3
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$$A = \left(\begin{matrix} \lambda - 1 & -1 & -1 \\ 1 & \lambda - 3 & 1 \\ -3 & 1 & \lambda + 1 \\ \end{matrix}\right)$$

$$\det A = \begin{vmatrix}A\end{vmatrix} = (\lambda - 1) \begin{vmatrix} \lambda - 3 & 1 \\ 1 & \lambda + 1 \\ \end{vmatrix} + 1\begin{vmatrix} 1 & 1 \\ -3 & \lambda + 1 \\ \end{vmatrix} - 1\begin{vmatrix} 1 & \lambda - 3 \\ -3 & 1 \\ \end{vmatrix}\\ = ((\lambda - 1)(\lambda - 3)(\lambda + 1)) - (1 + (\lambda + 1) + 3) + (1 +3\lambda - 9) \\ = (\lambda - 1)(\lambda - 3)(\lambda + 1) + (\lambda + 1) + 2(\lambda - 3)$$

The solution says that it is $(\lambda - 2)(\lambda + 2)(\lambda - 3)$. I feel like I am so close, but I don't get what I am supposed to do to get to the solution.

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3
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What is wrong with this solution? Here we line the two copy of the same matrix side by side and draw blue and red lines and add all products of numbers on the blue lines and substract products of numbers on the red lines. The solution is off by 4.

enter image description here

The determinant is thus

$$(\lambda-1)(\lambda-3)(\lambda+1)\\+(-3)(1)(-1)\\+(-1)(1)(1)\\-(-1)(\lambda-3)(-3)\\-(-1)(1)(\lambda+1)\\-(\lambda-1)(1)(1)$$

That is

$$(\lambda-1)(\lambda-3)(\lambda+1)+3-1-3(\lambda-3)+(\lambda+1)-(\lambda-1)\\ =(\lambda-1)(\lambda-3)(\lambda+1)+2-3(\lambda-3)+2\\ =(\lambda-1)(\lambda-3)(\lambda+1)+4-3(\lambda-3)\\ =(\lambda-3)\left[(\lambda-1)(\lambda+1)-3\right]+4\\ =(\lambda-3)(\lambda^2-1-3)+4\\ =(\lambda-3)(\lambda^2-4)+4\\ =(\lambda-3)(\lambda-2)(\lambda+2)+4$$

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  • $\begingroup$ You are using the rule of Sarrus, and your solution is actually correct. But you should write this in LaTeX. $\endgroup$ – Jean-Claude Arbaut Dec 5 '18 at 21:57
  • $\begingroup$ Really nice, maybe it is difficult make the lines with MathJax! I don't know how to obtain that :( $\endgroup$ – gimusi Dec 5 '18 at 22:07
  • $\begingroup$ Yes, very nice. This was very painful to keep track of, but I got the same answer and was able to follow along the math. $\endgroup$ – Evan Kim Dec 5 '18 at 22:20
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    $\begingroup$ I reworked your answer in LaTeX (except the matrices, made with Excel), can you check and confirm it's ok? $\endgroup$ – Jean-Claude Arbaut Dec 5 '18 at 22:27
  • $\begingroup$ Great! thank you soo much! The red line is not correct there should be one for (-1)(lambda-3)(-3) and there is none for (-1)(lambda-3)(-3). $\endgroup$ – mathpadawan Dec 6 '18 at 2:17
4
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$$=(\lambda - 1)\Big((\lambda - 3)(\lambda + 1)-1\Big) + ((\lambda + 1) + 3) -(1 +3\lambda - 9) $$

$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) + (\lambda + 4) -(3\lambda - 8) $$

$$=(\lambda - 1)\Big(\lambda ^2 - 2\lambda -4\Big) - 2\lambda +12 $$

$$=\lambda ^3 - 3\lambda^2 -2\lambda +4 - 2\lambda +12 $$

$$=\lambda ^3 - 3\lambda^2 -4\lambda +16$$

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  • $\begingroup$ I am trying to find the characteristic polynomial of A. I just posted this part of the problem because I thought the characteristic polynomial was just finding the determinant. I think I did |lambda I - A| wrong...let me check $\endgroup$ – Evan Kim Dec 5 '18 at 21:21
  • $\begingroup$ @EvanKim Thanks for your kind inputs in your question post and here. Just to let you know, it's good to contain any questions on the initial question form. Regarding to your questions, the eigenvalues are given by the $\lambda$s that nullify the determinant, simply meaning that you have to find the solutions to the equation $\det(\lambda I - A) = 0$. On the initial question, greedoid's answer follows exactly where you were mistaken. $\endgroup$ – Rebellos Dec 5 '18 at 21:36
  • $\begingroup$ so $λ^3−3λ^2−4λ+16$ is in fact $(λ−2)(λ+2)(λ−3)?$ $\endgroup$ – Evan Kim Dec 5 '18 at 21:47
  • $\begingroup$ @EvanKim No, if only because the constant term of your expression after expansion is obviously $12$. $\endgroup$ – Jean-Claude Arbaut Dec 5 '18 at 22:07
1
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HINT

We can simplify a little bit summing the third to the first row

$$\begin{vmatrix} \lambda - 1 & -1 & -1 \\ 1 & \lambda - 3 & 1 \\ -3 & 1 & \lambda + 1 \\ \end{vmatrix}= \begin{vmatrix} \lambda -4 & 0 & \lambda \\ 1 & \lambda - 3 & 1 \\ -3 & 1 & \lambda + 1 \\ \end{vmatrix}$$

and using Laplace expansion in the first row to obtain

$$\begin{vmatrix} \lambda -4 & 0 & \lambda \\ 1 & \lambda - 3 & 1 \\ -3 & 1 & \lambda + 1 \\ \end{vmatrix} =(\lambda -4) \cdot \begin{vmatrix} \lambda - 3 & 1 \\ 1 & \lambda + 1 \\ \end{vmatrix} +\lambda\cdot \begin{vmatrix} 1 & \lambda - 3 \\ -3 & 1 \\ \end{vmatrix}$$

Can you proceed from here?

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    $\begingroup$ Posts should be posted in final form. Comments should be limited to giving information relevant to the post. Discussions about votes are mostly off-topic. Excessive "+1 great posts" as a matter of fact can lead to suspensions. $\endgroup$ – quid Dec 5 '18 at 21:52
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    $\begingroup$ The text in the comment box explicitly says that these comments should be avoided and they are also discouraged in other ways. To do it sometimes will not cause issues, to do it all the time/often can lead to problems. $\endgroup$ – quid Dec 5 '18 at 21:59
  • $\begingroup$ Why the downvote for a correct answer? $\endgroup$ – gimusi Dec 7 '18 at 13:02

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