Evaluating a determinant for eigenvalues

Question

I need to evaluate

$$\left| {\matrix{ {3 - \lambda } & 1 & 1 \cr 2 & {4 - \lambda } & 2 \cr 1 & 1 & {3 - \lambda } \cr } } \right|$$

A direct computation became relatively complicated. What direction should I take to make the solution more neat?

By the way, I was able to see that for $\lambda =2$ the determinant is $0$ so $2$ is an eigenvalue.

Answer 1

Applying $C_1'=C_1-C_2$

$$\left| {\matrix{ {3 - \lambda } & 1 & 1 \cr 2 & {4 - \lambda } & 2 \cr 1 & 1 & {3 - \lambda } \cr } } \right|$$

$$=\left| {\matrix{ {3 - \lambda-1 } & 1 & 1 \cr 2-( 4 - \lambda)& {4 - \lambda } & 2 \cr 1-1 & 1 & {3 - \lambda } \cr } } \right|$$

$$=(2-\lambda)\left| {\matrix{ 1 & 1 & 1 \cr -1 & {4 - \lambda } & 2 \cr 0 & 1 & {3 - \lambda } \cr } } \right|$$

Applying $C_3'=C_3-C_2,$

$$\left| {\matrix{ 1 & 1 & 1 \cr -1 & {4 - \lambda } & 2 \cr 0 & 1 & {3 - \lambda } \cr } } \right|=\left| {\matrix{ 1 & 1 & 1-1 \cr -1 & {4 - \lambda } & 2-(4 - \lambda) \cr 0 & 1 & {3 - \lambda-1 } \cr } } \right|$$

$$=\left| {\matrix{ 1 & 1 & 0 \cr -1 & {4 - \lambda } & \lambda-2 \cr 0 & 1 & {-(\lambda-2) } \cr } } \right|$$

$$=(\lambda-2)\left| {\matrix{ 1 & 1 & 0 \cr -1 & {4 - \lambda } & 1 \cr 0 & 1 & {-1 } \cr } } \right|$$

and so on

Answer 2

In fact, for $\lambda=2$ all three columns become equal, hence $2$ is a double eigenvalue. Then the third eigenvalue can be found from the fact thet the trace $3+4+3$ of the original matrix is the sum of eigenvalues.

Answer 3

$\lambda=2$ is a double eigenvalue, since the rank of the matrix $$ A-2 I = \left(\begin{array}{ccc} 1& 1 & 1\\ 2 & 2 & 2 \\ 1 & 1 & 1 \end{array}\right) $$ is equal to one - it has only one linearly independent row.

Then Trace$(A)=10$. Thus the third eigenvalue is equal to $10-2-2=6$.

Therefore

$$ \det (A-\lambda I)=(2-\lambda)^2(6-\lambda). $$