Question
Find $b_1$ and $b_2$ so that the equation $Ax = b$ has solutions where $$A = \begin{bmatrix} 1 & 2 \\ 0 & 1\\ -1 & 2 \end{bmatrix};\quad b = \begin{bmatrix} b_1 \\ b_2 \\ 0 \end{bmatrix}.$$ Can this equation have a unique solution? Why or why not?
Work
$$\begin{bmatrix}1&2 \\0&1\\-1&2\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}b_1\\b_2\\0\end{bmatrix}$$
$$\begin{bmatrix} x_1 + 2x_2 \\ x_2 \\ -x_1 + 2x_2 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ 0 \end{bmatrix}$$
$$\left.\begin{array}{ll} x_1 + 2x_2 = b_1 \\ x_2 = b_2 \\ 2x_2 = x_1 \end{array}\right\} \Rightarrow \left.\begin{array}{ll}x_1 = b_1/2 \\ x_2 = b_1/4 \end{array}\right\} \Rightarrow b_1/4 = b_2 \Rightarrow b_1 = 4b_2$$
Hint: the equation $Ax=b$ has solutions $ \iff rank(A|b)=rank(A)$.