For what values a and b will the system have infinitely many solutions and have no solutions?

Question

Consider the following linear system $A\vec{x}=\vec{b}$ where,

$A= \begin{bmatrix} 1&1&3\\ 1&2&4\\ 1&3&a \end{bmatrix}$ and $\vec{b}= \begin{bmatrix} 2\\ 3\\ b \end{bmatrix}$ for constants a and b.

I reduced the matrix:

$A= \begin{bmatrix} 1&1&3&|2\\ 0&1&1&|1\\ 0&2&a-3&|b-2 \end{bmatrix}$ $\rightarrow \begin{bmatrix} 1&1&3&|2\\ 0&1&1&|1\\ 0&0&a-5&|b-4 \end{bmatrix}$

Does the system has infinitely many solutions if I get $0=0$ in the bottom row? and get that it has no solutions for a system where $a=5$ and $b \neq 4$?

Did I mess this up by not including part of the linear system? \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}

Answer 1

Yes, you are right. Always remember that the system of equations will have :

$1.$ a unique solution if Rank $_A$ = Rank $_{A:b} = n$ where $n$ is the order of the matrix $A$.
$2.$ no solutions if Rank $_A \neq $ Rank$_{A:b}$.
$3.$ infinite solutions if Rank $_A$ = Rank $_{A:b} < n$

Hope it helps.