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I am trying to calculate this determinant in order to find a characteristic polynomial of a matrix, however I couldn't see any operation that I can do to make my life easier calculating this, so I tried to calculate the determinant according to column $3$ since it has a zero, but it was such a long journey and in the end I got a polynomial with degree of $3$, so I also needed to find its roots (eigen values) once again and it took alot of time.
So I'm wondering if there's any easy way to calculate this determinant since most likely I just couldn't see the right operation to do to calculate it and took the long road.
Here's the determinant: \begin{vmatrix} x-2 & -3 & -2 \\ 1 & x-3 & 0 \\ -1 & -1 & x+1\end{vmatrix}
Any feedback and tips are really appreciated, thanks in advance.

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  • $\begingroup$ You also getting $x^3-4x^2+2x+17$? I don't see a trick to factor this without using the general cubic formula. With no rational root test or Cardano Theorem I don't see a nicer way. Where is this problem coming from? Is it a typo? I don't understand the "hint" given below. $\endgroup$ Feb 5, 2021 at 17:32
  • $\begingroup$ @JohnSamples Yes, It's a problem from a practice test, there's no typo I checked it alot of times while calculating cuz it wasn't working but I noticed in the end that I was checking if this matrix is similar to another matrix, and I've already found the eigenvalues of the other matrix, so I could've found the polynomial and substitute the roots in it to check if the matrices are similar. I just found this out, sorry for wasting your time on this. I'm still unsure, should I delete the question or close or something atm? $\endgroup$
    – Pwaol
    Feb 5, 2021 at 17:42
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    $\begingroup$ It's fine, leave it up. Somebody might say something interesting about computing cubics of the above type, maybe. Also, most of my wasted time was me miscalculating the determinant haha, so I deserve it. $\endgroup$ Feb 5, 2021 at 17:43

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Hint: There a simple method, you have just to do: \begin{align} (x-2)\begin{vmatrix} x-3 & 0 \\ -1 & x+1 \end{vmatrix} + (-1)(-3)\begin{vmatrix} 1 & 0 \\ -1 & x+1 \end{vmatrix} + (-2)\begin{vmatrix} 1 & x-3 \\ -1 & -1 \end{vmatrix} \end{align}

Can you continue from here?

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  • $\begingroup$ I appreciate the hint, it opened my eye to also search for columns with none zeroes, but It's not really working for me in this case since in last small det I'm getting stuff that I can't factor out so I need to open the polynomials (might be because you wrote $x$ instead of $1$ in the left upper position of the 3rd det) or might be me messing up with calculations, anyway I appreciate the hint $\endgroup$
    – Pwaol
    Feb 5, 2021 at 17:17
  • $\begingroup$ I have corrected it. $\endgroup$ Feb 6, 2021 at 21:25

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