Show that $x_2 = 0$ without solving the linear system explicitly.

Question

Given the following matrix equation:
$$Ax = B$$
where
$$A=\begin{pmatrix}
6 & -9 & 31 & 5 & -2\\
3 & 20 & 42 & 2 & 0 \\
26 & 9 & 92 & 20 & -10 \\
1 & 2 & 81 & 4 & 22 \\
4 & -1 & 3 & 1 & -10
\end{pmatrix}, x = \begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{pmatrix} \quad \text{and}\quad B= \begin{pmatrix}
1 \\
0 \\
5 \\
-11 \\
5
\end{pmatrix}.$$

Question: If the linear system has a unique solution, show that $x_2 = 0$ without solving the linear system explicitly.

I have been puzzling on this problem for few days.

I attempted to solve the linear system but it is not allowed in this question.

Any hint is appreciated.

Answer 1

Let $C$ denote the last column of $A$. Then we have

$$-\frac{1}{2}C=B.$$

This shows that $(0,0,0,0, -\frac{1}{2})^T$ is a solution of the system $Ax=B$.

Since the system has a unique solution, $(0,0,0,0, -\frac{1}{2})^T$ is this solution, hence $x_2=0.$