Let's break down the problem from the image you've uploaded:

Given Matrices:

1. Matrix A: $A = \begin{bmatrix} 1 & 2 \end{bmatrix}$
2. Matrix B: $B = \begin{bmatrix} 2 & -3 \\ 1 & 5 \end{bmatrix}$
3. Matrix C: $C = \begin{bmatrix} 6 & 4 & 0 \\ 8 & 2 & -10 \end{bmatrix}$
4. Matrix D: $D = \begin{bmatrix} 3 & 5 \\ 1 & 2 \\ 0 & -4 \end{bmatrix}$
5. Matrix E: $E = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 3 & 4 \\ 1 & 2 & 3 \end{bmatrix}$

The problem is asking for the result of $C + D^T$, where $D^T$ is the transpose of matrix $D$.

Step 1: Transpose of Matrix $D$

Transpose of a matrix is obtained by flipping it over its diagonal. For matrix $D$, we get: $D^T = \begin{bmatrix} 3 & 1 & 0 \\ 5 & 2 & -4 \end{bmatrix}$

Step 2: Adding $C$ and $D^T$

Now, we can add matrix $C$ and matrix $D^T$, element by element:

$C = \begin{bmatrix} 6 & 4 & 0 \\ 8 & 2 & -10 \end{bmatrix}, \quad D^T = \begin{bmatrix} 3 & 1 & 0 \\ 5 & 2 & -4 \end{bmatrix}$

Performing the element-wise addition:

$C + D^T = \begin{bmatrix} 6+3 & 4+1 & 0+0 \\ 8+5 & 2+2 & -10+(-4) \end{bmatrix} = \begin{bmatrix} 9 & 5 & 0 \\ 13 & 4 & -14 \end{bmatrix}$

Thus, the result of $C + D^T$ is: $\boxed{\begin{bmatrix} 9 & 5 & 0 \\ 13 & 4 & -14 \end{bmatrix}}$

Let me know if you need further clarifications or details.

Here are 5 related questions for further practice:

1. What is the transpose of matrix $B$?
2. Can you find the product $A \times B$?
3. What is the determinant of matrix $E$?
4. How do you find the inverse of matrix $B$, if it exists?
5. How would you calculate $D \times E$?

Tip:

Always remember that matrix addition requires the matrices to have the same dimensions!