Determine all values b1, b2 such that the following system has no solution

Question

Determine all values $b_1$, $b_2$ such that

$x_1+2x_2-x_3=b_1$

$-2x_1-4x_2+2x_3=b_2$

$x_1-x_2+x_3=2$

has no solution.

I know i want to do row echelon form after i put the system into a matrix

i ended up with:

$\begin{pmatrix}1 & 2 & -1 & b_1 \\0 & 1 & -1 & -2 \\0 & 0 & 0 & 2b_1+b_2\end{pmatrix}$ because i didn't know if i needed to keep going after getting a row of all zeros.

i don't know what a row of all zeros mean, does it mean no solution?

Answer 1

Think about what row 2 is telling you: $0*x + 0*y + 0*z = 2 b_1 + b_2$. For example if $b_1 = 1$ and $b_2 = 2$, then row 2 tells you that $0 = 4$. Hmmm, doesn't look good.

What if $b_1 = 1$ and $b_2 = -2$? Then we have $0 = 0$, that looks better...

I suggest reading pages 7-9 of Jim Hefferon's Linear Algebra book available here to get a better grasp on this.