# Find Eigenvalues of a Matrix to Diagonalize

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.

• Whoa! The determinant of $\lambda I-A$ isn’t $(\lambda+14)(\lambda-17)$; what should it be? – Brian M. Scott Dec 2 '14 at 7:43
• ...I really need sleep. Thank you! It should be $(\lambda+14)(\lambda-17)=-240$. It simplified to $x^2-3x+2$. Thanks for the kick in the pants! – rice2007 Dec 2 '14 at 7:54
• You’re welcome! (And get some sleep. :-)) – Brian M. Scott Dec 2 '14 at 7:55

## 2 Answers

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.

• Long night. I realized what the mistake was. Thank you. – rice2007 Dec 2 '14 at 7:55

your characteristic equation is not correct. Your mistake is on computing the determinate. Always remember trace of a matrix is equal to the summation of eigenvalues and the determinate is equal to the the multiplication.