linear algebra - Find the characteristic polynomial $P_A(lambda)$ of this matrix

Consider the following matrix A:

$$A= \begin{pmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \\ \end{pmatrix}$$

I have to find the characteristic polynomial $P_A(\lambda)$ using the following approach:

$$P_A(\lambda)=\det(A-\lambda I)$$

I worked out the first part:

$$\begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix}$$

But then I get stuck calculating the determinant with all those $\lambda$ floating around.

Help? :( The answer is supposed to be $P_A(\lambda)=-(\lambda-1)(\lambda+2)^2$

Answer

You could use properties of determinants to avoid having to factor a cubic afterwards; for example:

• subtract the last column from the first two;


• add the first two rows to the third:

$$\begin{vmatrix} -1-\lambda & 1 & 1 \\ 1 & -1-\lambda & 1 \\ 1 & 1 & -1-\lambda \\ \end{vmatrix}=\begin{vmatrix} -2-\lambda & 0 & 1 \\ 0 & -2-\lambda & 1 \\ 2+\lambda & 2+\lambda & -1-\lambda \\ \end{vmatrix}=\begin{vmatrix} -2-\lambda & 0 & 1 \\ 0 & -2-\lambda & 1 \\ 0 & 0 & 1-\lambda \\ \end{vmatrix}$$
This is the determinant of a diagonal matrix, so it is the product of the diagonal elements:
$$\left( -2-\lambda \right)^2\left( 1-\lambda \right) \color{blue}{ = 0 \iff \lambda = -2 \;\vee\; \;\lambda = 1}$$