# Science:Math Exam Resources/Courses/MATH152/April 2022/Question B3 (a)/Solution 1

Here is the argument for the standard procedure that computes the eigenvalues of a matrix. The equation

${\displaystyle Av=\lambda v}$
is equivalent to
${\displaystyle (A-\lambda \mathrm {Id} )v=0.}$
This latter equation has a non-zero solution if and only if the matrix ${\displaystyle A-\lambda \mathrm {Id} }$ is not invertible, which is equivalent to ${\displaystyle \det(A-\lambda \mathrm {Id} )=0}$. Therefore, we solve this last equation, since it is an equation of numbers (as opposed to an equation of vectors). We find the determinant to be

{\displaystyle {\begin{aligned}\det(A-\lambda \mathrm {Id} )&=\det \left[{\begin{array}{ccc}-\lambda &5&-5\\1&4-\lambda &1\\0&0&5-\lambda \end{array}}\right]\\&=(5-\lambda )\det \left[{\begin{array}{cc}-\lambda &5\\1&4-\lambda \end{array}}\right]\\&=(5-\lambda )(-\lambda (4-\lambda )-5)=(5-\lambda )(\lambda ^{2}-4\lambda -5)=(5-\lambda )(\lambda -5)(\lambda +1).\end{aligned}}}

From the factorization above, we see that there are two eigenvalues of ${\displaystyle A}$: -1 and 5, and 5 is called a repeated eigenvalue, since it shows up more than once in the factorization of ${\displaystyle \det(A-\lambda \mathrm {Id} )}$.