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(Poliedro) Matrizes

Enviado: Sáb 10 Mar, 2018 09:08
por leomaxwell
Se [tex3]A=\begin{bmatrix}
x & 0 \\
0 & y \\
\end{bmatrix}[/tex3] , então [tex3]A^n=\begin{bmatrix}
x^n & 0 \\
0 & y^n \\
\end{bmatrix}[/tex3] . Com base no resultado anterior, provar que:
[tex3]\sum_{i=0}^{n}A^i=\begin{bmatrix}
\frac{1-x^{n+1}}{1-x} & 0 \\
0 & \frac{1-y^{n+1}}{1-y} \\
\end{bmatrix}[/tex3]

:shock:

Re: (Poliedro) Matrizes

Enviado: Sáb 10 Mar, 2018 12:29
por lorramrj
Pela propriedade:

[tex3]A^0=\begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix}[/tex3]

[tex3]A^1=\begin{bmatrix}
x^1 & 0 \\
0 & y^1 \\
\end{bmatrix}[/tex3]

[tex3]A^2=\begin{bmatrix}
x^2 & 0 \\
0 & y^2 \\
\end{bmatrix}[/tex3]

[tex3]A^3=\begin{bmatrix}
x^3 & 0 \\
0 & y^3 \\
\end{bmatrix}[/tex3]
...
Logo:
[tex3]\sum_{i=0}^{n}A^i=\begin{bmatrix}
\frac{1-x^{n+1}}{1-x} & 0 \\
0 & \frac{1-y^{n+1}}{1-y} \\
\end{bmatrix} = \begin{bmatrix}
1 & 0 \\
0 & 1 \\
\end{bmatrix} + \begin{bmatrix}
x^1 & 0 \\
0 & y^1 \\
\end{bmatrix} + \begin{bmatrix}
x^2 & 0 \\
0 & y^2 \\
\end{bmatrix} + ... + \begin{bmatrix}
x^n & 0 \\
0 & y^n \\
\end{bmatrix} [/tex3]

# Na entrada [tex3]a_{11} \space \text{e} \space a_{22}[/tex3]:
[tex3]a_{11} = \sum_{i=0}^{n} x^n = 1 + x^1 + x^2 + x^3 + x^4 + ... + x^{n-1} + x^n = \frac{1-x^{n+1}}{x-1}[/tex3]
[tex3]a_{22} = \sum_{i=0}^{n} y^n = 1 + y^1 + y^2 + y^3 + y^4 + ... + y^{n-1} + y^n = \frac{1-y^{n+1}}{y-1}[/tex3]

Portanto:
[tex3]\sum_{i=0}^{n}A^i=\begin{bmatrix}
\frac{1-x^{n+1}}{1-x} & 0 \\
0 & \frac{1-y^{n+1}}{1-y} \\
\end{bmatrix}[/tex3]