Incidence Algebra Mobius Function/Matrix

I'm a little bit confused on how the Mobius Matrix came from, using the Hasse Diagram.

From the article, the Mobius function is defined as follows:

$$\mu(a,b)=\begin{cases} 1 & \text{if } a=b ;\\ 0 & \text{if } a\neq b \text{ and }\\ -\sum_{c:a \leq c <b} \mu(a,c) & \text{if } a<b \end{cases}$$

My attempt is:

\begin{bmatrix} 1 & -1 & -1 & -1\\ 0 & 1 & 1 & -1 \\ 0& 1 & 1 & -1 \\ 0& 0 & 0 &1 \end{bmatrix}

but i got it wrong in $\mu(a,d)=-1$, since from the article the Mobius matrix is:
I don't get to know why i am wrong, since based on the hasse diagram $a \leq d$.

The definition for the mobius function you provided is not correct (you have a "$a\neq b$" when it should be "$a\not\leq b$", and the "$a<c$" should be $a\leq c$"); as seen in the source article,

$$\mu(x,y)=\begin{cases} 1&\text{if }x=y\\0&\text{if } x\not\leq y\\ -\displaystyle\sum_{z:x\leq z<y}\mu(x,z)&\text{if }x<y\end{cases}$$

Now, we want to find $\mu(a,d)$. Since $a<d$, we need to find all $z$ such that $a\leq z<d$. Its clear that $z=a$, $b$, and $c$ satisfy that inequality. Therefore,

$$\mu(a,d)=-(\mu(a,a)+\mu(a,b)+\mu(a,c))=-(1+\mu(a,b)+\mu(a,c))$$

To find $\mu(a,b)$ and $\mu(a,c)$, we do the same thing as above. We can find that $$\mu(a,b)=-(\mu(a,a)) = -1\quad\text{ and }\quad\mu(a,c)=-(\mu(a,a)) = -1.$$

Therefore,

$$\mu(a,d)=-(1-1-1)=-(-1)=1.$$