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What is the determinant of the following matrix:

$$A=\begin{bmatrix}1&1&...&0\\1&...&0&1\\.&.&.&.\\0&1&...&1\end{bmatrix}$$

The size is $n\times n$ and the entries are ones except for zeros on the anti-diagonal.

My answer is $(-1)^{n-1}(n-1)$ but I am not sure.

hardmath
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T.fizuty
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    I'm not sure what the general entry pattern is; is it just zeros on the off diagonal and ones elsewhere? – Dave Jul 10 '19 at 16:00
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    Sorry, can't figure out the pattern! Is it all ones except the "anti-diagonal"? – David Jul 10 '19 at 16:01
  • Welcome to MSE. Please use MathJax; I don't understand your formula. – saulspatz Jul 10 '19 at 16:04
  • yes @Dave it just zeros on the off diagonal and ones elsewhere – T.fizuty Jul 10 '19 at 16:08
  • For n=3 I get $-2$. For n=4, I get $-3$ so I think your conjecture does not hold – David Jul 10 '19 at 16:18
  • I'm going to edit the Question, since you agree that "main diagonal" does not describe the location of the zero entries. The phrase "anti-diagonal" is more usual. Feel free to rollback my edit if you think it unsuitable, and please give some indication of what you've already done to solve the problem (e.g. some small examples or search for similar problems). – hardmath Jul 10 '19 at 16:42
  • @hardmath I'm used to hearing "minor diagonal". – saulspatz Jul 10 '19 at 16:44
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    Note that, after $\lfloor \frac{n}{2}\rfloor$ row swaps, we get $\det(A)=(-1)^{\lfloor n/2\rfloor}\det(J-I)$ where $J$ is the $n\times n$ matrix of all $1$'s and $I$ is the $n\times n$ identity matrix (so this just swaps the zeros to be on the main diagonal; maybe this makes it easier to visualize). – Dave Jul 10 '19 at 16:50

2 Answers2

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As I noted in my comment, we have $$\det(A)=(-1)^{\lfloor n/2\rfloor}\det(J-I)$$ where $J$ is the $n\times n$ matrix of all $1$'s and $I$ is the $n\times n$ identity matrix. Now you can use (for instance) this question to finish (the answer in this one provides the solution for $n=101$, but I think it is clear how to generalize from the answer).

The final answer can take the form $$\det(A)=(-1)^{\lfloor n/2\rfloor}(-1)^{n+1}(n-1).$$

Dave
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The determinant of $A$ is given by $(-1)^{\frac{2n+1+(-1)^{n+1}}{4}+1}(n-1)$ for $n>1$.

When $n=2,3,4,5,6,7,8,9,...$ then the determinant is $1,-2,-3,4,5,-6,-7,8,9,...$ respectively.

The part $(-1)^{\frac{2n+1+(-1)^{n+1}}{4}+1}$ tells us that the sign alternates every two terms.

Hussain-Alqatari
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  • The answer should make sence – T.fizuty Jul 10 '19 at 16:35
  • @T.fizuty I have edited my answer. Hope this helps. – Hussain-Alqatari Jul 10 '19 at 16:41
  • The sign is $+1$ when $n\equiv1,2\pmod{}4$ and $-1$ when $n\equiv0,3\pmod{4}$. Do you have a general proof, or just a few examples? – saulspatz Jul 10 '19 at 16:43
  • @saulspatz consider the numerator that appears in the exponent, for $n=1,2,3,4,5,6,7,8,...$, then when that numerator is divided by $4$, then we get $1,1,2,2,3,3,4,4,...$. The $+1$ is just to adjust the sign when $(-1)$ is raised to that power. By finite mathematical induction, it is easy to prove. You can try it. – Hussain-Alqatari Jul 10 '19 at 16:50