Vrftsjtryk

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

Attempt:

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

$v,u,w$ 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

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

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?

$$det A_n=begin{vmatrix}
0 & cdots & 1& 1 & 1 \
vdots & ddots & vdots & vdots & vdots \
1 & cdots & 1 & 1 & 0
end{vmatrix}=detleft(begin{pmatrix}
1 & cdots & 1& 1 & 1 \
vdots & ddots & vdots & vdots & vdots \
1 & cdots & 1 & 1 & 1
end{pmatrix}-I_nright)=det(B_n-I_n)$$
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 $(-1)^{(n-1)}(n-1)$.

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 $(-1)^{n-1}(n-1)$.

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.