[Math] How to find the eigenvalues of a matrix without a factorable trinomial

eigenvalues-eigenvectorslinear algebra

I need to find the eigenvalues/eigenvectors of the following matrix:

$$A= \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$$

I first used the equation for eigenvectors, and sorted out the determinant:

$$P_A(x) = \det (xI – A)$$
$$= \det\begin{bmatrix}x & 0\\0 & x\end{bmatrix} – \begin{bmatrix}-26 & 20\\-30 & 24\end{bmatrix}$$
$$ = \det\begin{bmatrix}x + 20 & -20\\ 30 & x – 24\end{bmatrix}$$
$$= (x + 20)(x – 24) – (-600)$$
$$= x^2 – 4x + 120$$

The problem at this point, is that the trinomial $ x^2 – 4x + 120$ can't be factored into $(x + a)(x + b)$. At least I don't think. The computer on Wolfram hasn't been able to do it, so I don't think it's factorable, which would mean I can't find the two x values that would give me the eigenvalues.

So how can I get the x values from a trinomial that isn't factoable? How else could I find the matrix's eigenvalues?

Best Answer

You didn't compute the polynomial correctly, it should be $x^2+2x-24 = (x+6)(x-4)$.