I am given matrices:
\begin{pmatrix}1&2\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ \:1&0\end{pmatrix}\begin{pmatrix}1&0\\ a\:&-2\end{pmatrix}
and asked to determine the value of $a$ such that the above matrices are linearly independent. I know how to work with vectors: put them together into a matrix and use row reduction, if no free columns then the vectors are linearly independent. But how to do this with matrices? Thank you.
Your question is the same as asking
For which $a$ will $x,y,z=0$ be the only solutions to the equation $$x\begin{pmatrix}1&2\\ \:0&1\end{pmatrix}+y\begin{pmatrix}1&0\\ \:1&0\end{pmatrix}+z\begin{pmatrix}1&0\\ \:a&-2\end{pmatrix}=0 \,?$$
Looking at the top-right entry, $x$ must be zero already. Then looking at the bottom-right entry, $z$ must be zero. So $y$ is zero too, and we did all this without even considering $a$, so $a$ can be anything.
You can consider the matrix:
$$ A=\left[ \begin{array}{cccc} 1&2&0&1\\ 1&0&1&0\\ 1&0&a&-2\\ \end{array} \right] $$ and because $A$ is of size $3\times4$, then
$$ A \quad\text{is of full rank} \quad \text{if and only if}\ \ \det(AA^{t})\neq 0. $$
