Finding x so that the rank of matrix A is 2

Question

Find all values of $x$ for which ${\rm rank}(A)=2$.

$$A=\begin{bmatrix}-2&1&0&5\\ 2&4&x&17\\1&3&6&12\end{bmatrix}$$

I've tried row reducing the matrix but I can't seem to get the right answer.

Answer 1

If $\text{rk}(A)=2$, the second row is a linear combination of the first row and the third row. That implies that $(2,4,17)$ is a linear combination of $(-2,1,5)$ and $(1,3,12)$, but since $$\det\begin{pmatrix}2 & 4 & 17 \\ -2 & 1 & 5 \\ 1 & 3 & 12 \end{pmatrix}\equiv \det \begin{pmatrix}0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 0 \end{pmatrix}\equiv 1\pmod{2}$$ the matrix on the left is invertible and there is no value of $x$ for which $\text{rk}(A)=2$.