How do I find k so that this rank is true? $$
        \begin{bmatrix}
        3 & 2 & -2 \\
        1 & 1 & -1 \\
        -1 & 2 & k \\
        \end{bmatrix}
$$
Find the values k for which rank(A) = 3
From what I have tried myself, I have brought it down to row-echelon form where it looks something like this:
$$
        \begin{bmatrix}
        1 & 1 & -1 \\
        0 & 1 & -1 \\
        0 & 0 & k+2 \\
        \end{bmatrix}
$$
Through various row operations, but I seem to be stuck on whether or not I'm doing the right thing. Thanks!
To start:
$$
        \begin{bmatrix}
        3 & 2 & -2 \\
        1 & 1 & -1 \\
        -1 & 2 & k \\
        \end{bmatrix}
$$
Interchanging Rows 1 and 2.
$$
        \begin{bmatrix}
        1 & 1 & -1 \\
        3 & 2 & -2 \\
        -1 & 2 & k \\
        \end{bmatrix}
$$
then, Row 2 minus 3 times Row 1 and Row 3 plus Row 1.
$$
        \begin{bmatrix}
        1 & 1 & -1 \\
        0 & -1 & 1 \\
        0 & 3 & k-1 \\
        \end{bmatrix}
$$
then,  multiply Row 2 by (-1)
$$
        \begin{bmatrix}
        1 & 1 & -1 \\
        0 & 1 & -1 \\
        0 & 3 & k-1 \\
        \end{bmatrix}
$$
Finally, Row 3 minus 3 times Row 2.
$$
        \begin{bmatrix}
        1 & 1 & -1 \\
        0 & 1 & -1 \\
        0 & 0 & k+2 \\
        \end{bmatrix}
$$

Hint:-
Rank$=3$ iff $\det(A)\ne 0$
