2023-M2-11

1. For $(E)$ has a unique solution, we have

   $\begin{array}{rcl}
   \begin{vmatrix} 1 & a & a+1 \\ x & a+4 & 2a+4 \\ 2 & 3 & 5 \end{vmatrix} & \neq & 0 \\
   (1)\begin{vmatrix} a+4 & 2a+4 \\ 3 & 5 \end{vmatrix} -(1)\begin{vmatrix} a & a+1 \\ 3 & 5 \end{vmatrix}+(2)\begin{vmatrix} a & a+1 \\a+4 & 2a+4 \end{vmatrix} & \neq & 0 \\
   5(a+4)-3(2a+4)-5a+3(a+1)+2(2a^2+4a)-2(a^2+5a+4) & \neq & 0 \\
   2a^2-5a+3 & \neq & 0 \\
   (2a-3)(a-1) & \neq & 0
   \end{array}$

   Therefore, $a\neq 1$ and $a\neq \dfrac{3}{2}$.

2. For $a=1$, we have
   

   $\begin{array}{l}
   \left(\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 1 & 5
   6 & b+1 \\ 2 & 3 & 5 & b \end{array}\right) \\
   \xrightarrow[R_3-2R_1 \to R_3]{R_2-R_1\to R_2} \left(\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 0 & 4 & 4 & b-1 \\ 0 & 1 & 1 & b-4 \end{array}\right) \\
   \xrightarrow{R_3-4R_2 \to R_3} \left(\begin{array}{ccc|c} 1 & 1 & 2 & 2 \\ 0 & 4 & 4 & b-1 \\ 0 & 0 & 0 & 3b-15 \end{array}\right)
   \end{array}$

   For $(E)$ is consistent, we have

   $\begin{array}{rcl}
   3b-15 & = & 0 \\
   b & = & 5
   \end{array}$

3. For $a\neq 1$ and $(E)$ is inconsistent, $a=\dfrac{3}{2}$. Hence, we have

   $\begin{array}{l}
   \left(\begin{array}{ccc|c} 1 & \dfrac{3}{2} & \dfrac{5}{2} & 2 \\
   1 & \dfrac{11}{2} & 7 & b+1 \\
   2 & 3 & 5 & b \end{array}\right) \\
   \xrightarrow[R_3-2R_1 \to R_3]{R_2-R_1\to R_2} \left(\begin{array}{ccc|c} 1 & \dfrac{3}{2} & \dfrac{5}{2} & 2 \\ 0 & 4 & \dfrac{9}{2} & b-1 \\ 0 & 0 & 0 & b-4 \end{array}\right)
   \end{array}$

   For $(E)$ is inconsistent, $b\neq 4$.

$\begin{array}{rcl}
x & = & \dfrac{\Delta_x}{\Delta} \\
x & = & \dfrac{\begin{vmatrix} 2 & 2 & 3 \\ s+1 & 6 & 8 \\ s & 3 & 5 \end{vmatrix}}{2(2)^2-5(2)+3} \\
x & = & \dfrac{(2)\begin{vmatrix} 6 & 8 \\ 3 & 5 \end{vmatrix} -(2)\begin{vmatrix} s+1 & 8 \\ s & 5 \end{vmatrix} +(3)\begin{vmatrix} s+1 & 6 \\ s & 3 \end{vmatrix}}{1} \\
x & = & 12-2(-3s+5)+3(-3x+3) \\
x & = & -3s+11
\end{array}$

$\begin{array}{rcl}
y & = & \dfrac{\Delta_y}{\Delta} \\
y & = & \dfrac{\begin{vmatrix} 1 & 2 & 3 \\ 1 & s+1 & 8 \\ 2 & s & 5 \end{vmatrix}}{2(2)^2-5(2)+3} \\
y & = & \dfrac{(1)\begin{vmatrix} s+1 & 8 \\ s & 5 \end{vmatrix} -(1)\begin{vmatrix} 2 & 3 \\ s & 5 \end{vmatrix} +(2)\begin{vmatrix} 2 & 3 \\ s+1 & 8 \end{vmatrix}}{1} \\
y & = & -3s+5-(10-3s)+2(13-3s) \\
y & = & -6s+21
\end{array}$

$\begin{array}{rcl}
z & = & \dfrac{\Delta_z}{\Delta} \\
z & = & \dfrac{\begin{vmatrix} 1 & 2 & 2 \\ 1 & 6 & s+1 \\ 2 & 3 & s \end{vmatrix}}{2(2)^2-5(2)+3} \\
z & = & \dfrac{(1)\begin{vmatrix} 6 & s+1 \\ 3 & s \end{vmatrix}-(1)\begin{vmatrix} 2 & 2 \\ 3 & s \end{vmatrix}+(2)\begin{vmatrix} 2 & 2 \\ 6 & s+1 \end{vmatrix}}{1} \\
z & = & 3s-3-(2s-6)+2(2s-10) \\
z & = & 5s-17
\end{array}$

Sub. $x=-3s+11$, $y=-6s+21$ and $z=5s-17$ into $mx+ny+z=-2$, we have

$\begin{array}{rcl}
m(-3s+11)+n(-6s+21)+(5s-17) & = & -2 \\
-3ms+11m-6ns+21n+5s-17 & = & -2 \\
(-3m-6n+5)s+(11m+21n-17) & = & -2
\end{array}$

By comparing the coefficients of the both sides, we have

$\left\{ \begin{array}{ll}
-3m-6n+5=0 & \ldots \unicode{x2460} \\
11m+21n-17=-2 & \ldots \unicode{x2461}
\end{array}\right.$

$7\times \unicode{x2460}+2\times \unicode{x2461}$, we have

$\begin{array}{rcl}
m +1 & = & -4 \\
m & = & -5
\end{array}$

Sub. $m=-5$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
-3(-5)-6n+5 & = & 0 \\
-6n+20 & = & 0 \\
n & = & \dfrac{10}{3}
\end{array}$

Therefore, there exist a pair of real constant $m$ and $n$ satisfying $mx+ny+z=-2$.