Define the $n$ x $n$ array $(a_{ij})$ by $a_{ij}$=$-1$+$2 \mbox{max}(i,j)$. Set up array $b_i$ in such a way that the solution of the system $Ax=b$ is $x_i$ =$1$ for $1 \leq i \leq n$.
My attempt: I tried looking at this in concrete cases to find a pattern. Note the answer was given to be $b_i=n^2+2(i-1)$. I just have no idea how to derive this for generally.
I consider $n=3$. And get A:You have for example $\begin{bmatrix}1&3&5\\3&3&5\\5&5&5\end{bmatrix}\circ \begin{bmatrix} 1\\1\\1\end{bmatrix} = \begin{bmatrix}9\\11\\15\end{bmatrix}$.
However, based on the answer given, it should be:\begin{bmatrix}9\\11\\13\end{bmatrix}.
Thus, I'm having 2 problems: 1. Is the answer given, wrong, or am I making a mistake that I am not seeing? 2. How was the answer derived, that is how was a formula obtained for $b_i$? Thank you.
$b_i = -n + 2 \sum max(i,j) = -n + 2\sum_{j \le i} max(i,j) + 2\sum_{j >i} max(i,j)$
$= -n + 2\sum_{j \le i}i + 2\sum_{j=i+1}^nj = -n + 2[i^2 + \frac {n(n+1) - i(i+1)}2] = n^2 + i^2-i $