Matrices trouble, making a,b,c no sol, a unique, and infin solutions

[mr_coffee]

Hello everyone. I'm having troubles understanding if I'm doing this right.
I have the matrix
\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ {-1} & 1 & 2 & b \\ 1 & {-3} & 0 & c \\ \end{array} } \right)

I row reduced it to:
\left( {\begin{array}{*{20}c} 2 & {-3} & {-3} & a \\ 0 & {-1} & 1 & 2b+a \\ 0 & {0} & 0 & {-3b-2a+c} \\ \end{array} } \right)

I'm suppose to find, In each case find if possible conditions on the numbers a, b, and c that the given syhstem has no solution, a unique solution, or infintnety many s9lutions. So would i let b = 0, a = 0 and c equal 0 to make it have an infinit many solutions. and then let a, b and c be any number so the expression: -3b-2a+c wil not equal 0, so 0 = 4, no solution. How would i find a unique solution? if the last row is already 0 = -3b-2a+c? thanks!

 

[TD]

In order to have at least one solution, the last row cannot by all 0 and a non-zero final element. So in order to have solutions, you have to let -3b-2a+c = 0.

Can you take it from here?