Give a square matrix $A$ = ${(a_{ij})_{n \times n}}$, evaluate the $\det(A)$

Question

Give a square matrix $A$ = ${(a_{ij})_{n \times n}}$, evaluate the $\det(A)$ if:

${a_{ij}}$ = $(-1)^{|i-j|}$ if $i \neq j$, and ${a_{ij}}$ = 2 if $i=j$

I may have found the solution to the question is that $$\det(A) = n +1$$

But I don't know how to prove this answer. At first, I tried to use mathematical induction and co-factor expansion but it's don't work, and I read a similar question on the site but it doesn't give me a clear method for proving this. So can anyone fact-check my finding and help with this, thank you in advance.

the reference question: " https://math.stackexchange.com/questions/3166880/evaluate-det-a-for-an-n-times-n-matrix-a-a-ij-when"

Answer 1

Gaussian elimination, the definition of determinant and the fact for a upper-triangular matrix the determinant is the product of the values on the principal diagonal, give other way in order to find the $|A|$ where $A=[a_{ij}]\in M_{n\times n}({\bf R})$ is a matrix defined by the law $$a_{ij}=\begin{cases}(-1)^{|i-j|},\quad i\not=j,\\2,\quad i=j\end{cases}$$

Firstly, the control of the row exchanges (permutations $\pi$) can be done with the signum, which we denote as ${\rm sgn}(\pi)$. Now, the determinant of the $n\times n$ matrix is given by \begin{align*}&=\begin{vmatrix}2&-1&1&-1&1&\cdots\\-1&2&-1&1&-1&\cdots\\1&-1&2&-1&1&\cdots\\-1&1&-1&2&-1&\cdots\\1&-1&1&-1&2&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&2\end{vmatrix}={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\2&-1&1&-1&1&\cdots\\1&-1&2&-1&1&\cdots\\-1&1&-1&2&-1&\cdots\\1&-1&1&-1&2&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&2\end{vmatrix}\\ &={\rm sgn}(\pi)\begin{vmatrix}\boxed{-1}&2&-1&1&-1&\cdots\\2&-1&1&-1&1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&\boxed{3}&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1\end{vmatrix}\\ &={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&\boxed{3}&-1&1&-1&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}={\rm sgn}(\pi) \begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&\boxed{-4}&1&-1&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}\\ &={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&\boxed{-4}&1&-1&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&\boxed{5}&-1&\cdots\\0&0&0&1&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}\\ &={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\0&0&0&\boxed{5}&-1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\0&0&0&0&\boxed{-6}&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&1 \end{vmatrix}\\ &={\rm sgn}(\pi)\begin{vmatrix}-1&2&-1&1&-1&\cdots\\0&1&1&0&0&\cdots\\0&0&1&1&0&\cdots\\0&0&0&1&1&\cdots\\0&0&0&0&1&\cdots\\\vdots&\vdots&\vdots&\vdots&\vdots&\boxed{(-1)^{n}(n+1)} \end{vmatrix}\\&={\rm sgn}(\pi)(-1)(-1)^{n}(n+1)\\&={\rm sgn}(\pi)(-1)^{n+1}(n+1) \end{align*} Thus $\det(A)={\rm sgn}(\pi)(-1)^{n+1}(n+1)$.

• If $n$ is odd number, we have ${\rm sgn}(\pi)=1$. Thus, $\det(A)=n+1$.
• If $n$ is even number, we have ${\rm sgn}(\pi)=-1$. Thus. $\det(A)=n+1$.

Therefore, $\det(A)=n+1$.

Answer 2

The matrix $((-1)^{i-j})_{1\le i,j\le n}$ is easy to diagonalize, and its eigenvalues are $0$ with multiplicity $n-1$ and $n$ with multiplicity $1.$ This proves your conjecture.