I'm holding this eqauation about 5 hours, but I can't handle it.
It's on the Applied Mathematics and comptation's article titled 'interval weight generation approaches based on consistency test and interval comparison matrices'.
It's about Interval AHP, using by consistency index.
This article says formula (13) is the expansion of formula (11)
how can I drive from (13) to (11)??
I attach the equation.
If it has a problem with copyright, plz tell me to delete it.
For some fixed $i$, your equation (13) is just the equality of (11), but looked at from the perspective of the $i$'th component: Equation (11) states that two vectors are equal, while equation (13) states that some number is equal to $0$, where this number depends on $i$. Let me repeat equation (11) here:
$$\widehat{A}W=\lambda_{\text{max}}W,$$
where
$$\lambda_{\text{max}} = n+(n-1)\text{RI}\cdot\text{CI}.$$
So let us look at the $i$'th component of equation (11). Oh, you didn't say so, but it looks like the diagonal of the matrix $A$ consists of all $1$'s, so that $a_{ii}=1$ for all $i=1,\cdots,n$. We are going to need this. Consider first the left hand side $\widehat{A}W$ of the equation.
Remember how you multiply a matrix by a vector? You do this "row by column". If I want to find the $i$'th component of the vector $\widehat{A}W$, then I take the $i$'th row of $\widehat{A}$ and multiply it component by component with $W$, and then add all the products. Denoting by $[\widehat{A}W]_i$ the $i$'th component of this vector, I get:
$$[\widehat{A}W]_i = \sum_{j=1}^n \hat{a}_{ij}w_j.$$
But you said so yourself that if the upper triangular entries of $\widehat{A}$ are $\hat{a}_{ij}$, then the lower triangular entries are $1/\hat{a}_{ij}$. To illustrate, this is the $4\times 4$ version of $\widehat{A}$: $$\begin{bmatrix} 1 & \hat{a}_{12} & \hat{a}_{13} & \hat{a}_{14}\\ 1/\hat{a}_{12} & 1 & \hat{a}_{23} & \hat{a}_{24}\\ 1/\hat{a}_{13} & 1/\hat{a}_{23} & 1 & \hat{a}_{34}\\ 1/\hat{a}_{14} & 1/\hat{a}_{24} & 1/\hat{a}_{34} & 1 \end{bmatrix}$$ So my $i$th component of $\widehat{A}W$ from above actually turns into $$[\widehat{A}W]_i = \sum_{j=1}^{i-1}\frac{w_j}{\hat{a}_{ij}}+w_i+\sum_{j=i+1}^nw_j\hat{a}_{ij}.$$ What happened? I looked at the $i$'th row of $\widehat{A}$, and realized that I can group the elements of $\widehat{A}$ such that the first sum above comes from all the terms below the diagonal, the middle term corresponds to the diagonal element $(i,i)$, and the last sum comes from all the terms of row $i$ that are above the diagonal.
Now I want to consider the right hand side $\lambda_{\text{max}}W$ of equation (11). You know that $$\lambda_{\text{max}} = n+(n-1)\text{RI}\cdot\text{CI},$$ and I will just rewrite this slightly as follows. $$\lambda_{\text{max}} = n+(n-1)\text{RI}\cdot\text{CI} = 1+(n-1)+(n-1)\text{RI}\cdot\text{CI}=1+(n-1)(1+\text{RI}\cdot\text{CI}).$$ The expression $\lambda_{\text{max}}W$ is really just a number $\lambda_{\text{max}}$ times a vector $W$, and I can calculate this by multiplying each entry $w_i$ of $W$ by $\lambda$. Using my new expression for $\lambda$, I get: $$[\lambda W]_i = (1+(n-1)(1+\text{RI}\cdot\text{CI}))w_i=w_i+(n-1)(1+\text{RI}\cdot\text{CI})w_i.$$ So I have now calculated the $i$'th component of the left hand side and of the right hand side of equation (11), and letting these two expression be equal, I obtain: $$\sum_{j=1}^{i-1}\frac{w_j}{\hat{a}_{ij}}+w_i+\sum_{j=i+1}^nw_j\hat{a}_{ij}=w_i+(n-1)(1+\text{RI}\cdot\text{CI})w_i.$$ Notice how I can cancel the term $w_i$ from both sides. Doing this, I get: $$\sum_{j=1}^{i-1}\frac{w_j}{\hat{a}_{ij}}+\sum_{j=i+1}^nw_j\hat{a}_{ij}=(n-1)(1+\text{RI}\cdot\text{CI})w_i.$$ This is practically equation (13)! Moving the left hand side over to the right hand side (and changing the sign), I finally obtain: $$\sum_{j=1}^{i-1}\frac{w_j}{\hat{a}_{ij}}-(n-1)(1+\text{RI}\cdot\text{CI})w_i+\sum_{j=i+1}^nw_j\hat{a}_{ij}=0,$$ which is what I set out to show.