Let's work through the math problems from the image. Here's a step-by-step solution for each question:

1. Find the result of $A = \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix}$ and $C = \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix}$:

(a) Calculate $A + C$:

$A + C = \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 + 1 & 1 + 0 \\ -2 + (-1) & 4 + 2 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ -3 & 6 \end{bmatrix}$

(b) Calculate $2A - 3C$:

$2A = 2 \times \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ -4 & 8 \end{bmatrix}$

$3C = 3 \times \begin{bmatrix} 1 & 0 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ -3 & 6 \end{bmatrix}$

$2A - 3C = \begin{bmatrix} 6 & 2 \\ -4 & 8 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ -3 & 6 \end{bmatrix} = \begin{bmatrix} 6 - 3 & 2 - 0 \\ -4 - (-3) & 8 - 6 \end{bmatrix} = \begin{bmatrix} 3 & 2 \\ -1 & 2 \end{bmatrix}$

2. Find the result of $B = \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix}$:

(a) Calculate $B^T$:

The transpose of $B$ is:

$B^T = \begin{bmatrix} 1 & 3 \\ -2 & 0 \end{bmatrix}$

(b) Calculate $3A - 2B$:

$3A = 3 \times \begin{bmatrix} 3 & 1 \\ -2 & 4 \end{bmatrix} = \begin{bmatrix} 9 & 3 \\ -6 & 12 \end{bmatrix}$

$2B = 2 \times \begin{bmatrix} 1 & -2 \\ 3 & 0 \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 6 & 0 \end{bmatrix}$

$3A - 2B = \begin{bmatrix} 9 & 3 \\ -6 & 12 \end{bmatrix} - \begin{bmatrix} 2 & -4 \\ 6 & 0 \end{bmatrix} = \begin{bmatrix} 9 - 2 & 3 - (-4) \\ -6 - 6 & 12 - 0 \end{bmatrix} = \begin{bmatrix} 7 & 7 \\ -12 & 12 \end{bmatrix}$

3. Given matrix $D = \begin{bmatrix} 4 & 2 \\ 0 & 5 \end{bmatrix}$, find $D^{-1}$:

To find $D^{-1}$:

1. Calculate the determinant of $D$:

$\text{Det}(D) = (4)(5) - (2)(0) = 20$

2. Find the adjoint of $D$:

$\text{Adj}(D) = \begin{bmatrix} 5 & -2 \\ 0 & 4 \end{bmatrix}$

3. Divide by the determinant:

$D^{-1} = \frac{1}{20} \times \begin{bmatrix} 5 & -2 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} \frac{5}{20} & \frac{-2}{20} \\ \frac{0}{20} & \frac{4}{20} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} & -\frac{1}{10} \\ 0 & \frac{1}{5} \end{bmatrix}$

Let me know if you need any further details or explanations!

Related Questions:

1. How do you find the inverse of a matrix if the determinant is zero?
2. What are the properties of matrix addition and subtraction?
3. How do you multiply two matrices together?
4. What is the significance of the transpose of a matrix?
5. How do you find the determinant of a 3x3 matrix?

Tip: Always check if a matrix is invertible by ensuring its determinant is not zero.