Science:Math Exam Resources/Courses/MATH152/April 2016/Question A 03/Solution 1

Let $A=\left[{\begin{array}{cc}1&4\\3&2\end{array}}\right].$ The eigenvalues of $A$ then are the roots of the characteristic equation $\det(A-\lambda I)=0$. Straightforward computations yield

$0=\det(A-\lambda I)=\det {\begin{bmatrix}1-\lambda &4\\3&2-\lambda \end{bmatrix}}=(1-\lambda )(2-\lambda )-12=\lambda ^{2}-3\lambda -10=(\lambda -5)(\lambda +2).$

It is easy to see that the roots are $-2$ and $5$. Therefore, the eigenvalues of the matrix $A$ are $\color {blue}-2{\mbox{ and }}5$.