linear algebra - Making a determinant easier to solve to find characteristic polynomial

I have to find a characteristic polynomial for the follwoing matrix:

$$A=\left( \begin{matrix}0 & -1 & 1\\ -1 & 0 & 2 \\ 1 & 2 & 0 \end{matrix}\right)$$

My goal to find the roots of the characteristic polynomial.

The characteristic polynomial is given by solving $det(\lambda I -A)$, and then we get:

$$det(\lambda I -A)=det\left( \begin{matrix}t & 1 & -1\\ 1 & t & -2 \\ -1 & -2 & t \end{matrix}\right)$$

The final polynomial I get is $$P_A=t^3-6t+4$$

While there is a way to solve this polynomial I believe there is a way to make this determinant more "comfortable" to solve.

Can I add the third column to the second column?

$$det(\lambda I -A)=det\left( \begin{matrix}t & 0 & -1\\ 1 & t-2 & -2 \\ -1 & t-2 & t \end{matrix}\right)$$

And Then, can I do the following (and why?)

$$ det(\lambda I -A)=(t-2)*det\left( \begin{matrix}t & 0 & -1\\ 1 & 1 & -2 \\ -1 & 1 & t \end{matrix}\right)$$
What kind of elementary operations, or other arithmetic ones I can do, to make the determinant easier to solve?

Thanks,

Alan

Answer

Personally, I prefer this method
$$|A-\lambda I|=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\

-1 & -\lambda & 2 \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=-\lambda\left| \begin{array}{rr}
-\lambda & 2\\
2 & -\lambda \\
\end{array}\right|
+\left| \begin{array}{rr}
-1 & 2\\
1 & -\lambda \\

\end{array}\right|
+\left| \begin{array}{rr}
-1 & -\lambda\\
1 & 2 \\
\end{array}\right|$$
$$=-\lambda^3+6\lambda-4 =(2-\lambda)(\lambda^2+2\lambda-2)$$
$$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$
You can also triangularize the matrix first. This makes the determinant trivial to compute
$$|A-\lambda I|=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\

-1 & -\lambda & 2 \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
1 & 2 & -\lambda \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\

0 & 2-\lambda & 2-\lambda \\
\lambda & 2\lambda & -\lambda^2 \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\
0 & 2-\lambda & 2-\lambda \\
0 & 2\lambda-1 & 1-\lambda^2 \\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
-\lambda & -1 & 1\\

0 & 2-\lambda & 2-\lambda \\
0 & 0 & -\lambda^2-2\lambda+2\\
\end{array}\right|$$
$$=\left| \begin{array}{rrr}
1 & \lambda^{-1} & -\lambda^{-1}\\
0 & 2-\lambda & 2-\lambda \\
0 & 0 & \lambda^2+2\lambda-2\\
\end{array}\right|$$
$$=(2-\lambda)(\lambda^2+2\lambda-2)$$
$$=(2-\lambda)\left(\lambda-\sqrt3+1\right)\left(\lambda+\sqrt3+1\right)$$

In any case, I suggest you look up the properties of the determinant.