# Finding Eigenvalues of 3 by 3 matrix

I am trying to find the eigenvalues of the following matrix $A$:

$$A={\begin{bmatrix} 5 & 0 & 1 \\ -2 & 1 & 0 \\ -3 & 0 & 1 \end{bmatrix}}.$$

Since $Ax = \lambda x$, $(\lambda I - A)x = 0$ which means $\det(\lambda I-A) =0$.

From that i tried to get the determinant of $(\lambda I-A)$, $$\det{\begin{bmatrix} \lambda-5 & 0 & 1 \\ -2 & \lambda-1 & 0 \\ -3 & 0 & \lambda-1 \end{bmatrix}}.$$

Expanding along the first row gives me

$$(\lambda-5)(\lambda-1)(\lambda-1)+(-1)(-3\lambda+3),$$

but when I tried to solve it, I don't seem to get the correct eigenvalues 1,2,4. I am trying to understand the process better. Can someone point out where did I went wrong in my calculation?

• You haven't computed $\lambda I - A$ correctly: all the non-diagonal entries should be the negative of what you've written. – Alex Zorn Nov 22 '17 at 19:57
• my silly mistake.. thanks for pointing out – Chris Aung Nov 22 '17 at 20:09

$$\det(\lambda I - A) = \det\left[\begin{array}{ccc}\lambda -5 & 0 & -1\\2 & \lambda-1 & 0\\3 & 0 & \lambda-1\end{array}\right] = (\lambda-5)(\lambda-1)^2 + (-1)(-3)(\lambda-1).$$
Now just factor the polynomial; start by noticing the common $(\lambda-1)$ term: $$(\lambda-5)(\lambda-1)^2 + (-1)(-3)(\lambda-1) = (\lambda-1)(\lambda^2 -6\lambda+8)=(\lambda-1)(\lambda-2)(\lambda-4).$$
You can expand the determinant by the second column, getting $$(\lambda-1)\bigl((\lambda-5)(\lambda-1)+3\bigr)=(\lambda-1)(\lambda^2-6\lambda+8)=(\lambda-1)\bigl((\lambda-3)^2-1\bigr)),$$ whence the roots: $\lambda=1$, $\lambda=3\pm1$.