I am given a set of $2\times 2$ matrices and am asked if the set forms a basis for $M_{22}$. Now, I know to show this, I must show that the set spans $M_{22}$ and is linearly independent.
Show that $\begin{pmatrix} 3&6\\ 3&-6\\ \end{pmatrix}$ $\begin{pmatrix} 0&-1\\ -1&0\\ \end{pmatrix}$ $\begin{pmatrix} 0&-8\\ -12&-4\\ \end{pmatrix}$ $\begin{pmatrix} 1&0\\ -1&2\\ \end{pmatrix}$ forms a basis for $M_{22}$.
This implies we must show the system $$k_1M_1+k_2M_2+k_3M_3+k_4M_4=\vec{0}$$ possesses only the trivial solution.
Then we must also show any $M_{22}$ matrix $B$ can be written as a linear combination of $M_1, M_2, M_3, M_4$ $$k_1M_1+k_2M_2+k_3M_3+k_4M_4=B$$
My approach: $$k_1(3,6,3,-6)+k_2(0,-1,-1,0)+k_3(0,-8,-12,-4)+k_4(1,0,-1,2) = \vec{0}$$Form the $4\times4$ matrix: $$\begin{pmatrix} 3&6&3&-6\\ 0&-1&-1&0\\ 0&-8&-12&-4\\ 1&0&-1&2\\ \end{pmatrix}$$ I say we can simultaneously show linear independence and spanning by showing that the determinant of the above matrix is nonzero. My question is: Is this method valid?