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 $x_2=0.$

Answer 2

Hint:

Use Cramer's formula for $x_2$ (which supposes the determinaant of the linear system is not $0$).

Answer 3

$x$ is a (unique) linear combination of the columns of $A$. $B$ and the last column of $A$ are linear dependent. Thus the unique solution must be in $\{(0,0,0,0,\lambda) | \lambda \in \mathbb R\}$.