Using the determinant to find an eigenvector

Question

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?

Answer 1

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