(a) If A = [1, 0, 2; 0, 2, 1; 2, 0, 3], then show that A^3-6 A^2+7 A +2 I = O.

getting, $\displaystyle \mathrm{A}^{2}=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]$

$$\text { getting, } \mathrm{A}^{$\displaystyle 3$}=\left[\begin{array}{lll} 21 & 0 & 34 12 & 8 & 23 34 & 0 & 55 \end{array}\right]

$$$$ \left[\begin{array}{lll} 21 & 0 & 34 12 & 8 & 23 34 & 0 & 55 \end{array}\right]-\left[\begin{array}{ccc} 30 & 0 & 48 12 & 24 & 30 48 & 0 & 78 \end{array}\right]+\left[\begin{array}{ccc} 7 & 0 & 14 0 & 14 & 7 14 & 0 & 21 \end{array}\right]+\left[\begin{array}{lll} 2 & 0 & 0 0 & 2 & 0 0 & 0 & 2 \end{array}\right]

$$(b)adj $\displaystyle \mathrm{A}=\left[\begin{array}{rr} -7 & -2 \\ -5 & 3 \end{array}\right]$

$$\begin{aligned} & |\mathrm{A}|=-$\displaystyle 31$ & \mathrm{A}^{-$\displaystyle 1$}=\frac{-1}{31}\left[\begin{array}{rr} -7 & -2 -5 & 3 \end{array}\right] \end{aligned}

$$Given system of equations is

$$\left\lfloor\begin{array}{rr} $\displaystyle 3$ & $\displaystyle 5$ $\displaystyle 2$ & -$\displaystyle 7$ \end{array}\right\rfloor\left[\begin{array}{l} \mathrm{x} \mathrm{y} \end{array}\right]=\left[\begin{array}{c} 11 -3 \end{array}\right]

$$which is $\displaystyle \mathrm{A}^{\prime} \mathrm{X}=\mathrm{B}$, where $\displaystyle \mathrm{X}=\left[\begin{array}{l}\mathrm{x} \\ \mathrm{y}\end{array}\right], \mathrm{B}=\left[\begin{array}{c}11 \\ -3\end{array}\right]$