Krdytkyu

Calculate the inverse of the matrix

begin{bmatrix}
-1& 1& ...& ...&1 \
1& -1& 1& ... &1 \
...& ...& ...& ...&1 \
1&1 &1 & ...&1 \
1& 1 &1 &1 &-1
end{bmatrix}

$-1$ on the diagonal and $1$ on the rest.

The key I think is to perform a sequence of elementary transformations on

[ A | $I_{n}$ ] until we get [ $I_{n}$ | $A^{-1}$ ] but that seems to be complicated.

Let $e$ be the all one vector. We have

$$A=-2I+ee^T$$
By Sheman-Morrison formula:
begin{align}A^{-1}&=(-2I+ee^T)^{-1}\&=-frac12I-frac{-left(frac12Iright)ee^Tleft(-frac12Iright)}{1+e^Tleft( -frac12Iright)e} \
&=-frac12I-frac{frac14ee^T}{1-frac{n}2} \
&=-frac12I-frac{ee^T}{4-2n}
end{align}

Hence, the off diagonal entries are $-frac1{4-2n}$ and the diagonal entries are $-frac12-frac1{4-2n}$.

Remark: If $n=2$, the matrix is not invertible.

The inverse also has a constant $a$ on the main diagonal and $b$ everywhere else. The constants do depend on the matrix size $n;$ as noted, for $n=2$ there is no inverse. Just try the thing for $n=3$ and $n=4$ and $n=5.$ By that time you should have it, for $n geq 3$

for $n=3:$
$$
left(
begin{array}{rrr}
-1 & 1&1 \
1 & -1&1 \
1 & 1&-1 \
end{array}
right)
left(
begin{array}{ccc}
a & b&b \
b & a&b \
b & b&a \
end{array}
right)=
left(
begin{array}{rrr}
1 & 0&0 \
0 & 1&0 \
0 & 0&1 \
end{array}
right)
$$

for $n=4,$ different $a,b:$
$$
left(
begin{array}{rrrr}
-1 & 1&1&1 \
1 & -1&1&1 \
1 & 1&-1&1 \
1 & 1&1&-1 \
end{array}
right)
left(
begin{array}{cccc}
a & b&b&b \
b & a&b&b \
b & b&a&b \
b & b&b&a \
end{array}
right)=
left(
begin{array}{rrrr}
1 & 0&0&0 \
0 & 1&0&0 \
0 & 0&1&0 \
0 & 0&0&1 \
end{array}
right)
$$

for $n=5,$ still different $a,b:$
$$
left(
begin{array}{rrrrr}
-1 & 1&1&1&1 \
1 & -1&1&1&1 \
1 & 1&-1&1&1 \
1 & 1&1&-1&1 \
1&1&1&1&-1 \
end{array}
right)
left(
begin{array}{ccccc}
a & b&b&b &b\
b & a&b&b&b \
b & b&a&b &b \
b & b&b&a &b \
b&b&b&b&a \
end{array}
right)=
left(
begin{array}{rrrrr}
1 & 0&0&0 &0 \
0 & 1&0&0&0 \
0 & 0&1&0 &0 \
0 & 0&0&1 &0 \
0&0&0&0&1
end{array}
right)
$$

It's not too bad...
$$begin{array}{c}-1\-1\vdots\*
end{array}left[begin{array}{cccc|cccc}
-1&1&cdots&1 &1\
1&-1&cdots&1 &&1\
vdots&vdots&ddots&vdots &&&ddots\
1&1&cdots&-1 &&&&1
end{array}right] implies$$

$$begin{array}{c}*\*\vdots\small 1/2end{array}
left[begin{array}{cccc|cccc}
-2&&&2 &1&&&-1\
&-2&&2 &&1&&-1\
&&ddots& &&&ddots\
1&1&&-1 &&&&1
end{array}right] implies$$

$$begin{array}{c}*small -1/2\*small -1/2\vdots\* small ^1!/_{!n-2}end{array}
left[begin{array}{cccc|cccc}
-2&&&2 &1&&&-1\
&-2&&2 &&1&&-1\
&&ddots& &&&ddots\
&&&n-2 &small 1/2&small 1/2&&small-^{(n-3)!}/_{!2}
end{array}right] implies$$

$$begin{array}{c}1\1\vdots\*end{array}
left[begin{array}{cccc|cccc}
1&&&-1 &small -1/2&&&small 1/2\
&1&&-1 &&small -1/2&&small 1/2\
&&ddots& &&&ddots\
&&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
end{array}right] implies$$

$$begin{array}{c} \ \ \ end{array}
left[begin{array}{cccc|cccc}
1&&& &small -^1!/_{!2}+ ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
&1&& &small ^1!/_{!2(n-2)}&small -^1!/_{!2}+ ^1!/_{!2(n-2)}&cdots&small ^1!/_{!2(n-2)}\
&&ddots& &vdots&vdots&ddots&vdots\
&&&1 &small ^1!/_{!2(n-2)}&small ^1!/_{!2(n-2)}&cdots&small -^1!/_{!2}+ ^1!/_{!2(n-2)}
end{array}right]$$