Question: 1) If $A = \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} -5 & -2 \\ 4 & 1 \end{pmatrix}$
Find:- a) $A^2$ b) Adjoint $A$ c) $B^{-1}$
1. **Problem Statement:**
Find the following for matrices $A$ and $B$:
a) $A^2$
b) Adjoint of $A$
c) Inverse of $B$
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2. **Recall formulas and rules:**
- Matrix multiplication: $A^2 = A \times A$
- Adjoint of a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$
- Inverse of a $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is $M^{-1} = \frac{1}{|M|} \times \text{adj}(M)$ where $|M| = ad - bc$
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3. **Calculate $A^2$:**
$$
A = \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}
$$
$$
A^2 = A \times A = \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix} \times \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}
$$
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,
$$
A^2 = \begin{pmatrix} 26 & 14 \\ 35 & 19 \end{pmatrix}
$$
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4. **Find Adjoint of $A$:**
For $A = \begin{pmatrix} 4 & 2 \\ 5 & 3 \end{pmatrix}$,
$$
\text{adj}(A) = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}
$$
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5. **Find $B^{-1}$:**
Calculate determinant of $B$:
$$
|B| = (-5)(1) - (-2)(4) = -5 + 8 = 3
$$
Calculate adjoint of $B$:
$$
\text{adj}(B) = \begin{pmatrix} 1 & 2 \\ -4 & -5 \end{pmatrix}
$$
Therefore,
$$
B^{-1} = \frac{1}{3} \times \begin{pmatrix} 1 & 2 \\ -4 & -5 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} & \frac{2}{3} \\ -\frac{4}{3} & -\frac{5}{3} \end{pmatrix}
$$
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**Final answers:**
- a) $A^2 = \begin{pmatrix} 26 & 14 \\ 35 & 19 \end{pmatrix}$
- b) $\text{adj}(A) = \begin{pmatrix} 3 & -2 \\ -5 & 4 \end{pmatrix}$
- c) $B^{-1} = \begin{pmatrix} \frac{1}{3} & \frac{2}{3} \\ -\frac{4}{3} & -\frac{5}{3} \end{pmatrix}$