Diagonal entries are zero, others are $1$. Find the determinant.

Question

$\det\begin{vmatrix} 0 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 0 \end{vmatrix}=?$

Attempt:

First I tried to use linearity property of the determinants such that $$\det\binom{ v+ku }{ w }=\det\binom{v }{ w }+k\det \binom{ u }{ w }$$

$v,u,w$ are vectors $k$ is scalar.

I have tried to divide it into $n$ parts and tried to compose with sense but didn't acomplish.

Second I tried to make use of "Row Reduction" i.e. adding scalar multiple of some row to another does not change the determinant, so I added the all different rows into other i.e. adding $2, 3,4,\dots,n$th row to first row and similarly doing for all rows we got

$$\det\begin{vmatrix} 0 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 0 \end{vmatrix}=\det\begin{vmatrix} n-1 & \cdots & n-1& n-1 & n-1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ n-1 & \cdots & n-1 & n-1 & n-1 \end{vmatrix}=0$$

The last determinant is zero (I guess) so the given determinant is zero?

I don't have the answer this question, so I am not sure. How to calculate this determinant?

Answer 1

$$\det A_n=\begin{vmatrix} 0 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 0 \end{vmatrix}=\det\left(\begin{pmatrix} 1 & \cdots & 1& 1 & 1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ 1 & \cdots & 1 & 1 & 1 \end{pmatrix}-I_n\right)=\det(B_n-I_n)$$ Now, $B_n$ has rank $1$, so $0$ is a $(n-1)$ fold eigenvalue of $B_n$ Hence, $(-1)$ is a $(n-1)$ fold eigenvalue of $A_n$. The other eigenvalue is $(n-1)$, since it's the sum of elements in each row. Hence, the determinant is $(-1)^{(n-1)}(n-1)$.

Answer 2

Notice that

$$\begin{bmatrix} 0 & 1 & \cdots & 1\\ 1 & 0 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 &\cdots & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = (n-1)\begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix}$$

$$\begin{bmatrix} 0 & 1 & \cdots & 1\\ 1 & 0 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 &\cdots & 0 \end{bmatrix}\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0 \end{bmatrix} = -\begin{bmatrix} 1 \\ -1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$ $$\begin{bmatrix} 0 & 1 & \cdots & 1\\ 1 & 0 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 &\cdots & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0 \end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ -1 \\ \vdots \\ 0 \end{bmatrix}$$ $$\vdots $$ $$\begin{bmatrix} 0 & 1 & \cdots & 1\\ 1 & 0 & \cdots & 1\\ \vdots & \vdots & \ddots & \vdots\\ 1 & 1 &\cdots & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1 \end{bmatrix} = -\begin{bmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ -1 \end{bmatrix}$$

All eigenvectors here are linearly independent so the eigenvalues are $-1$ with multiplicity $n-1$ and $n-1$ with multiplicity one.

Therefore the determinant is $(-1)^{n-1}(n-1)$.

Answer 3

For $n = 1$, we have $$ \det \left[ \begin{matrix} 0 \end{matrix} \right] = 0.$$

For $n=2$, we have $$ \det \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] = -1. $$

For $n = 3$, we have $$ \det \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] = 0 \det \left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] - 1 \det \left[ \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right] + 1 \det \left[ \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right] = 2. $$

For $n = 4$, we have $$ \det \left[ \begin{matrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{matrix} \right] = -3. $$

For $n = 5$, we have $$ \det \left[ \begin{matrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{matrix} \right] = 4. $$

So, in general, if $A$ is your $n \times n$ matrix with all diagonal entries equal to $0$ and all off-diagonal entries equal to $1$, then we have $$ \det A = (-1)^{n-1} (n-1). $$

Hope this helps.

For calculation of determinants, I've used this online tool.