Proving Determinant of Mirror-Image Identity Matrix

[A.Magnus]

I was given this $n \times n$ matrix $A$ which is a mirror-image of identity matrix, ie., its non-main diagonal consists of entries of $1$, the rest of entries are $0$. I need to find out the determinant of $A$. Having experimented with $n = 2, 3, ...,$ I observed that for $n = 2 + 4k$ or $n = 3 + 4k$, then $det(A) = -1$. Otherwise $det(A) = 1$. But observation alone is not enough, I need to prove it to $n$. I was told that using induction will do it, but I don't know how to do it. Any helping hand would be very much appreciated, thank you before hand for your graciousness. ~MA

 

[Evgeny.Makarov]

Let $d_n$ be the determinant of the $n\times n$ matrix. Then $d_1=1$ and $d_{n+1}=(-1)^nd_n$. We want to prove that
\[
d_n=\begin{cases}1,&n\equiv0,1\pmod{4}\\-1,&n\equiv2,3\pmod{4}.\end{cases}
\]
Denote this statement by $P(n)$. First we check $P(1)$. Then we have to prove that for all $n$, $P(n)$ implies $P(n+1)$. Here we have to consider four cases that correspond to four possible remainders when $n$ is divided by 4.

 

[HallsofIvy]

Whenever you "swap" two rows of a determinant, you multiply it by -1. It should be easy to see that you can go from this "mirror-image" to the identity matrix by a series of swaps of two rows, starting with swapping the first and last rows, etc. If the number of rows is even, say n= 2k, there will be k such swaps. If the number of rows is odd, say n= 2k+ 1, there are still k such swaps since the middle row stays where it is.

 

[A.Magnus]

Whenever you "swap" two rows of a determinant, you multiply it by -1. It should be easy to see that you can go from this "mirror-image" to the identity matrix by a series of swaps of two rows, starting with swapping the first and last rows, etc. If the number of rows is even, say n= 2k, there will be k such swaps. If the number of rows is odd, say n= 2k+ 1, there are still k such swaps since the middle row stays where it is.

Thank you! ~MA

 

[A.Magnus]

Let $d_n$ be the determinant of the $n\times n$ matrix. Then $d_1=1$ and $d_{n+1}=(-1)^nd_n$. We want to prove that
\[
d_n=\begin{cases}1,&n\equiv0,1\pmod{4}\\-1,&n\equiv2,3\pmod{4}.\end{cases}
\]
Denote this statement by $P(n)$. First we check $P(1)$. Then we have to prove that for all $n$, $P(n)$ implies $P(n+1)$. Here we have to consider four cases that correspond to four possible remainders when $n$ is divided by 4.

Got it now, thank you for your gracious helping hand, and time. ~MA