I have the following question here:
Find all vectors $\begin{pmatrix}a\\ b \end{pmatrix}$ $\in \mathbb{R}^2$ such that the solution to the linear system with augmented matrix $\left(\begin{array}{cc|c} 1&2\;&\;a\\ 3&2\;&\;b\end{array}\right)$ is the vector $\begin{pmatrix}-a\\ -b \end{pmatrix}$.
This question really confuses me. So my approach was to do row reduction and make the solution equal to $\begin{pmatrix}-a\\ -b \end{pmatrix}$.
So row reducing, we get:
$$\left(\begin{array}{cc|c} 1 & 0 \;&\; a-2\left(\frac{b-3a}{4} \right)\\ 0 & 1 \;&\; \frac{b-3a}{4} \end{array}\right)$$
So that means we have $a-2\left(\frac{b-3a}{4} \right)=-a$ and $\frac{b-3a}{4}=-b$. Simplifying, we get: $$ \left\{ \begin{aligned} \frac{7}{2}a - \frac{1}{2}b &= 0 \\[3pt] -\frac{3}{4}a + \frac{5}{4}b &= 0 \end{aligned} \right. $$ So this gives $a=0,\, b=0$ as the only solution. Does that seem right? It feels off to get a trivial answer as the solution.
Although your approach should work, you've made an arithmetic error. Here's a slightly different approach.
In order for $\bigl(\begin{smallmatrix} -a \\ -b \end{smallmatrix}\bigr)$ to be a solution to the linear system, we must be able to substitute in $x=-a$ and $y=-b$: $$ \left\{ \begin{aligned} 1(-a) + 2(-b) &= a \\[2pt] 3(-a) + 2(-b) &= b \end{aligned} \right. $$ i.e. $$ \left\{ \begin{aligned} -2a - 2b &= 0 \\[2pt] -3a - 3b &= 0 \end{aligned} \right. $$
This is a singular system whose solution set is a $1$-dimensional subspace. Solve it yourself before revealing.