1. Aratati ca:
\(
\left| {\begin{array}{c}
x & y & z \\
{x^2 } & {y^2 } & {z^2 } \\
{yz} & {xz} & {xy} \\
\end{array}} \right| = (xy + yz + zx)(x - y)(y - z)(z - x)
\)
2. Demonstrati prin inductie(stiu ca e Vandermonde, dar demonstratia asta...):
\(
\left| {\begin{array}{c}
1 & 1 & 1 & \cdots & 1 \\
{a_1 } & {a_2 } & {a_3 } & \cdots & {a_n } \\
{a_1^2 } & {a_2^2 } & {a_3^2 } & \cdots & {a_n^2 } \\
\vdots & \vdots & \vdots & \cdots & \vdots \\
{a_1^{n - 1} } & {a_2^{n - 1} } & {a_3^{n - 1} } & \cdots & {a_n^{n - 1} } \\
\end{array}} \right| = \prod\limits_{1 \le j \triangleleft i \le n} {(a_i - a_j )}
\)
La primul am incercat eu asa:\(
\begin{array}{l}
\left| {\begin{array}{c}
x & y & z \\
{x^2 } & {y^2 } & {z^2 } \\
{yz} & {xz} & {xy} \\
\end{array}} \right| = \frac{1}{{xyz}}\left| {\begin{array}{c}
{x^2 } & {y^2 } & {z^2 } \\
{x^3 } & {y^3 } & {z^3 } \\
{xyz} & {xyz} & {xyz} \\
\end{array}} \right| = \frac{{xyz}}{{xyz}}\left| {\begin{array}{c}
{x^2 } & {y^2 } & {z^2 } \\
{x^3 } & {y^3 } & {z^3 } \\
1 & 1 & 1 \\
\end{array}} \right| \\
= \left| {\begin{array}{c}
1 & 1 & 1 \\
{x^2 } & {y^2 } & {z^2 } \\
{x^3 } & {y^3 } & {z^3 } \\
\end{array}} \right| \\
\end{array}
\) dar...
2 determinanti; demonstratie prin inductie si Vandermonde
- Beniamin Bogosel
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Pentru 1:
Scazi a doua coloana din prima si a treia din a doua si obtii
\( \Delta=(x-y)(y-z)\left|\begin{matrix} 1 & 1 & z \\ x+y & y+z & z^2 \\ -z & -x & xy \end{matrix} \right| \).
Acum scazi a doua coloana din prima si dai factor \( (z-x) \) si obtii
\( \Delta=(x-y)(y-z)(z-x)\left|\begin{matrix} 0 & 1 & z \\ -1 & y+z & z^2 \\ -1 & -x & xy \end{matrix} \right| \). Dezvolti determinantul asta pe prima coloana si vezi ca este egal chiar cu \( xy+yz+zx \).
Scazi a doua coloana din prima si a treia din a doua si obtii
\( \Delta=(x-y)(y-z)\left|\begin{matrix} 1 & 1 & z \\ x+y & y+z & z^2 \\ -z & -x & xy \end{matrix} \right| \).
Acum scazi a doua coloana din prima si dai factor \( (z-x) \) si obtii
\( \Delta=(x-y)(y-z)(z-x)\left|\begin{matrix} 0 & 1 & z \\ -1 & y+z & z^2 \\ -1 & -x & xy \end{matrix} \right| \). Dezvolti determinantul asta pe prima coloana si vezi ca este egal chiar cu \( xy+yz+zx \).
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
- Beniamin Bogosel
- Co-admin
- Posts: 710
- Joined: Fri Mar 07, 2008 12:01 am
- Location: Timisoara sau Sofronea (Arad)
- Contact:
Ai relatia de recurenta \( V(x_1,x_2,...,x_{n+1})=\prod_{k=1}^n(x_{n+1}-x_k)V(x_1,x_2,...,x_n) \), care se demonstreaza facand operatii cu linii si coloane:
-se scade ultima coloana din celelalte si se dezvolta dupa prima linie;
-se dau factori pe coloane \( (x_{n+1}-x_1),... \);
-etc...
-se scade ultima coloana din celelalte si se dezvolta dupa prima linie;
-se dau factori pe coloane \( (x_{n+1}-x_1),... \);
-etc...
Yesterday is history,
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog
Tomorow is a mistery,
But today is a gift.
That's why it's called present.
Blog