4 & -8 \end{bmatrix}$

2. Calculate $BDC$: $BDC = \begin{bmatrix} -\frac{3}{2} & 3 \\ \frac{5}{2} & -5 \\ 4 & -8 \end{bmatrix} \begin{bmatrix} -1 & 3 & 4 \\ 2 & 4 & 2 \end{bmatrix}$ Calculating each element:

   • Row 1: $(- \frac{3}{2})(-1) + (3)(2), (- \frac{3}{2})(3) + (3)(4), (- \frac{3}{2})(4) + (3)(2)$
   • Row 2: Same steps apply

After