Science:Math Exam Resources/Courses/MATH152/April 2011/Question B 05 (b)
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Question B 05 (b) |
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Consider the matrix below:
It is known that is an eigenvalue of A. Find the other two eigenvalues, and 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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How can we relate eigenvalues to the determinant and trace of the matrix? |
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution 1 |
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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 our matrix is with one eigenvalue being, . Also recall that the product of the eigenvalues is equal to the determinant of the matrix and the sum is equal to the trace. We have that tr(A)=3 and det(A)=2. Therefore, Therefore we have that We can rearrange for to get which has solution where . If we were to plug this in to get we'd get This is completely unsurprising since we know that complex eigenvalues occur in complex conjugate pairs. Therefore we take one of them for and one for to get that the three eigenvalues are |
Solution 2 |
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Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. If you are unfamiliar with the relationship between the determinant and trace of a matrix and its eigenvalues, then you can compute the remaining eigenvalues using the characteristic polynomial. That is, for the given matrix we can solve the polynomial equation . We find that We already know that is a solution, so we can long divide the polynomial by to get an equation for the remaining eigenvalues. If we do this long division, we get so the other eigenvalues, and , must solve the equation , which is the same equation we found in Solution 1. Therefore we know that the solution is giving the three eigenvalues |