Calculating eigenvalues for a $3 \times 3$ matrix without solving a cubic

Question

---

I am trying to find the eigenvalue for question (h), however I am unable to factor out and find the eigenvalues(roots) after I take the determinant of the characteristic equation.

Let $x$ be an eigenvalue. I am left with $$(-2-x)((1-x)(6-x)-4)-2(2(6-x)-2)+(4-(1-x))=0.$$

As for hint: says do not expand to try and solve the resulting cubic, I am looking for help on how to factorize this question so I do not need to solve a cubic polynomial

Anyway the correct answer for the eigenvalues are $x_1=1$, $x_2=-3$ and $x_3=7$.

Answer 1

Since the equation is $$ \left( { - 2 - x} \right)\left[ {\left( {1 - x} \right)\left( {6 - x} \right) - 4} \right] - 2\left[ {2\left( {6 - x} \right) - 2} \right] + 4 - \left( {1 - x} \right) = 0 $$ you get $$ \left( { - 2 - x} \right)\left( {1 - x} \right)\left( {6 - x} \right) - 4\left( { - 2 - x} \right) - 4\left( {6 - x} \right) + 8 - \left( {1 - x} \right) = 0 $$ thus $$ \left( { - 2 - x} \right)\left( {1 - x} \right)\left( {6 - x} \right) - 4\left[ {\left( { - 2 - x} \right) + \left( {6 - x} \right) - 2} \right] - (1 - x) = 0 $$ and so $$ \left( { - 2 - x} \right)\left( {1 - x} \right)\left( {6 - x} \right) - 9\left( {1 - x} \right) = 0 $$ and finally $$ \left( {1 - x} \right)\left[ {\left( { - 2 - x} \right)\left( {6 - x} \right) - 9} \right] = 0 $$ Now it is easy since you have $x^2-4x-21=0$ which has the solutions $x=7$ and $x=-3$.

Answer 2

$$A= \left[ \begin{array}{} -2&2&1\\ 2&1&2\\ 1&2&6 \end{array} \right] \iff det(A-\lambda I) = \left| \begin{array}{} -2-\lambda&2&1\\ 2&1-\lambda&2\\ 1&2&6-\lambda \end{array} \right|$$

$\\$

$$\left| \begin{array}{} -2-\lambda&2&1\\ 2&1-\lambda&2\\ 1&2&6-\lambda \end{array} \right| \iff \left| \begin{array}{} -2-\lambda&2&1\\ 0&-3-\lambda&-10+2\lambda\\ 1&2&6-\lambda \end{array} \right|$$

$\\$

$$ \iff \left| \begin{array}{} 0&6+2\lambda&1+(6-\lambda)(2+\lambda)\\ 0&-3-\lambda&-10+2\lambda\\ 1&2&6-\lambda \end{array} \right| \iff \left| \begin{array}{} 6+2\lambda&1+(6-\lambda)(2+\lambda)\\ -3-\lambda&-10+2\lambda \end{array} \right|$$

$\\$

$$ \iff 4(3+\lambda)(-5+\lambda)+(3+\lambda)+(3+\lambda)(6-\lambda)(2+\lambda)$$

$\\$

$$ \iff (3+\lambda)[4(-5+\lambda)+1+(6-\lambda)(2+\lambda)]$$

$\\$

$$ \iff (3+\lambda)[-20+4\lambda+1+12+6\lambda-2\lambda-\lambda^2]$$

$\\$

$$ \iff (3+\lambda)[-7+8\lambda-\lambda^2]$$

$\\$

$$ \iff -(3+\lambda)(\lambda-7)(\lambda-1)$$

Now it is easy to solve: $$-(3+\lambda)(\lambda-7)(\lambda-1)=0,$$

i.e we have $\lambda_1=-3, \lambda_2=7, \lambda_3=1$.