Inverse of this $6 \times 6$ matrix?

Question

Let $A= \begin{pmatrix} 1 & 1 & 1&0&0&1 \\ 1 & -1&1&-1&0&1\\ 1&-1&-1&0&-1&1\\ 1&-1&0&1&1&0\\ 1&1&0&-1&1& -1\\ 1&1&-1&1&-1&0 \end{pmatrix} $.

Then the inverse given in the paper I have is $A^{-1}=\frac16\begin{pmatrix} 1 & 1 & 1&1&1&1 \\ 1 & -1&-1&-1&1&1\\ 2&2&-2&0&-2&0\\ 0&0&-2&2&-2&2\\ 0&-2&0&2&2& -2\\ 2&-2&2&0&0&-2 \end{pmatrix}$

but when I find it by online calculator here, I get $A^{-1}=$

Shouldn't they be same as matrix inverse is unique?

Answer 1

Your inverse is wrong, if you calculate $A^{-1}\cdot A$ with your inverse the value of the upper right number will be $\frac{1}{2}$ and not zero.