1

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.

SADBOYS
  • 1,219

3 Answers3

3

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(\frac12I\right)ee^T\left(-\frac12I\right)}{1+e^T\left( -\frac12I\right)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.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
2

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

Will Jagy
  • 139,541
1

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

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

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

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

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

Klaas van Aarsen
  • 5,908