[latex] \[
\textbf{ Divide the Square Matrix A by the Square Matrix B, where:}
\]
\[
A = \begin{pmatrix} 8 & 4 \\ 2 & 6 \end{pmatrix}, \quad B = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}
\]
\[
\textbf{Solution: Find the inverse of B, then multiply A by B^{-1}.}
\]
\[
\text{Step 1: Compute the determinant of B.}
\]
\[
\text{det}(B) = (2)(2) – (1)(1) = 4 – 1 = 3
\]
\[
\text{Step 2: Compute the inverse of B.}
\]
\[
B^{-1} = \frac{1}{\text{det}(B)} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}
\]
\[
\text{Step 3: Multiply A by B^{-1}.}
\]
\[
A \div B = A \cdot B^{-1} = \begin{pmatrix} 8 & 4 \\ 2 & 6 \end{pmatrix} \cdot \frac{1}{3} \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix}
\]
\[
A \div B = \frac{1}{3} \begin{pmatrix} (8)(2) + (4)(-1) & (8)(-1) + (4)(2) \\ (2)(2) + (6)(-1) & (2)(-1) + (6)(2) \end{pmatrix}
\]
\[
A \div B = \frac{1}{3} \begin{pmatrix} 16 – 4 & -8 + 8 \\ 4 – 6 & -2 + 12 \end{pmatrix}
\]
\[
A \div B = \frac{1}{3} \begin{pmatrix} 12 & 0 \\ -2 & 10 \end{pmatrix} = \begin{pmatrix} 4 & 0 \\ -\frac{2}{3} & \frac{10}{3} \end{pmatrix}
\]
\[
\textbf{Answer: } \begin{pmatrix} 4 & 0 \\ -\frac{2}{3} & \frac{10}{3} \end{pmatrix}
\]