I am working on finding a matrix to diagonalize matrix $A$. Matrix $A$ is given:
$$ \begin{pmatrix} -14 & 12 \\ -20 & 17 \\ \end{pmatrix} $$
I was taught that finding the characteristic equation is done by finding the solution to $\det(\lambda I-A) = 0$. Doing so gave me the matrix
$$ \begin{pmatrix} \lambda+14 & -12 \\ 20 & \lambda-17 \\ \end{pmatrix} $$
This leads to $(\lambda+14)(\lambda-17) = 0$ which yields eigenvalues of $\lambda=-14$ and $\lambda=17$. However, I checked these values on W|A, and they give $x^2-3x+2$ as the characteristic equation and eigenvalues of $\lambda=2$ and $\lambda=1$. Can someone tell me what step I am overlooking? Thanks in advanced.
The determinant of a 2x2 matrix $\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$ is not $ad$, but $ad-bc$. If the matrix is 3x3 the determinant has six terms, and this number grows quickly with the size of the matrix.