Find the characteristic polynomial and eigenvalues of $A=\left[\begin{matrix} 3 & 1 & 1 \\ 0 & 5 & 0 \\ -2 & 0 & 7\end{matrix}\right]$

$$ \begin{align*} \det(A-\lambda I) &= \begin{vmatrix} 3-x & 1 & 1 \\ 0 & 5-x & 0 \\ -2 & 0 & 7-x \end{vmatrix}\\ &= (3-x) \begin{vmatrix} 5-x & 0 \\ 0 & 7-x \end{vmatrix} +1(-1) \begin{vmatrix} 0 & 0 \\ -2 & 7-x \end{vmatrix} +1\begin{vmatrix} 0 & 5-x \\ -2 & 0 \end{vmatrix}\\ &=(3-x)(5-x)(7-x)+1[0-(-2(5-x))]\\ &=(3-x)(35-5x-7x-x^2)+1(0-(-10+2x))\\ &=(3-x)(35-5x-7x-x^2)+1(0+10-2x))\\ &=(3-x)(35-5x-7x-x^2)+11-2x\\ &=105-36x+3x^2-35x+12x^2-x^3+11-2x\\ &=116-73x+15x^2-x^3\\ &=-x^3 +15x^2 -73x +116 = 0\\ \end{align*} $$


closed as off-topic by user99914, Demosthene, Najib Idrissi, Christopher, hardmath May 19 '15 at 11:47

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    $\begingroup$ Pls use MathJax! Here you will find a basic tutorial meta.math.stackexchange.com/questions/5020/… $\endgroup$ – aGer May 19 '15 at 8:44
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    $\begingroup$ @JessePFrancis, your edit has removed the work shown by the OP. $\endgroup$ – Joel Reyes Noche May 19 '15 at 8:48
  • $\begingroup$ Compute $det(A-\lambda I)=0$. This will give you the characteristic polynomial. From it you will get the eigenvalues as well. $\endgroup$ – Buzi May 19 '15 at 8:50
  • $\begingroup$ DIdn't notice! Please revert! $\endgroup$ – Jesse P Francis May 19 '15 at 8:50
  • $\begingroup$ Your question was put on hold, the message above (and possibly comments) should give an explanation why. You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote to reopen this. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) $\endgroup$ – Martin Sleziak May 19 '15 at 14:13


As there are a number of $0$s, compute $\det(A-\lambda I)$ using the rule of Sarrus. You'll have at once $\lambda -5\,$ factored out in the determinant, and you'll only have to solve a quadratic equation to find the remaining eigenvalues.


Using Laplace expansion for the second row we get:

$\begin{vmatrix} 3-x & 1 & 1 \\ 0 & 5-x & 0 \\ -2 & 0 & 7-x \end{vmatrix}= (5-x)\begin{vmatrix} 3-x & 1 \\ -2 & 7-x \end{vmatrix}= (5-x)(x^2-10x+23)$

(We used second row, because two elements of that row are zero, so the expansion has only one non-zero term.)

Can you find roots (eigenvalues) from there?

You can check your results on WA: http://www.wolframalpha.com/input/?i=%5B%5B3%2C1%2C1%5D%2C%5B0%2C5%2C0%5D%2C%5B-2%2C0%2C7%5D%5D http://www.wolframalpha.com/input/?i=x%5E2-10x%2B23%3D0

Two comments on your attempted solution (which was posted in the original version of your post and then several times removed or added back - see the revision history):

  • You made a mistake that when you replaced $1(10-2x)$ by $11-2x$.
  • If you notice that all summands contain the common fact $5-x$, you could get the result in simpler form.

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