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i have the following homework problem:

Find the determinant of X(s):

X(s) = [s, 1, 1, 1], [1, s, 1, 1], [1, 1, s, 1], [1, 1, 1, s]

I know i can exploit the fact that the product of the diagonal gives me the determinant if the matrix is upper, lower or both triangular- but i don't know how i can turn it into a triangular?

If i try regular row operations, i get: X(s) = [s, 1, 1, 1], [0, s-1, 1, 1], [0, 0, s-1, 1], [1-s, 0, 0, s-1]

This is by first subtracting R1 from R2-4, and then subtracting R4 from R2-3. But i can't get any further, how do i solve this?

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marked as duplicate by Dietrich Burde, Misha Lavrov, GNUSupporter 8964民主女神 地下教會, Cesareo, YiFan Feb 10 at 1:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head $\endgroup$ – melbil Jan 30 at 19:53
  • $\begingroup$ Come on, this one is really elementary. Just what you need. You started like this yourself, so it is not "way above your head", I think.Please search before posting. $\endgroup$ – Dietrich Burde Jan 30 at 19:54
  • $\begingroup$ I think your problem is that you don't know how to properly apply the Gaussian elimination algorithm to get your matrix into an upper triangle form. Also see $(\mathrm{D}5)$ and $(\mathrm{D}6)$ of my answer to this question. I also don't know how much you already know, but using the LaPlace Expansion can be really helpful in this case $\endgroup$ – user635162 Jan 30 at 19:55
  • $\begingroup$ @melbil Please on this site, never use offending terms like "are you serious ?" The link that Dietrich Burde has given you, especially its solution by "Paul" is very understandable (juste take $a=s$ and $b=1$). $\endgroup$ – Jean Marie Jan 30 at 20:06
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Here's something simple: first exploit linearity of the determinant w.r.t. rows: $$ D=\begin{vmatrix} s&1&1&1\\1&s&1&1\\1&1&s&1\\1&1&1&s \end{vmatrix}= \begin{vmatrix} s-1&0&0&1-s\\0&s-1&0&1-s\\0&0&s-1&1-s\\1&1&1&s \end{vmatrix}= (s-1)^3 \begin{vmatrix} 1&0&0&-1\\0&1&0&-1\\0&0&1&-1\\1&1&1&s \end{vmatrix}$$ Next expand by the first row: $$D=(s-1)^3\left(1\cdot\begin{vmatrix} 1&0&-1\\ 0&1&-1\\1 &1&s \end{vmatrix}-(-1) \begin{vmatrix} 0&1&0\\0&0&1\\1&1&1 \end{vmatrix}\,\right)$$ Now the two $3\times3$ determinants are: \begin{align} &\begin{vmatrix} 1&0&-1\\ 0&1&-1\\1 &1&s \end{vmatrix}=\begin{vmatrix} 1&0&0\\ 0&1&0\\1 &1&s+2 \end{vmatrix}=s+2\\[1ex] &\begin{vmatrix} 0&1&0\\0&0&1\\1&1&1 \end{vmatrix}=1\qquad\text{(expanding by the first column)} \end{align} and ultimately, we obtain $$D=(s-1)^3(s+3).$$

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  • $\begingroup$ Thank you very much for taking the time to explain that. $\endgroup$ – melbil Jan 30 at 21:37
  • $\begingroup$ You still have to find out what linear combinations were done ;o) $\endgroup$ – Bernard Jan 30 at 21:39

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