I'd like some verification of my solution method since I don't have any solution at hand. I'm given the following matrix and asked to determine the rank of it in function of $x$:
$\begin{pmatrix} 2 & 2 & -6 & 8 \\ 3 & 3 & -9 & 8 \\ 1 & 1 & x & 4 \\ \end{pmatrix}$
I started with the fact that for any given matrix $A \in \mathbb{R}^{m \times n}$ the column space $C(A) = \mathbb{R}^m$. In this case, the column space is given by:
$$C(A) = vct \{\begin{pmatrix} 2 \\ 3 \\ 1 \\ \end{pmatrix}, \begin{pmatrix} 2 \\ 3 \\ 1 \\ \end{pmatrix}, \begin{pmatrix} -6 \\ -9 \\ x \\ \end{pmatrix}, \begin{pmatrix} 8 \\ 8 \\ 4 \\ \end{pmatrix}\}$$ Since the column space $C(A)$ is $\mathbb{R}^3$ or $\mathbb{R}^4$ depending on $x$, I set out to find for which value of $x$ the vector $(-6, -9, x)^T$ would be a linear combination of the other vectors indicating that the column space would be equal to $\mathbb{R}^3$.
Solving the following system of linear equations:
$$ \begin{pmatrix} 2 & 2 & 8 & | & -6 \\ 3 & 3 & 8 & | & -9 \\ 1 & 1 & 4 & | & x \\ \end{pmatrix} \rightarrow \begin{pmatrix} 1 & 1 & 4 & | & x \\ 0 & 0 & -4 & | & -9 - 3x \\ 0 & 0 & 0 & | & -6-2x \\ \end{pmatrix}$$
That the matrix is unsolvable when $-6-2x \neq 0$ or when $x \neq -3$. Thus I came to the conclusion that
$$ rank(A) = 3 \iff x = -3$$ $$ rank(A) = 4 \iff x \neq -3$$
I have thus no idea if this solution is correct. If not, could someone tell me how I should solve it?
It can't be correct: the rank of $A$ can never be $4$ since the column space is a subspace of $\mathbb R^3$. The first and last column are linearly independent, so the rank is at least $2$.
Depending on $x$ it might be equal to $2$ or $3$ and your calculation shows that the rank is $3$ as long as $x\neq -3$ and $2$ when $x=-3$.