Question

Find a fundamental matrix for x'=(-3/4, 1/8, 1/2, -3/4)x. (I got (-3/4-r)^2-1/16=0, how do I find the roots from here?)

Answer 1

Is that a 2x2 matrix? And if it is, should I interpret that as the values left to right (row at a time), or top to bottom (column at a time)?

Answer 2

Yes, -3/4 and 1/8 on the left, 1/2 and -3/4 on the right.

Answer 3

How do I find the roots?

Answer 4

Do you mean the eigenvalues?

Answer 5

Yes.

Answer 6

So then it's just a matter of solving for \(r\):
\[\begin{align*}\left(-\frac{3}{4}-\lambda\right)^2-\frac{1}{16}&=0\\
\left(\frac{3}{4}+\lambda\right)^2&=\frac{1}{16}&&\text{factored out }(-1)^2\\
\frac{3}{4}+\lambda=\pm\frac{1}{4}
\end{align*}\]

Answer 7

What's next?

Answer 8

Well, you get \(\lambda_1=-\dfrac{1}{2}\) and \(\lambda_2=-1\), right?

Next thing to do is find the associated eigenvectors.

Answer 9

Got it.