# basic question about linear system? . Solve the system $Ax=b$

solve the linear system $Ax=b$ where $b= (1,1,1,1,1,1,1,1,1,1)$ and $A$ is the matrix given by

$$a_{ij} = i+j \ \ , \ if \ \ \ i = j+1$$ $$a_{ij} = i-j \ \ ,if \ \ \ i = j-1$$ $$a_{ij} = ij \ \ ,if \ \ \ i=j$$

i,j = 1,...,10. I cant see the trick ... someone can give me a help ?

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@MaisamHedyelloo in my exercise I dont know if $a_{i,j} = 0$ if $i-j > p$ or if $j-i > q$ .... Do you think my professor make a mistake ? – math student Jun 24 '13 at 6:06
No matrix $A$ is "given" by those equations since, for example, they don't tell us what $a_{13}$ is. I would guess that $a_{ij}$ is meant to be zero if the three given cases don't apply. – Gerry Myerson Jun 24 '13 at 12:42

If the other entries are indeed $0$, we can find this solution comptuationally:
$$\begin{bmatrix} 1 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 4 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 5 & 9 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 7 & 16 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 9 & 25 & -1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 11 & 36 & -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 13 & 49 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 15 & 64 & -1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 17 & 81 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 19 & 100 \\ \end{bmatrix} x=b$$ has the solution $$x=\frac{1}{3038618579743}\begin{bmatrix} 2277107646977 \\ -761510932766 \\ 746660630124 \\ -126227572457 \\ 168364671813 \\ 34450063469 \\ 53595095084 \\ 35391904470 \\ 30389732597 \\ 24612136604\end{bmatrix}$$ where $$b=\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1 \\ 1\end{bmatrix}.$$