This is my system of linear equations $$ \left\{ \begin{array}{c} 1x_1+2x_2-1x_3=2 \\ -3x_1+1x_2-3x_3=1\\ 4x_1+ax_2-4x_3=b \end{array} \right. $$
My Rank matrix looks like this:
$$ \begin{bmatrix} 1 & 2 & -1 & |& 2\\ -3 & 1 & 3 & |& 1\\ 4 & a & -4 & |& b\\ \end{bmatrix} $$
I have to find a and b such that the system (i) has an infinite number of solution (ii)does not have any solution (iii) does the system has a unique solution?
I know that I have to find the rank to find a and b but don't understand how to do it. I have try to reduce it by row echelon but I am not able to do so since the last row is becoming entirely zero. Could someone please guide me on solving it
Note here that the $3^{rd}$ column, is exactly $-1$ times the first one. So, taking both $x_1$ and $x_3$ as $\lambda$, you would get,
$\begin{bmatrix} 1& 2 & -1\\ -3& 1 & 3\\ 4& a &-4 \end{bmatrix} \begin{bmatrix} \lambda\\ x_2\\ \lambda \end{bmatrix}=\begin{bmatrix} 2\\ 1\\ b \end{bmatrix}$
Which you can write as
$\begin{bmatrix} 1\\ -3\\ 4 \end{bmatrix}\lambda+ \begin{bmatrix} 2\\ 1\\ a \end{bmatrix}x_2+ \begin{bmatrix} -1\\ 3\\ -4 \end{bmatrix}\lambda=\begin{bmatrix} 2\\ 1\\ b \end{bmatrix}$
Which can be simplified to
$ \begin{bmatrix} 2\\ 1\\ a \end{bmatrix}x_2+ \textbf{0}=\begin{bmatrix} 2\\ 1\\ b \end{bmatrix}$
So, if you take $x_2$ to be equal to $1$, and then $a=b$, the system has infinite solutions, i.e
$\begin{bmatrix} \lambda\\ 1\\ \lambda \end{bmatrix}\forall\lambda\in\mathbb{R}$
This answers your first question without the headache of Gaussian elimination. And I'm sure you can figure the rest of them out with the fact that the first and third columns are linearly dependent.
Hope it helps!