Cfrgtkky

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!

Continue reducing!

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

Can you prove that for $;a=4;$ you get an incongruent system (no solution)? What happens if $;aneq4;$ ?

After subtracting $a-2$ times row 2 from row 3, the (3,3) entry is $2-(a-2)=4-a.$ That is zero iff $a=4.$ So if $a=4$ there will be 0 or infinitely many solutions. In row 3 column 4 it will be $(3-a)-(3-a)(a-2)=(3-a)[1-(a-2)]=(3-a)^2$ which isn't $0$ when $a=4.$ So if your calculation so far is right, no solutions.

Thanks to Don Antonio for fix.