1. **State the problem:** Given matrix $A = \begin{bmatrix}5 & 0 \\ -6 & 2\end{bmatrix}$, find $A^2$ (the square of matrix $A$). 2. **Formula used:** To find $A^2$, we multiply matrix $A$ by itself: $$A^2 = A \times A$$ 3. **Matrix multiplication rule:** For two matrices $X = \begin{bmatrix}x_{11} & x_{12} \\ x_{21} & x_{22}\end{bmatrix}$ and $Y = \begin{bmatrix}y_{11} & y_{12} \\ y_{21} & y_{22}\end{bmatrix}$, their product $XY$ is: $$XY = \begin{bmatrix}x_{11}y_{11} + x_{12}y_{21} & x_{11}y_{12} + x_{12}y_{22} \\ x_{21}y_{11} + x_{22}y_{21} & x_{21}y_{12} + x_{22}y_{22}\end{bmatrix}$$ 4. **Calculate $A^2$:** $$A^2 = \begin{bmatrix}5 & 0 \\ -6 & 2\end{bmatrix} \times \begin{bmatrix}5 & 0 \\ -6 & 2\end{bmatrix} = \begin{bmatrix}5\times5 + 0\times(-6) & 5\times0 + 0\times2 \\ -6\times5 + 2\times(-6) & -6\times0 + 2\times2\end{bmatrix}$$ 5. **Simplify each element:** - Top-left: $5\times5 + 0 = 25$ - Top-right: $0 + 0 = 0$ - Bottom-left: $-30 + (-12) = -42$ - Bottom-right: $0 + 4 = 4$ 6. **Final result:** $$A^2 = \begin{bmatrix}25 & 0 \\ -42 & 4\end{bmatrix}$$ This is the square of matrix $A$.