Por meio de escalonamento é possível resolver o conjunto de sistemas representado pela matriz aumentada \([A_n|I_n]\). A solução do problema é exatamente a matriz \(A^{-1}\).
\[[A|I] = \left[
\begin{array}{cccccc|cccccc}
n & 1 & 1 & \dots & 1 & 1 & 1 & 0 & 0 & \dots & 0 & 0 \\
1 & n & 1 & \dots & 1 & 1 & 0 & 1 & 0 & \dots & 0 & 0 \\
1 & 1 & n & \dots & 1 & 1 & 0 & 0 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
1 & 1 & 1 & \dots & n & 1 & 0 & 0 & 0 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 1 & n & 0 & 0 & 0 & \dots & 0 & 1 \\
\end{array}
\right]
\]
1°: Até a penúltima linha, deve-se subtrair, de cada linha, a linha subsequente.
\[\left[
\begin{array}{cccccc|cccccc}
n-1 & 1-n & 0 & \dots & 0 & 0 & 1 & -1 & 0 & \dots & 0 & 0 \\
0 & n-1 & 1-n & \dots & 0 & 0 & 0 & 1 & -1 & \dots & 0 & 0 \\
0 & 0 & n-1 & \dots & 0 & 0 & 0 & 0 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & n-1 & 1-n & 0 & 0 & 0 & \dots & 1 & -1 \\
1 & 1 & 1 & \dots & 1 & n & 0 & 0 & 0 & \dots & 0 & 1 \\
\end{array}
\right]
\]
2°: Até a antepenúltima linha, deve-se somar cada linha com o conjunto das demais que a sucedem (exceto pela última linha).
\[\left[
\begin{array}{cccccc|cccccc}
n-1 & 0 & 0 & \dots & 0 & 1-n & 1 & 0 & 0 & \dots & 0 & -1 \\
0 & n-1 & 0 & \dots & 0 & 1-n & 0 & 1 & 0 & \dots & 0 & -1 \\
0 & 0 & n-1 & \dots & 0 & 1-n & 0 & 0 & 1 & \dots & 0 & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & n-1 & 1-n & 0 & 0 & 0 & \dots & 1 & -1 \\
1 & 1 & 1 & \dots & 1 & n & 0 & 0 & 0 & \dots & 0 & 1 \\
\end{array}
\right]
\]
3°: Multiplica-se a última linha por (n-1).
\[\left[
\begin{array}{cccccc|cccccc}
n-1 & 0 & 0 & \dots & 0 & 1-n & 1 & 0 & 0 & \dots & 0 & -1 \\
0 & n-1 & 0 & \dots & 0 & 1-n & 0 & 1 & 0 & \dots & 0 & -1 \\
0 & 0 & n-1 & \dots & 0 & 1-n & 0 & 0 & 1 & \dots & 0 & -1 \\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots &
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & \dots & n-1 & 1-n & 0 & 0 & 0 & \dots & 1 & -1 \\
n-1 & n-1 & n-1 & \dots & n-1 & n(n-1) & 0 & 0 & 0 & \dots & 0 & n-1 \\
\end{array}
\right]
\]
4°: Subtrai-se, da última linha, cada uma das demais.*
\[\left[
\begin{array}{ccccc|ccccc}
n-1 & 0 & \dots & 0 & 1-n & 1 & 0 & \dots & 0 & -1 \\
0 & n-1 & \dots & 0 & 1-n & 0 & 1 & \dots & 0 & -1 \\
\vdots &\vdots & \ddots&\vdots &\vdots &\vdots &\vdots &\ddots &\vdots & \vdots \\
0 & 0 & \dots & n-1 & 1-n & 0 & 0 & \dots & 1 & -1 \\
0 & 0 & \dots & 0 & (n-1)(2n-1) & -1 & -1 & \dots & -1 & 2(n-1) \\
\end{array}
\right]
\]
5°: Divide-se a última linha por (2n-1).
\[\left[
\begin{array}{ccccc|ccccc}
n-1 & 0 & \dots & 0 & 1-n & 1 & 0 & \dots & 0 & -1 \\
0 & n-1 & \dots & 0 & 1-n & 0 & 1 & \dots & 0 & -1 \\
\vdots &\vdots & \ddots&\vdots &\vdots &\vdots &\vdots & \ddots &\vdots & \vdots \\
0 & 0 & \dots & n-1 & 1-n & 0 & 0 & \dots & 1 & -1 \\
0 & 0 & \dots & 0 & n-1 &\frac{-1}{2n-1}&\frac{-1}{2n-1}& \dots &\frac{-1}{2n-1}& \frac{2(n-1)}{2n-1} \\
\end{array}
\right]
\]
6°: Soma-se cada linha à última.
\[\left[
\begin{array}{ccccc|ccccc}
n-1 & 0 & \dots & 0 & 0 & \frac{2(n-1)}{2n-1} & \frac{-1}{2n-1} & \dots & \frac{-1}{2n-1} & \frac{-1}{2n-1} \\
0 & n-1 & \dots & 0 & 0 & \frac{-1}{2n-1} & \frac{2(n-1)}{2n-1} & \dots & \frac{-1}{2n-1} & \frac{-1}{2n-1} \\
\vdots &\vdots & \ddots &\vdots &\vdots &\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & n-1 & 0 & \frac{-1}{2n-1} & \frac{-1}{2n-1} & \dots & \frac{2(n-1)}{2n-1} & \frac{-1}{2n-1} \\
0 & 0 & \dots & 0 & n-1 & \frac{-1}{2n-1} & \frac{-1}{2n-1} & \dots & \frac{-1}{2n-1} & \frac{2(n-1)}{2n-1} \\
\end{array}
\right]
\]
7°: Divide-se todas as linhas por n-1.
\[\left[
\begin{array}{ccccc|ccccc}
1 & 0 & \dots & 0 & 0 & \frac{2}{2n-1} & \frac{-1}{(2n-1)(n-1)} & \dots & \frac{-1}{(2n-1)(n-1)} & \frac{-1}{(2n-1)(n-1)} \\
0 & 1 & \dots & 0 & 0 & \frac{-1}{(2n-1)(n-1)} & \frac{2}{2n-1} & \dots & \frac{-1}{(2n-1)(n-1)} & \frac{-1}{(2n-1)(n-1)} \\
\vdots &\vdots & \ddots &\vdots &\vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & 0 & \frac{-1}{(2n-1)(n-1)} & \frac{-1}{(2n-1)(n-1)} & \dots & \frac{2}{2n-1} & \frac{-1}{(2n-1)(n-1)} \\
0 & 0 & \dots & 0 & 1 & \frac{-1}{(2n-1)(n-1)} & \frac{-1}{(2n-1)(n-1)} & \dots & \frac{-1}{(2n-1)(n-1)} & \frac{2}{2n-1} \\
\end{array}
\right]
\]
- A dimensão foi reduzida para caber no espaço do editor de texto.