Dissertação 8 - ITA - 2ª Fase Dia 1 - ITA 2025

$$\begin{gathered} {\mathrm{detA}}_1=\left|\begin{array}{c}1\end{array}\right|=1 \\ {\mathrm{detA}}_2=\left|\begin{array}{cc}1& 2\\ 2& 2\end{array}\right|=-2 \\ {\mathrm{detA}}_3=\left|\begin{array}{ccc}1& 2& 3\\ 2& 2& 3\\ 3& 3& 3\end{array}\right|=3 \end{gathered}$$

${\mathrm{detA}}_{\mathrm{k}}=\left|\begin{array}{cccccc}1& 2& 3& \cdots & \mathrm{k}-1& \mathrm{k}\\ 2& 2& 3& \cdots & \mathrm{k}-1& \mathrm{k}\\ 3& 3& 3& \cdots & \mathrm{k}-1& \mathrm{k}\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ \mathrm{k}-1& \mathrm{k}-1& \mathrm{k}-1& \cdots & \mathrm{k}-1& \mathrm{k}\\ \mathrm{k}& \mathrm{k}& \mathrm{k}& \cdots & \mathrm{k}& \mathrm{k}\end{array}\right|=\mathrm{k}\left|\begin{array}{cccccc}1& 2& 3& \cdots & \mathrm{k}-1& 1\\ 2& 2& 3& \cdots & \mathrm{k}-1& 1\\ 3& 3& 3& \cdots & \mathrm{k}-1& 1\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ \mathrm{k}-1& \mathrm{k}-1& \mathrm{k}-1& \cdots & \mathrm{k}-1& 1\\ \mathrm{k}& \mathrm{k}& \mathrm{k}& \cdots & \mathrm{k}& 1\end{array}\right|$

Multiplicando a coluna k por ($-\mathrm{n}$) e somando com cada coluna n, tem-se:

${\mathrm{detA}}_{\mathrm{k}}=\mathrm{k}\left|\begin{array}{cccccc}0& 0& 0& \cdots & 0& 1\\ 1& 0& 0& \cdots & 0& 1\\ 2& 1& 0& \cdots & 0& 1\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ \mathrm{k}-2& \mathrm{k}-3& \mathrm{k}-4& \cdots & 0& 1\\ \mathrm{k}-1& \mathrm{k}-2& \mathrm{k}-3& \cdots & 1& 1\end{array}\right|$

Fazendo $\mathrm{k}-1$ trocas de colunas paralelas, tem-se:

$$\begin{gathered} {\mathrm{detA}}_{\mathrm{k}}=\mathrm{k}{\left(-1\right)}^{\mathrm{k}-1}\left|\begin{array}{cccccc}1& 0& 0& \cdots & 0& 0\\ 1& 1& 0& \cdots & 0& 0\\ 1& 2& 1& \cdots & 0& 0\\ ⋮& ⋮& ⋮& \ddots & ⋮& ⋮\\ 1& \mathrm{k}-2& \mathrm{k}-3& \cdots & 1& 0\\ 1& \mathrm{k}-1& \mathrm{k}-2& \cdots & 2& 1\end{array}\right| \\ {\mathrm{detA}}_{\mathrm{k}}=\mathrm{k}{\left(-1\right)}^{\mathrm{k}-1} \end{gathered}$$

Logo, $\sum _{\mathrm{k}=1}^{2025}{\mathrm{detA}}_{\mathrm{k}}=1-2+3-4+...-2024+2025=\boxed{1013}$.