Determine the values of $k$ so that the following linear system has unique, infinite or no solutions.

Question

Determine the values of $k$ so that the following linear system has a unique solution, a infinite solutions and no solution.

$$ x + y -z + w = 1 \\ 2y + z + w = 0 \\ kz + (k-1)w = 2 \\ (k^2-k)w = k-1 $$

I've written the augmented matrix , but I'm stuck.

\begin{bmatrix} 1 & 1 & -1 & 1 & 1 \\ 0 & 2 & 1 & 1 & 0 \\ 0 & 0 & k & k-1 &2 \\ 0 & 0 & 0 & k²-k & k-1 \end{bmatrix}

I've gotten the following solution from a sotware:

$$ w = 1/k+1\\ x = 2k^2 + 4k + 9 / 2k^2+ 2k\\ y = -2k - / 2k² + 2k \\ z = k + 3 / k^2 + k$$ when $$(k \neq 1, k \neq 0, k \neq -1 )$$

But I start getting confused about how to solve the augmented matrix.

And I saw on internet similar problems beeing solved in diferent ways, but I'm not getting what I really should do here.

Answer 1

You don't need the determinant. Note that the 4th row of the REF you computed says $k(k-1)z = (k-1)$, and think about which $k$ will give you in the 4th row:

1. $0z = 0$ which is infinite solution of $z$,
2. or $0z = b$, for some $b \neq 0$ which is no solution for $z$,
3. or $az = b$ where $a \neq 0$, which gives a unique solution of $z$.

Try going from there.