Systems of equations with unknown constant

Question

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.

Answer 1

Excellent.

The second row of your matrix says that $(7-a)x_1=0$

So either $a=7$ or $x_1=0$

The fourth row of your matrix says that $(7-a)x_3=0$

So either $a=7$ or $x_3=0$

The third row says that $(7-a)x_2+x_3=-2$

If $a=7$, then $x_3=-2$

If $a \ne 7$, then you have $x_3=0$ from the fourth row, so $x_2={2 \over a-7}$

The first row says that $x_1+2x_2+2x_3+ax_4=2$

If $a \ne 7$, then $0+{4 \over a-7}+0+ax_4=2$, so $a_4=2-{4 \over a-7}$ (single solution $x_1=0$, $x_2={2 \over a-7}$, $x_3=0$, $x_4={2a-18 \over a-7}$)

If $a=7$, then $x_1+2x_2-4+7x_4=2 \Rightarrow x_1+2x_2+7x_4=6$ (infinite number of solutions)

Answer 2

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

$$\begin{bmatrix} 0\\ 0\\ -1\\ \frac{5}{7}\end{bmatrix} + \gamma_1 \begin{bmatrix} 1\\ 0\\ 0\\ -\frac{1}{7}\end{bmatrix} + \gamma_2 \begin{bmatrix} 0\\ 1\\ 0\\ -\frac{2}{7}\end{bmatrix}$$