It should now be plausible that \begin{equation*} Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{2a+b+3c+4d=0,\,-a+3b-5c-2d=0} \end{equation*} is a subspace of the vector space $M_{22}$ (Example VSM). (It is.) To find the dimension of $Z$ we must first find a basis, though any old basis will do.
First concentrate on the conditions relating $a,\,b,\,c$ and $d$. They form a homogeneous system of two equations in four variables with coefficient matrix \begin{equation*} \begin{bmatrix} 2 & 1 & 3 & 4\\ -1 & 3 & -5 & -2 \end{bmatrix} \end{equation*}
We can row-reduce this matrix to obtain \begin{equation*} \begin{bmatrix} \leading{1} & 0 & 2 & 2\\ 0 & \leading{1} & -1 & 0 \end{bmatrix} \end{equation*}
Rewrite the two equations represented by each row of this matrix, expressing the dependent variables ($a$ and $b$) in terms of the free variables ($c$ and $d$), and we obtain, \begin{align*} a&=-2c-2d\\ b&=c \end{align*}
We can now write a typical entry of $Z$ strictly in terms of $c$ and $d$, and we can decompose the result, \begin{equation*} \begin{bmatrix}a&b\\c&d\end{bmatrix}= \begin{bmatrix}-2c-2d&c\\c&d\end{bmatrix}= \begin{bmatrix}-2c&c\\c&0\end{bmatrix}+ \begin{bmatrix}-2d&0\\0&d\end{bmatrix}= c\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ d\begin{bmatrix}-2&0\\0&1\end{bmatrix} \end{equation*}
This equation says that an arbitrary matrix in $Z$ can be written as a linear combination of the two vectors in \begin{equation*} S=\set{\begin{bmatrix}-2&1\\1&0\end{bmatrix},\,\begin{bmatrix}-2&0\\0&1\end{bmatrix}} \end{equation*} so we know that \begin{equation*} Z=\spn{S}= \spn{\set{ \begin{bmatrix}-2&1\\1&0\end{bmatrix},\, \begin{bmatrix}-2&0\\0&1\end{bmatrix} }} \end{equation*}
Are these two matrices (vectors) also linearly independent? Begin with a relation of linear dependence on $S$, \begin{align*} a_1\begin{bmatrix}-2&1\\1&0\end{bmatrix}+ a_2\begin{bmatrix}-2&0\\0&1\end{bmatrix}&=\zeromatrix\\ \begin{bmatrix}-2a_1-2a_2&a_1\\a_1&a_2\end{bmatrix}&= \begin{bmatrix}0&0\\0&0\end{bmatrix} \end{align*}
From the equality of the two entries in the last row, we conclude that $a_1=0$, $a_2=0$. Thus the only possible relation of linear dependence is the trivial one, and therefore $S$ is linearly independent (Definition LI). So $S$ is a basis for $Z$ (Definition B). Finally, we can conclude that $\dimension{Z}=2$ (Definition D) since $S$ has two elements.