# Finding determinant (using row reduction)

I'm trying to calculate the following

$\det \begin{pmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{pmatrix}$

and I'm sure there must be a way to get this as an upper/lower triangular matrix. However whichever way I try looking at it, I can't see how to make it into one, if you minus the bottom row from each of the other rows you're still left with the bottom row and the same goes for columns. Is there some clever trick to note here? I've also briefly entertained the idea of proving a general form (since I'm sure one exists for a matrix of this form) but it seems a bit overkill...

Many thanks!

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Is row reduction a must? – Git Gud Mar 24 '13 at 23:49
@GitGud Nope, I just thought it'd be easier than using minors. – Noble. Mar 24 '13 at 23:54
in general, you might try strategies with smaller matrices, and see which ones start to give you strategy as you increase the size of the matrix. that is, try to consider a simpler and similar problem. This can help with several types of problems encountered in a linear algebra course. – Brady Trainor Mar 25 '13 at 3:30

Hints: make zeros below the entry 2-1 on the first column and develop wrt it and etc.:

$$\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ -1 & -1 & x-1 & -1 & -1 \\ -1 & -1 &-1 & x-1 & -1 \\ -1 & -1 & -1 & -1 & x-1 \end{vmatrix}=\begin{vmatrix} x-1 & -1 & -1 & -1 & -1\\ -1 & x-1 & -1 & -1 & -1 \\ 0 & -x & x & 0 & 0 \\ 0 & -x &0 & x & 0 \\ 0 & -x & 0 & 0 & x \end{vmatrix}=$$

$$=(x-1)\begin{vmatrix} x-1 & -1 & -1 & -1 \\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\ -x & 0 & 0 & x \end{vmatrix}+\begin{vmatrix} -1 & -1 & -1 & -1\\ -x & x & 0 & 0 \\ -x &0 & x & 0 \\ -x & 0 & 0 & x \end{vmatrix}\stackrel{\text{develop by Col. 2}}=$$

$$=(x-1)\left(\,\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}+x\begin{vmatrix} x-1 & -1 & -1 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}\,\right)+\begin{vmatrix} -x & 0 & 0 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}+x\begin{vmatrix} -1 & -1 & -1 \\ -x & x & 0 \\ -x & 0 & x \end{vmatrix}=$$

$$=(x-1)\left(-x^3+x(x^2(x-1)-2x^2)\right)+(-x^4-3x^3)=$$

$$=(x-1)\left(-x^3+x(x^3-3x^2)\right)-x^3(x+3)=(x-1)(x^4-4x^3)-x^3(x+3)=$$

$$=x^3(x-1)(x-4)-x^3(x+3)=\ldots$$

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I believe we have conflicting answers. Mine yields $0$ with multiplicity $4$. – Git Gud Mar 25 '13 at 0:14
Knowing myself I'd say there's a good probability my calculations are wrong, but it doesn't really matter imo: it's up to the OP to follow hints, advices (this was my target) and find out the actual answer... – DonAntonio Mar 25 '13 at 0:16
..and I think you're right: the given matrix is $\,\det(xI-A)\,$ , with $\,A=(a_{ij})\;,\;\;a_{ij}=-1\,\,\,\forall i\,,\,j\,$ , so it actually is the characteristic pol. of a $\,5\times 5\,$ matrix of rank $1$... – DonAntonio Mar 25 '13 at 0:19
@DonAntonio Thank you Don, the combination of row reduction and minors has worked well here! – Noble. Mar 25 '13 at 2:37
@DonAntonio - I also forgot to mention, the mistake in your solution is you're missing an $x$ in front of the last matrix when you developed by the second column. As you said though, it isn't a big deal (but I thought I'd tell you in case you wanted to correct the post). – Noble. Mar 25 '13 at 2:47

By Gaussian elimination, using the the first entry of the fifth row as the first pivot, we get that $$\begin{vmatrix} -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \\ -1 & -1 & -1 & -1 & -1+x \end{vmatrix} = \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ -1+x & -1 & -1 & -1 & -1 \\ -1 & -1+x & -1 & -1 & -1 \\ -1 & -1 & -1+x & -1 & -1 \\ -1 & -1 & -1 & -1+x & -1 \end{vmatrix} = \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & x & 0 & 0 & -x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix} = \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & x & 0 & -x \\ 0 & 0 & 0 & x & -x \end{vmatrix} = -\begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & x & -x \end{vmatrix} = \begin{vmatrix} -1 & -1 & -1 & -1 & -1+x \\ 0 & -x & -x & -x & x^2-2x \\ 0 & 0 & -x & -x & x^2-3x \\ 0 & 0 & 0 & -x & x^2-4x \\ 0 & 0 & 0 & 0 & x^2-5x \end{vmatrix} = (-1)(-x)^3(x^2-5x) = x^4(x-5).$$

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Indeed, thank you. – Branimir Ćaćić Mar 25 '13 at 17:01