1. **State the problem:** Given matrices $$A = \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix}, \quad B = \begin{pmatrix}-5 & -2 \\ 4 & 1\end{pmatrix}$$ Find: a) $A^2$ b) Adjoint of $A$ c) Inverse of $B$ 2. **Find $A^2$: Multiply matrix $A$ by itself** $$A^2 = A \times A = \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix} \times \begin{pmatrix}4 & 2 \\ 5 & 3\end{pmatrix}$$ Calculate each element: - Top-left: $4 \times 4 + 2 \times 5 = 16 + 10 = 26$ - Top-right: $4 \times 2 + 2 \times 3 = 8 + 6 = 14$ - Bottom-left: $5 \times 4 + 3 \times 5 = 20 + 15 = 35$ - Bottom-right: $5 \times 2 + 3 \times 3 = 10 + 9 = 19$ So, $$A^2 = \begin{pmatrix}26 & 14 \\ 35 & 19\end{pmatrix}$$ 3. **Find the adjoint of $A$:** The adjoint of a 2x2 matrix $\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $$\text{adj}(A) = \begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$$ For $A$, $a=4$, $b=2$, $c=5$, $d=3$, so $$\text{adj}(A) = \begin{pmatrix}3 & -2 \\ -5 & 4\end{pmatrix}$$ 4. **Find the inverse of $B$:** The inverse of a 2x2 matrix $B = \begin{pmatrix}a & b \\ c & d\end{pmatrix}$ is $$B^{-1} = \frac{1}{\det(B)} \begin{pmatrix}d & -b \\ -c & a\end{pmatrix}$$ where $$\det(B) = ad - bc$$ Calculate determinant of $B$: $$\det(B) = (-5)(1) - (-2)(4) = -5 + 8 = 3$$ Calculate adjoint of $B$: $$\text{adj}(B) = \begin{pmatrix}1 & 2 \\ -4 & -5\end{pmatrix}$$ So, $$B^{-1} = \frac{1}{3} \begin{pmatrix}1 & 2 \\ -4 & -5\end{pmatrix} = \begin{pmatrix}\frac{1}{3} & \frac{2}{3} \\ -\frac{4}{3} & -\frac{5}{3}\end{pmatrix}$$ 5. **Summary of answers:** - $A^2 = \begin{pmatrix}26 & 14 \\ 35 & 19\end{pmatrix}$ - $\text{adj}(A) = \begin{pmatrix}3 & -2 \\ -5 & 4\end{pmatrix}$ - $B^{-1} = \begin{pmatrix}\frac{1}{3} & \frac{2}{3} \\ -\frac{4}{3} & -\frac{5}{3}\end{pmatrix}$