In Example SSM22 we discovered that \begin{equation*} Q=\set{ \begin{bmatrix}-3&1\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\-4&1\end{bmatrix} } \end{equation*} is a spanning set for the subspace \begin{equation*} Z=\setparts{\begin{bmatrix}a&b\\c&d\end{bmatrix}}{a+3b-c-5d=0,\ -2a-6b+3c+14d=0} \end{equation*} of the vector space of all $2\times 2$ matrices, $M_{22}$. If we can also determine that $Q$ is linearly independent in $Z$ (or in $M_{22}$), then it will qualify as a basis for $Z$.
Let us begin with a relation of linear dependence. \begin{align*} \begin{bmatrix}0&0\\0&0\end{bmatrix} &= \alpha_1\begin{bmatrix}-3&1\\0&0\end{bmatrix}+ \alpha_2\begin{bmatrix}1&0\\-4&1\end{bmatrix}\\ &=\begin{bmatrix} -3\alpha_1 +\alpha_2 & \alpha_1\\ -4\alpha_2 & \alpha_2 \end{bmatrix} \end{align*}
Using our definition of matrix equality (Definition ME) we equate entries and get a homogeneous system of four equations in two variables, \begin{align*} -3\alpha_1 +\alpha_2&=0\\ \alpha_1&=0\\ -4\alpha_2&=0\\ \alpha_2&=0 \end{align*}
We could row-reduce the coefficient matrix of this homogeneous system, but it is not necessary. The second and fourth equations tell us that $\alpha_1=0$, $\alpha_2=0$ is the only solution to this homogeneous system. This qualifies the set $Q$ as being linearly independent, since the only relation of linear dependence is trivial (Definition LI). Therefore $Q$ is a basis for $Z$ (Definition B).