Science:Math Exam Resources/Courses/MATH221/December 2007/Question 07 (a)

MATH221 December 2007

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Question 07 (a)
Consider the matrix $A={\begin{pmatrix}0&-4&-6\\-1&0&-3\\1&2&5\end{pmatrix}}$. Verify that 1 is an eigenvalue of A.

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

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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. Recall that 1 is an eigenvalue of A if and only if $\det(A-I)=0$. Therefore it suffices to show that $\det(A-I)=0$. We have $A-I={\begin{pmatrix}-1&-4&-6\\-1&-1&-3\\1&2&4\end{pmatrix}}$ We would prefer to avoid doing lots of computation by finding the determinant right away here. Instead, recall that adding multiples of one row to another does not change the determinant. Therefore, adding the second row to the third, and then subtracting the first row from the second, yields the matrix ${\begin{pmatrix}-1&-4&-6\\0&3&3\\0&1&1\end{pmatrix}}$ which has the same determinant as A - I. But note that two rows are just multiples of each other. Therefore the determinant of the above matrix is zero, and hence $\det(A-I)=0$.

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