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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?

Flimzy_Programmer
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  • What, are you serious? I'm just starting out with linear algebra and the answers in those "duplicates" are WAY above my head – Flimzy_Programmer Jan 30 '19 at 19:53
  • 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. – Dietrich Burde Jan 30 '19 at 19:54
  • 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 –  Jan 30 '19 at 19:55
  • @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 $b=1$). – Jean Marie Jan 30 '19 at 20:06

1 Answers1

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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 $$D=(s-1)^3(s+3).$$

Bernard
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