# Thread: Finding the eigenvalues of a 3x3 matrix

1. ## Finding the eigenvalues of a 3x3 matrix

Is there a method for determining the eigenvalues of a 3x3 matrix? The only examples my book does are already in upper triangular form, so det(1-tI) isn't exactly difficult (just multiply the diagonal). Any help on this? I've provided an example below!

$\left(\begin{array}{ccc}0&-2&-3\\-1&1&-1\\2&2&5\end{array}\right)$

2. ...

3. I would think that you already know that to find the eigenvalues of any matrix, A, you find the solutions to the equation det(A- $\lambda$I)= 0.

Here, $A- \lambda I$= $
\left(\begin{array}{ccc}-\lambda &-2 & -3\\-1&1-\lambda&-1\\2&2&5-\lambda\end{array}\right)
$
.

So the equation is $\left|A- \lambda I\right|= \left|\begin{array}{ccc}-\lambda &-2 & -3\\-1&1-\lambda&-1\\2&2&5-\lambda\end{array}\right|= 0$.

It is probably simplest to expand that on the first column:
$\left|\begin{array}{ccc}-\lambda &-2 & -3\\-1&1-\lambda&-1\\2&2&5-\lambda\end{array}\right|$ $= -\lambda\left|\begin{array}{cc}1-\lambda & -1 \\ 2 & 5-\lambda\end{array}\right|$ $+ \left|\begin{array}{cc}-2 & -3 \\ 2 & 5-\lambda\end{array}\right|$ $+ 2\left|\begin{array}{cc}-2 & -3 \\ 1-\lambda & -1 \end{array}\right|= 0$.

4. That's what I was afraid it was. It just seems so messy to me... Oh well!