# Basis of matrices with a variable

So I have these bunch of matrices I want to find the value of a to find the basis

$$\begin{pmatrix} 2 & 2 \\ 1 & -2 \\ \end{pmatrix}$$

$$\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$$

$$\begin{pmatrix} 1 & a \\ 2 & -2 \\ \end{pmatrix}$$

$$\begin{pmatrix} 1 & a \\ 1 & -1 \\ \end{pmatrix}$$

What I could do is write them in a different way

$$A\pmatrix{2\\2\\1\\-2}+B\pmatrix{0\\0\\1\\1}+C\pmatrix{1\\a\\2\\-2}+D\pmatrix{1\\a\\1\\-1}$$

Now I can find the RREF but since those are letter "A"s i dont know what to do. Someone told me I should use the determinant test. How do I use the determinant test in this situation?

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Are you trying to find the value of $a$ that makes the collection above a basis? The columns form a basis iff $\det$ of the columns is non zero. –  copper.hat Jan 31 '13 at 19:57
A slightly simpler way would be to notice that $v_1,..,v_4$ is a basis iff $v_1,v_2,v_3, v_3-v_4$ is also a basis. This eliminates one of the $a$s. –  copper.hat Jan 31 '13 at 20:01

Put up your vectors as columns in a matrix. These are linearly independent (and thus a basis) iff the determinant of this matrix is nonzero. You should be able to solve this with a quadratic equation in a.

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I want to find the set of values for $a$ that would make this a basis
? Anyways, $a$ is just a real number; don't be intimidated by the fact it's a variable rather than a decimal constant. Just do what you would normally do to test if those four vectors form a basis.
The only real complication is things like the fact $a$ is not a non-zero variable, so, e.g., you can't divide by $a$. If you get to a point where it matters whether $a$ is zero or not, you'd have to split the domain into the domain where $a$ is zero and the domain where $a$ is nonzero and treat the two cases separately.