# Using the determinant to find an eigenvector

I am studying maths as a hobby. I have got to the subject of linear algebra and in particular eigenvectors. I know how to find the determinant of a 3 x 3 matrix but am stumped at the following worked example in the text book.

Find the eigenvectors and corresponding eigenvalues of

$$\begin{pmatrix} 2 & -1 & 1\\ 0 & 2 & 0\\ 1 & 3 & 2\\ \end{pmatrix}$$

If $$\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}$$ is an eigenvector and $$\lambda$$ the corresponding eigenvalue then:

$$\begin{pmatrix} 2 & -1 & 1\\ 0 & 2 & 0\\ 1 & 3 & 2\\ \end{pmatrix}$$ $$\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}$$ = $$\lambda\begin{pmatrix} x\\ y\\ z\\ \end{pmatrix}$$

So the characteristic equation is

$$\begin{vmatrix}\begin{pmatrix} 2 & -1 & 1\\ 0 & 2 & 0\\ 1 & 3 & 2\\ \end{pmatrix}-\lambda\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}\end{vmatrix}=0$$

$$\Rightarrow \begin{vmatrix} 2-\lambda & -1 & 1\\ 0 & 2-\lambda & 0\\ 1 & 3 & 2-\lambda \end{vmatrix}= 0$$

Expanding the determinant gives

$$(2-\lambda)(2-\lambda)(2-\lambda)+1(-1)(2-\lambda)=0$$

This is the part I don't understand. When I calculate the determinant I get

$$(2-\lambda)(2-\lambda)(2-\lambda) + (2-\lambda)$$

Am I misunderstanding the term "expanding the determinant" or am I making some simple mistake somewhere?

• It is a simple mistake, you forgot the minus sign.
– F_M_
Jan 25, 2022 at 15:19

$$\begin{vmatrix} 2-\lambda & -1 & 1\\ 0 & 2-\lambda & 0\\ 1 & 3 & 2-\lambda \end{vmatrix}= (2-\lambda)\begin{vmatrix} 2-\lambda & 0\\ 3 & 2-\lambda \end{vmatrix} -(-1)\begin{vmatrix} 0 & 0\\ 1 & 2-\lambda \end{vmatrix}+1\begin{vmatrix} 0 & 2-\lambda \\ 1 & 3 \end{vmatrix}=$$ $$=(2-\lambda)\cdot(2-\lambda)^2+1\cdot 0+1\cdot(-1(2-\lambda))$$