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

MATH152 April 2016

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Question A 03
Find the eigenvalues of the $2\times 2$ matrix below. It is not necessary to find the corresponding eigenvectors. $\left[{\begin{array}{cc}1&4\\3&2\end{array}}\right]$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Let $A$ be a $2\times 2$ matirx. One can find the eigenvalues of $A$ by solving the following characteristic equation: $det(A-\lambda I)=0.$

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $\color {blue}-2{\mbox{ and }}5$.