2022-M2-08

$\begin{array}{rcl}
\Delta & \neq & 0 \\
\begin{vmatrix} a & 2 & -1 \\ -1 & a & 2 \\ 2 & -1 & a \end{vmatrix} & \neq & 0 \\
a\begin{vmatrix} a & 2 \\ -1 & a \end{vmatrix}-2\begin{vmatrix} -1 & 2 \\ 2 & a \end{vmatrix}+(-1)\begin{vmatrix} -1 & a \\ 2 & -1 \end{vmatrix} & \neq & 0 \\
a(a^2+2)-2(-a-4)-(1-2a) & \neq & 0 \\
a^3+2a+2a+8-1+2a & \neq & 0 \\
a^3+6a+7 & \neq & 0 \\
a^3+1+6a+6 & \neq & 0 \\
(a+1)(a^2-a+1)+6(a+1) & \neq & 0 \\
(a+1)(a^2-a+7) & \neq & 0
\end{array}$

By Cramer’s rule, we have

$\begin{array}{rcl}
y & = & \dfrac{\Delta_y}{\Delta} \\
y & = & \dfrac{\begin{vmatrix} a & 4k & -1 \\ -1 & 4 & 2 \\ 2 & k^2 & a \end{vmatrix}}{(a+1)(a^2-a+7)} \\
y & = & \dfrac{a\begin{vmatrix} 4 & 2 \\ k^2 & a \end{vmatrix}-4k\begin{vmatrix} -1 & 2 \\ 2 & a \end{vmatrix}+(-1)\begin{vmatrix} -1 & 4 \\ 2 & k^2 \end{vmatrix}}{(a+1)(a^2-a+7)} \\
y & = & \dfrac{a(4a-2k^2)-4k(-a-4)-(-k^2-8)}{(a+1)(a^2-a+7)} \\
y & = & \dfrac{4a^2-2ak^2+4ak+16k+k^2+8}{(a+1)(a^2-a+7)}
\end{array}$

$\begin{array}{rcl}
\Delta & = & 0 \\
(a+1)(a^2-a+7) & = & 0
\end{array}$

Therefore, $a=-1$.

Consider the augmented matrix of $(E)$, we have

$\begin{array}{l}
\left(\begin{array}{ccc|c} -1 & 2 & -1 & 4k \\ -1 & -1 & 2 & 4 \\ 2 & -1 & -1 & k^2 \end{array}\right) \\
\xrightarrow[R_3+2R_1 \to R_3]{R_2-R_1\to R_2} \left(\begin{array}{ccc|c} -1 & 2 & -1 & 4k \\ 0 & -3 & 3 & 4-4k \\ 0 & 3 & -3 & k^2+8k \end{array}\right) \\
\xrightarrow[R_1\times -1 \to R_1]{R_3+R_2 \to R_3} \left(\begin{array}{ccc|c} 1 & -2 & 1 & -4k \\ 0 & -3 & 3 & 4-4k \\ 0 & 0 & 0 & k^2+4k+4 \end{array}\right)
\end{array}$

For $(E)$ has infinitely many solutions,

$\begin{array}{rcl}
k^2+4k+4 & = & 0 \\
(k+2)^2 & = & 0
\end{array}$

Therefore, $k=-2$ (repeated).

Hence, we have

$\begin{array}{l}
\left(\begin{array}{ccc|c} 1 & -2 & 1 & -4(-2) \\ 0 & -3 & 3 & 4-4(-2) \\ 0 & 0 & 0 & 0 \end{array}\right) \\
\xrightarrow{R_2\times \frac{-1}{3} \to R_2} \left(\begin{array}{ccc|c} 1 & -2 & 1 & 8 \\ 0 & 1 & -1 & -4 \\ 0 & 0 & 0 & 0 \end{array}\right)
\end{array}$

Let $z=t$, $t\in \mathbb{R}$.

$\begin{array}{rcl}
y-(t) & = & -4 \\
y & = & t-4
\end{array}$

Also,

$\begin{array}{rcl}
x-2(t-4)+t & = & 8 \\
x-2t+8+t & = & 8 \\
x & = & t
\end{array}$

Therefore, the general solution of $(E)$ is $\{(t,t-4,t)|t\in\mathbb{R}\}$.