If $\vec{b} \neq \vec{0}$, then the solution set of the system of equations $[A|\vec{b}]$ cannot be a plane through the origin.

Question

Is my proof complete? I feel like something is missing.

What I have written:

Consider $\vec{b} \in \mathbb{R}^3$, without loss of generality.

Then $[A|\vec{b}] = \begin{bmatrix} a_1 & ... & a_{1n}|&b_1 \\ a_2 & ... & a_{2n}|&b_2 \\ a_3 & ... & a_{3n}|&b_3 \end{bmatrix}$

In order for the solution set to be a plane through the origin, $b_1 = b_2 = b_3 = 0$. However, $\vec{b}\neq\vec{0}$, so then $[A|\vec{b}]$ cannot be a plane through the origin.

Answer 1

There's two significant problems I see:

• You're essentially just stating the contrapositive. This is not a proof.
• You're assuming that $\vec{b} \in \mathbb{R}^3$. So this would only prove a special case of the result you're asked to prove.

I suggest you start here:

If the solution set is a plane through the origin, then there is a solution to the system of equations, namely the all zero solution. This implies that $A\vec{0}=\vec{b}$. Do you see a problem with this?