Value of a such that the system has no solution/infinitely many solutions

Question

I have this system of equations here:

$Ax=b$, where:

$A=$$
\left(
\begin{array}{ccc}
\ 1 & 2 & 3 \\
\ 1 & 3 & 4\\
\ 1 & a & 5\\
\end{array}
\right)
$

$b=$$
\left(
\begin{array}{c}
\ a\\
\ 3\\
\ 3\\
\end{array}
\right)
$

I have to find the values of $a$ in which the system has no solution and the value of a in which the system has infinitely many solutions.

I wrote my augmented matrix as:

$
\left(
\begin{array}{cccc}
\ 1 & 2 & 3 & a \\
\ 1 & 3 & 4 & 3\\
\ 1 & a & 5 & 3\\
\end{array}
\right)
$

I then attempted to row reduce and then do cases but after a certain point, I am stuck. So far, I've row reduced down to:

$
\left(
\begin{array}{cccc}
\ 1 & 2 & 3 & a \\
\ 0 & 1 & 1 & 3-a\\
\ 0 & a-2 & 2 & 3-a\\
\end{array}
\right)
$

I'm not really sure what to do after this point though. I should probably row reduce more but I'm stuck on how I would show the conditions for each case of a. If anyone could guide me in the right direction, that would be much appreciated!

Answer 1

Continue reducing!

$$\left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 0 & 1 & 1 & 3-a\\ \ 0 & a-2 & 2 & 3-a\\ \end{array} \right)\longrightarrow\left( \begin{array}{cccc} \ 1 & 2 & 3 & a \\ \ 0 & 1 & 1 & 3-a\\ \ 0 & 0 & 4-a & (a-3)^2\\ \end{array} \right)$$

Can you prove that for $\;a=4\;$ you get an incongruent system (no solution)? What happens if $\;a\neq4\;$ ?