Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
add comment

2 Answers

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.

share|improve this answer
add comment

Is your question?

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.