Eigenvalue method for homogeneous eq's

[cue928]

I am working on a problem and before I post the remaining questions on it, I want to be sure I calculated the eigenvector correctly. The eigenvalue I used was lambda = 3-4i.

<br /> \begin{bmatrix} 3-lambda &amp; -4\\ 4 &amp; 3-lambda\end{bmatrix}<br />
After substituting, the eigenvector I came up with is V = [1 i]

 

[Dick]

I am working on a problem and before I post the remaining questions on it, I want to be sure I calculated the eigenvector correctly. The eigenvalue I used was lambda = 3-4i.

<br /> \begin{bmatrix} 3-lambda &amp; -4\\ 4 &amp; 3-lambda\end{bmatrix}<br />
After substituting, the eigenvector I came up with is V = [1 i]

Fine so far.

 

[cue928]

Okay, so I have V = [1 i] and lambda = 3-4i
I came up with the following eq's that differed from the book only in the signs I placed in quotation marks...
x1(t) = e^3t[C1 cos(4t) "-" C2 sin(4t)]
x2(t) = e^3t[C1 cos(4t) "-" C2 sin(4t)]

In both cases, the book uses a "+" instead. But, I can't figure out where I dropped the sign.