Finding all values of k for exactly one, infinitely many, and no solutions

Question

I'm just looking for confirmation on whether I did this problem correctly or not.

My work:

$$ \left[ \begin{array}{ccc|c} 1&-1&0&k\\ 3&0&2&5\\ 2&-2&k^2-k&2\\ \end{array} \right]$$

R2 = R2 - 3R1

R3 = R3 - 2R1

$$ \left[ \begin{array}{ccc|c} 1&-1&0&k\\ 0&3&2&5-3k\\ 0&0&k^2-k&2-2k\\ \end{array} \right]$$

For no solution I know that $k^2 - k$ will equal to 0 while $2-2k$ is a real number

$k^2-k = 0$

$k(k-1)$

k = 0 when k = 0,1

Using this I see that $2-2k = 1$ when k = 0 so for there to be

No Solution: k = 0

Similarly, I see that both equations are 0 when k = 1 so

Infinitely Many Solutions: k = 1

With those cases out of the way when k =/= 0,1 it will have a unique solution

Unique Solution: k =/= 0,1

Is my work here correct?

Answer 1

Yes, well apart from the fact that there is no thing such as $-0$, you are correct. Cheers!

Answer 2

this is also what i have got $$(k^2-k)x_3=2(1-k)$$

if $k=1$ then we get $x_3$ as a arbitrary real number if $$k\ne 1$$ thus we get $$kx_3=2$$ so $$x_3=\frac{2}{k}$$ ($k=0$ is not possible)