<s>[INST] Systems of equations with unknown constant How do I solve this system? It says I must row reduce it to solve it (depending on parameter $a$).
$(8−a)x_1 + 2x_2 + 3x_3 + ax_4 = 2$ 
$x_1 + (9−a)x_2 + 4x_3 + ax_4 = 1$
$x_1 + 2x_2 + (10−a)x_3 + ax_4 = 2$
$x_1 + 2x_2 + 3x_3 + ax_4 = 2$ 
I have reduced it to the form 
$$\left[
        \begin{array}{cccc|c}
        1 & 2 & 3 & a & 2\\
        7-a & 0 & 0 & 0 & 0\\
        0 & 7-a & 1 & 0 & -1\\
        0 & 0 & 7-a & 0 & 0\\
        \end{array}
\right]$$
How do I proceed from here? If there were no $a$ then I would be able to put it in row reduced echelon form, but now because of the $a$ I don't know how to continue.
 [/INST] Permuting the columns,
$$\det \left[
  \begin{array}{ccc|c}
        1 & 2 & 3 & a\\
        \hline
        7-a & 0 & 0 & 0\\
        0 & 7-a & 1 & 0\\
        0 & 0 & 7-a & 0\\
  \end{array} 
\right] = \pm \det \left[
  \begin{array}{c|ccc}
        a & 1 & 2 & 3\\
        \hline
        0 & 7-a & 0 & 0\\
        0 & 0 & 7-a & 1\\
        0 & 0 & 0 & 7-a\\
  \end{array} 
\right] = \pm a (7-a)^3$$
Thus, if $a = 0$ or $a = 7$, then the matrices above are singular and the linear system being studied either has infinitely many solutions or it has none. We have three cases to consider.

$a \neq 0$ and $a \neq 7$
The linear system has the unique solution
$$\begin{bmatrix} 0\\ \dfrac{1}{a-7}\\ 0\\ \dfrac{2a-16}{a (a-7)}\end{bmatrix}$$

$a = 0$
The 4th column is zero, i.e., $x_4$ is free. However, the system is then inconsistent, as $-\frac{2}{7} \neq 2$.

$a = 7$
The southwest $3 \times 3$ block contains only zeros on the main diagonal. There are two degrees of freedom. The solution set is the $2$-dimensional affine space
$$\begin{bmatrix} 0\\ 0\\ -1\\ \frac{5}{7}\end{bmatrix} + \gamma_1 \begin{bmatrix} 1\\ 0\\ 0\\ -\frac{1}{7}\end{bmatrix} + \gamma_2 \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{2}{7}\end{bmatrix}$$
</s>