第３問
\begin{align}
\begin{aligned}
A^2 &= \begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{bmatrix}
\begin{bmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{bmatrix} \\
&= \begin{bmatrix}
a_{11}^2+a_{12}a_{21} & a_{11}a_{12}+a_{12}a_{22} \\
a_{11}a_{21}+a_{21}a_{22} & a_{12}a_{21}+a_{22}^2 \\
\end{bmatrix}\\
&= \begin{bmatrix}
a_{11}^2+a_{12}a_{21} & a_{12}(a_{12}+a_{22}) \\
a_{21}(a_{11}+a_{22}) & a_{12}a_{21}+a_{22}^2 \\
\end{bmatrix}
\end{aligned}
\end{align}
となります．
(2)
2×2行列の行列式の定義より，
\begin{align}
\begin{aligned}
|A| &= \begin{vmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{vmatrix}\\
&= a_{11}a_{22}-a_{12}a_{21}
\end{aligned}
\end{align}
となります．
(3)
2×2行列の逆行列の定義より，
\begin{align}
\begin{aligned}
A^{-1} &= \frac{1}{|A|}\begin{bmatrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{bmatrix}\\
&= \frac{1}{a_{11}a_{22}-a_{12}a_{21}}\begin{bmatrix}
a_{22} & -a_{12} \\
-a_{21} & a_{11} \\
\end{bmatrix}
\end{aligned}
\end{align}
となります．
(4)固有方程式$|A-\lambda I_2|=0$を解きます．
\begin{align}
\begin{aligned}
|A-\lambda I_2| &= \begin{vmatrix}
a_{11}-\lambda & a_{12} \\
a_{21} & a_{22}-\lambda \\
\end{vmatrix}\\
&= (a_{11}-\lambda)(a_{22}-\lambda)-a_{12}a_{21}\\
&= a_{11}a_{22}-\lambda a_{11}-\lambda a_{22}+\lambda^2-a_{12}a_{21}\\
&= \lambda^2-\lambda(a_{11}+a_{22})+a_{11}a_{22}-a_{12}a_{21} \\
&= 0
\end{aligned}
\end{align}
より，2次方程式の解の公式を用いて，
$\lambda_{1,2} = \frac{a_{11}+a_{22} \pm \sqrt{(a_{11}+a_{22})^2 – 4(a_{11}a_{22}-a_{12}a_{21})}}{2}$
を得ます．