Dimension of a subspace of $3\times 3$ matrices

Question

The question is asking to find the dimension of the subspace $W$, where, $V = M_{3,3}$, $$ W = \{A \in V: A\mathbf{x}=\mathbf{0}\} $$ where $\mathbf{x} = \langle 1, 2, 3\rangle^T$ (column vector).

I defined an arbitrary matrix $A$ which contains the coefficients $a,b,\dots,i$ such that $A\mathbf{x} = \mathbf{0}$ , which gives us a three equation with 9 unknowns.

As all the variables are unknown does that means we have 9 free variables and thus the dimension is 9?

Answer 1

HINT

If the columns of $A$ are $A_1, A_2, A_3$ your constraint says $$ 1A_1 +2A_2 + 3A_3 = 0 \iff A_1 = -2A_2 - 3A_3, $$ which means that your matrix looks like $$ A = \left[ -2A_2-3A_3,A_2,A_3 \right], $$ so how many free parameters do you really have?

Answer 2

Matrix $A$ is of the form

$\begin{pmatrix} a&b&(-a-2b)/3\\c&d&(-c-2d)/3\\e&f&(-e-2f)/3\\\end{pmatrix}$.

Observe that only $6$ free variables have been used.