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If I have this following matrix,

$$ \left( \begin{array}{ccc|c} a & 0 & b & 2 \\ a & a & 4 & 4 \\ 0 & a & 2 & b \\ \end{array} \right) $$

and $a=0$, the I am asked to find all values of $b$ for which the system has

$(i)$ a unique solution,

$(ii)$ no solutions,

$(iii)$ infinitely many solutions

So if a=0, then I have

$$ \left( \begin{array}{ccc|c} 0 & 0 & b & 2 \\ 0 & 0 & 4 & 4 \\ 0 & 0 & 2 & b \\ \end{array} \right) $$

Now from here, I would think if $b=0$ then we have no solutions. I am not sure about the other parts though.

The solution says this and I am a little confused as to why...

enter image description here

Why isn't there a unique solution? And why do we have infinitely many solutions when $b=2$? Is it because two rows are exactly the same?

If I pick $b$ as something other than 2, can't I make this unique as well?

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    If $b=2$, then 1, 2 and 3 are linearly dependent, so they contain the same information. Try to solve the system for this value $b=2$ and you'll find you don't have enough information to fully determine one solution, but there will be infinitely many solutions. – 2017-02-08
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    But that just tell me $x_3=1$ and $x_1=1$ right? – 2017-02-08
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    It just tells you $x_3=1$, but nothing about $x_1$ or $x_2$. – 2017-02-08
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    This is why there are infinitely many solutions, because no matter what values of $x_1,x_2$ you plug in, as long as $x_3=1$, the system will be satisfied. – 2017-02-08
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    Coefficient of $x_1$ and $x_2$ is $0$. Hence you can have any value for both of them. – 2017-02-08
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    Yeah I agree with you but the solutions tell me that $x_3=s$ not 1 so I think there is a typo. – 2017-02-08
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    Look at the expression for the infinitely many solutions again. The coefficient $s$ appears in front of $(0,1,0)$, not $(0,0,1)$, so it is $x_2$ that equals $s$ in the solution. – 2017-02-08
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    Ohh I see it now. They just wrote the order differently haha. Wow. Thanks. I can't believe I missed this. – 2017-02-08

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