# How to determine the value of a variable in a matrix to make it linearly independent of two other given matrices.

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.

• I am not entirely sure what you mean. Do you mean that you want $a$ such that $xA+yB=C$, where $x$ and $y$ are scalars and $A$, $B$ and $C$ correspond to your matrices? If that is the case, then for every $a$ there is no $x$ and $y$ such that the equality $xA+yB=C$ holds.
– EdG
Commented Nov 5, 2017 at 3:00

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

$$A \quad\text{is of full rank} \quad \text{if and only if}\ \ \det(AA^{t})\neq 0.$$

• Can $a$ be any number then? Commented Nov 5, 2017 at 3:03
• Any number $a$ such that $\det(AA^{t})\neq 0$ when you work in $\mathbb{R}$. When you work in $\mathbb{C}$ you consider $\det(AA^{*})\neq 0$. Commented Nov 5, 2017 at 3:07
• I don't know what C is, working in R so any will do. Thank you very much! Commented Nov 5, 2017 at 3:08