Let's work through the matrix problems from your question systematically.

Part (a)

Given: $A = \begin{pmatrix} 4 & 7 \\ 5 & 2 \end{pmatrix}, \quad B = \begin{pmatrix} 3 & -2 \\ 4 & -6 \end{pmatrix}$ We need to compute the following:

(i) $A - \frac{1}{2} B$

First, compute $\frac{1}{2} B$: $\frac{1}{2} B = \frac{1}{2} \begin{pmatrix} 3 & -2 \\ 4 & -6 \end{pmatrix} = \begin{pmatrix} \frac{3}{2} & -1 \\ 2 & -3 \end{pmatrix}$ Now subtract: $A - \frac{1}{2} B = \begin{pmatrix} 4 & 7 \\ 5 & 2 \end{pmatrix} - \begin{pmatrix} \frac{3}{2} & -1 \\ 2 & -3 \end{pmatrix} = \begin{pmatrix} 4 - \frac{3}{2} & 7 - (-1) \\ 5 - 2 & 2 - (-3) \end{pmatrix}$ $A - \frac{1}{2} B = \begin{pmatrix} \frac{5}{2} & 8 \\ 3 & 5 \end{pmatrix}$

(ii) $A^T + B$

First, find the transpose of $A$, denoted $A^T$: $A^T = \begin{pmatrix} 4 & 5 \\ 7 & 2 \end{pmatrix}$ Now add $A^T$ to $B$: $A^T + B = \begin{pmatrix} 4 & 5 \\ 7 & 2 \end{pmatrix} + \begin{pmatrix} 3 & -2 \\ 4 & -6 \end{pmatrix} = \begin{pmatrix} 7 & 3 \\ 11 & -4 \end{pmatrix}$

(iii) $A^2$

Compute the product $A \times A$: $A^2 = \begin{pmatrix} 4 & 7 \\ 5 & 2 \end{pmatrix} \times \begin{pmatrix} 4 & 7 \\ 5 & 2 \end{pmatrix}$ Using matrix multiplication: $A^2 = \begin{pmatrix} 4 \cdot 4 + 7 \cdot 5 & 4 \cdot 7 + 7 \cdot 2 \\ 5 \cdot 4 + 2 \cdot 5 & 5 \cdot 7 + 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 48 & 42 \\ 30 & 39 \end{pmatrix}$

Part (b)

Given: $A = \begin{pmatrix} 2 & 4 \\ 3 & 9 \end{pmatrix}, \quad B = \begin{pmatrix} 7 & 1 \\ -2 & 9 \\ 4 & 3 \end{pmatrix}$ We need to compute the following:

(i) $A \times B$

Since $A$ is a $2 \times 2$ matrix and $B$ is a $2 \times 3$ matrix, the result will be a $2 \times 3$ matrix: [ A \times B = \begin{pmatrix} 2 & 4 \ 3 & 9 \end{pmatrix} \times \begin{pmatrix} 7 & 1 \ -2 & 9 \ 4 & 3 \endalp