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

MATH221 December 2007

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Question 07 (b)
Consider the matrix $A={\begin{pmatrix}0&-4&-6\\-1&0&-3\\1&2&5\end{pmatrix}}$. Find all eigenvalues 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. We already verified that 1 is an eigenvalue of A in part (a). To determine other possible eigenvalues, we solve $\det(A-\lambda I)=0$ for $\lambda$. We have $A-\lambda I={\begin{pmatrix}-\lambda &-4&-6\\-1&-\lambda &-3\\1&2&5-\lambda \end{pmatrix}}$ Let us make use of the fact that adding multiples of one row to another does not change the determinant to avoid lots of computation. Adding the second row of $A-\lambda I$ to the third, while subtracting $\lambda$ times the second row from the first, yields the matrix $B={\begin{pmatrix}0&\lambda ^{2}-4&3\lambda -6\\-1&-\lambda &-3\\0&2-\lambda &2-\lambda \end{pmatrix}}={\begin{pmatrix}0&(\lambda +2)(\lambda -2)&3(\lambda -2)\\-1&-\lambda &-3\\0&2-\lambda &2-\lambda \end{pmatrix}}$ which has the same determinant has $A-\lambda I$. By expanding down the first column, the determinant is $\det(A-\lambda I)=\det B=(\lambda +2)(\lambda -2)(2-\lambda )-3(\lambda -2)(2-\lambda )=(\lambda -2)^{2}(1-\lambda )$ Therefore, the other eigenvalue of A is 2.

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