$$x_1 + 2x_2 + 3x_3 = b_1 \\ 2x_1 + 5x_2 + 3x_3 = b_2 \\ x_2 – 3x_3 = b_3$$
Use Gauss method: $-2p_1+p_2$ to produce $x_2-x_3=-2b_1+b_2$.
$p_1+p_3$ to produce $x_1+3x_2=b_1+b_3$
$-p_2+p_1$ to produce $-x_1-3x_2= b_1-b_2$
The above produces the system:
the first row $-x_1-3x_2=b_1-b_2$ second row $x_2-x_3=-2b_1+b_2$ third row $x_1+3x_2=b_1+b_3$
Do Gauss method again. $p_1+p_3$ to produce $0=2b_1-b_2+b_3$
Since I got a homogeneous equation from the above. I can just stop and rewrite the system.
first row $-x_1-3x_2=b_1-b_2$ second row $x_2-x_3=-2b_1+b_2$ third row $0 = 2b_1-b_2+b_3.$
Write the homogeneous equation in terms of $b_3. b_3 = -2b_1+b_2$
So I can say the system is consistent if and only if $b_3=-2b_1+b_2.$
May I get a verification of my answer?
Best Answer
In fact, you can directly check it yourself.
The condition means that LHS of $L_3$ is LHS of $-2 L_1 + L_2$, where $L_i$ is the i-th line of the system, and this is true. Moreover, since LHS of $L_1$ and $L_2$ are independant, there is exactly one linear condition on the $b_i$ in order to have a solution.
So your answer is correct.