Using augmented matrices to find a number

Question

There's this system of equations

$$(8 − a)x_1 + 2x_2 + 3x_3 + ax_4 = 2$$

$$x_1 + (9 − a)x_2 + 4x_3 + ax_4 = 1$$

$$x_1 + 2x_2 + (10 − a)x_3 + ax_4 = 2$$

$$x_1 + 2x_2 + 3x_3 + ax_4 = 2$$

Now I have managed to find $a$, $(a=7)$ using substitution. However if you were to use matrices to do it, how would you solve the system depending on the parameter $a$?

\begin{pmatrix} (8-a) & 2 & 3 & a & 2 \\ 1 & (9-a) & 4 & a & 1 \\ 1 & 2 & (10-a) & a & 2 \\ 1 & 2 & 3 & a & 2 \\ \end{pmatrix}

Answer 1

You can use Gauss or Gauss-Jordan elimination to determine the solutions: $$ A x = b \Rightarrow [A\vert b] \to \left[ \begin{array}{rrrr|r} (8-a) & 2 & 3 & a & 2 \\ 1 & (9-a) & 4 & a & 1 \\ 1 & 2 & (10-a) & a & 2 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 2 - 2(8-a) & 3 - 3(8-a) & a - a(8-a) & 2-2(8-a) \\ 0 & (9-a)-2 & 1 & 0 & -1 \\ 0 & 0 & (10-a)-3 & 0 & 0 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & -14 + 2 a & -21 + 3 a & -7a + a^2 & -14 + 2a \\ 0 & 7-a & 1 & 0 & -1 \\ 0 & 0 & 7-a & 0 & 0 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & -21 + 3 a + 2 & -7a + a^2 & -14 + 2a - 2 \\ 0 & 7-a & 1 & 0 & -1 \\ 0 & 0 & 7-a & 0 & 0 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 2 & -7a + a^2 & -14 + 2a - 2 \\ 0 & 7-a & 1 & 0 & -1 \\ 0 & 0 & 7-a & 0 & 0 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] $$

We now have to make a case distinction 1.) $a = 7 $ or 2.) $a \ne 7 $, because this determines if we can divide a row by $(7-a)$ or not, or if there is a leading non-zero entry or not, which affects the dimension of the solution space.

Case 1 ($a = 7$):

$$ [A\vert b] \to \left[ \begin{array}{rrrr|r} 0 & 0 & 2 & 0 & - 2 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 1 & 2 & 3 & 7 & 2 \\ \end{array} \right] \\ \to \left[ \begin{array}{rrrr|r} 1 & 2 & 3 & 7 & 2 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] \\ \to \left[ \begin{array}{rrrr|r} 1 & 2 & 0 & 7 & 5 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ \end{array} \right] $$

Case 2 ($a \ne 7$):

$$ [A\vert b] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 2 & -7a + a^2 & -14 + 2a - 2 \\ 0 & 1 & 1/(7-a) & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 2 & 3 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 0 & -7a + a^2 & -14 + 2a - 2 \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 2 & 0 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 0 & -a & -2 - 2/(7-a) \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 2 & 0 & a & 2 \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 0 & -a & -2 - 2/(7-a) \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 2 & 0 & 0 & -2/(7-a) \\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 0 & 0 & 0 & a & 2 + 2/(7-a) \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0\\ \end{array} \right] \to \\ \left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & a & 2 + 2/(7-a) \\ \end{array} \right] $$ The last row implies $a \ne 0$ for a solution, otherwise there would be a condition $0 = 16/7$ which is not to fulfill by any vector $x$.

Case 3 ($a\ne 0, a\ne 7$):

$$ [A\vert b] \to \to \\ \left[ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & -1/(7-a) \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & (2/a) + 2/((7-a)a) \\ \end{array} \right] $$

Result:

Thus the system has the unique solution $$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 0 \\ -1/(7-a) \\ 0 \\ (2/a) + 2/((7-a)a) \end{pmatrix} $$ for $a \ne 0, a \ne 7$.

For $a = 0$ there is no solution.

For $a = 7$ there are infinite many solutions $$ \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix} = \begin{pmatrix} 5 -2s - 7t \\ s \\ -1 \\ t \end{pmatrix} $$ for $s, t \in \mathbb{R}$.

Answer 2

$a=7$ is not a given, correct? Because I think I found some other solutions.
Do row reductions for the matrix. You will get a matrix that looks like this: $$\left[ \begin{array}{cccc|c} 1 & 2 & 3 & a & 2\\ 7-a & 0 & 0 & 0 & 0\\ 0 & 7-a & 1 & 0 & -1\\ 0 & 0 & 7-a & 0 & 0\\ \end{array} \right]$$ Although it is not a standard row-echelon from, you can still solve the system form here.
From row 4 you can see that $(7-a)x_3=0$, so either $a=7$ or $x_3=0$
From row 2 you can see that $(7-a)x_1=0$, so either $a=7$ or $x_1=0$

So you should solve for each case:

When $a=7$, with two leading 1's in column 1 & 3, you'll have two free variables $x_2$ and $x_4$. Set $x_2=r$ and $x_4=s$ and solve for $x_1$ and $x_3$ correspondingly.

When $a\not=7$, then $x_1=x_3=0$, solve for $x_2$ and $x_4$ in terms of a.