Formula Cauchy-Binet

Vom considera numerele naturale nenule m si n , m ≤ n si matricile A=\left ( a_{ij} \right )\in M_{m,n}\left ( \mathbb{C} \right ) , B=\left ( b_{ij} \right )\in M_{n,m}\left ( \mathbb{C} \right ). Pentru 1\leqslant j_{1}\leqslant j_{2}\leqslant ....\leqslant j_{m}\leqslant n facem notatiile:

A^{j_{1}j_{2}....j_{m}}=\left | \begin{matrix} a_{1j_{1}} &a_{1j_{2}} &.... &a_{1j_{m}} \\ a_{2j_{1}} &a_{2j_{2}} &.... &a_{2j_{m}} \\ .... &.... &.... &.... \\ a_{mj_{1}} &a_{mj_{2}} &.... &a_{mj_{m}} \end{matrix} \right | , B_{j_{1}j_{2}....j_{m}}=\left | \begin{matrix} b_{j_{1}1} &b_{j_{1}2} &.... &b_{j_{1}m} \\ b_{j_{2}1} &b_{j_{2}2} &.... &b_{j_{2}m} \\ .... &.... &.... &.... \\ b_{j_{m}1} &b_{j_{m}2} &.... &b_{j_{m}m} \end{matrix} \right |

si demonstram urmatoarea egalitate cunoscuta sub numele de formula Cauchy-Binet:

det(A\cdot B)=\sum_{1\leqslant j_{1}<j_{2}<....< j_{m}\leqslant n}^{ }A^{j_{1}j_{2}....j_{m}}\cdot B_{j_{1}j_{2}....j_{m}} .

Pentru demonstratia acestei formule, vom nota A\cdot B=C=\left ( c_{ij} \right )\in M_{m}\left ( \mathbb{C} \right ) si avem:
det(A\cdot B)=\sum_{\sigma \in S_{m}}^{ }\varepsilon\left ( \sigma \right )c_{1\sigma \left ( 1 \right )}c_{2\sigma \left ( 2 \right )}....c_{m\sigma \left ( m \right )}=
=\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )\left [ \sum_{k_{1}=1}^{n}a_{1k_{1}}b_{k_{1}\sigma \left ( 1 \right )} \right ]\left [ \sum_{k_{2}=1}^{n}a_{2k_{2}}b_{k_{2}\sigma \left ( 2 \right )} \right ]....\left [ \sum_{k_{m}=1}^{n}a_{mk_{m}}b_{k_{m}\sigma \left ( m \right )} \right ]=
=\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}b_{k_{1\sigma \left ( 1 \right )}}b_{k_{2}\sigma \left ( 2 \right )}....b_{k_{m}\sigma \left ( m \right )}=
=\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}\sum_{\sigma \in S_{m}}^{ }\varepsilon \left ( \sigma \right )b_{k_{1}\sigma \left ( 1 \right )}b_{k_{2}\sigma \left ( 2 \right )}....b_{k_{m}\sigma \left ( m \right )}=
=\sum_{k_{1},k_{2},....,k_{m}=1}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=
=\sum_{\begin{matrix} k_{1},k_{2},....,k_{m}=1\\ B_{k_{1},k_{2},....,k_{m}}\neq 0 \end{matrix}}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=
=\sum_{\begin{matrix} k_{1},k_{2},....,k_{m}=1\\ k_{1},k_{2},....,k_{m}\: distincte \end{matrix}}^{n}a_{1k_{1}}a_{2k_{2}}....a_{mk_{m}}B_{k_{1}k_{2}....k_{m}}=
=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }\sum_{\tau \in S_{m}}^{ }a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}B_{\tau \left ( j_{1} \right )\tau \left ( j_{2} \right )....\tau \left ( j_{m} \right )}=
=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }\sum_{\tau \in S_{m}}^{ }\varepsilon \left ( \tau \right )a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}B_{ j_{1} j_{2} .... j_{m} }=
=\sum_{1\leqslant j_{1}<j_{2}<....<j_{m}\leqslant n}^{ }B_{ j_{1} j_{2} .... j_{m} }\sum_{\tau \in S_{m}}^{ }\varepsilon \left ( \tau \right )a_{1\tau \left ( j_{1} \right )}a_{2\tau \left ( j_{2} \right )}....a_{m\tau \left ( j_{m} \right )}=
\sum_{1\leqslant j_{1}<j_{2}<....< j_{m}\leqslant n}^{ }A^{j_{1}j_{2}....j_{m}}\cdot B_{j_{1}j_{2}....j_{m}} .
Observatie. In cazul particular n = m, formula Cauchy-Binet ne conduce la egalitatea cunoscuta det\left ( A\cdot B \right )=det\left ( A \right )\cdot det\left ( B \right ) .
Aplicatie:Daca n ≥3 si a_{i},b_{i},c_{i}\in \mathbb{R},i=\overline{1,n}, atunci:
\left | \begin{matrix} a_{1}^{2}+....+a_{n}^{2} &a_{1}b_{1}+....+a_{n}b_{n} &a_{1}c_{1}+....+a_{n}c_{n} \\ a_{1}b_{1}+....+a_{n}b_{n} &b_{1}^{2}+....+b_{n}^{2} &b_{1}c_{1}+....+b_{n}c_{n} \\ a_{1}c_{1}+....+a_{n}c_{n} &b_{1}c_{1}+....+b_{n}c_{n} & c_{1}^{2}+....+c_{n}^{2} \end{matrix} \right |\geqslant 0 .
Solutie.Considerand matricile A=\left ( \begin{matrix} a_{1} &.... &a_{n} \\ b_{1} &.... &b_{n} \\ c_{1} &.... &c_{n} \end{matrix} \right ),B=\left ( \begin{matrix} a_{1} &b_{1} &c_{1} \\ .... &.... &.... \\ a_{n} &b_{n} &c_{n} \end{matrix} \right ), notand cu Δ determinantul din enunt si tinand cont de formula Cauchy-Binet obtinem:
\bigtriangleup =det\left ( A\cdot B \right )=\sum_{1\leqslant i<j<k\leqslant n}^{ }A^{ijk}\cdot B_{ijk}=
=\sum_{1\leqslant i<j<k\leqslant n}^{ } \left | \begin{matrix} a_{i} &a_{j} &a_{k} \\ b_{i} &b_{j} &b_{k} \\ c_{i} &c_{j} &c_{k} \end{matrix} \right |\cdot \left | \begin{matrix} a_{i} &b_{i} &c_{i} \\ a_{j} &b_{j} &c_{j} \\ a_{k} &b_{k} &c_{k} \end{matrix} \right |=\sum_{1\leqslant i<j<k\leqslant n}^{ } \left | \begin{matrix} a_{i} &a_{j} &a_{k} \\ b_{i} &b_{j} &b_{k} \\ c_{i} &c_{j} &c_{k} \end{matrix} \right |^{^{2}}\geqslant 0 .
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Reclame

Determinantul matricei adjuncte

Consideram matricea A=\begin{pmatrix} a_{11} &.... &a_{1j} &.... &a_{1n} \\ .... &.... &.... &.... &.... \\ a_{i1} &.... &a_{ij} &.... &a_{in} \\ .... &.... &.... &.... &.... \\ a_{n1} &.... &a_{nj} &.... &a_{nn} \end{pmatrix} cu elemente numere complexe si definim:

  • \Delta {_{ij}} determinantul matricei obtinuta din matricea A prin eliminarea liniei „i ”  si coloanei  „j ” , i,j\epsilon \left \{ 1,2,....,n \right \};
  • A {_{ij}}=(-1)^{i+j}\Delta {_{ij}} complementul algebric al elementului a {_{ij}}, i,j\epsilon \left \{ 1,2,....,n \right \};
  • A^{*}=\begin{pmatrix} A_{11} &.... &A_{j1} &.... &A_{n1} \\ .... &.... &.... &.... &.... \\ A_{1i} &.... &A_{ji} &.... &A_{ni} \\ .... &.... &.... &.... &.... \\ A_{1n} &.... &A_{jn} &.... &A_{nn} \end{pmatrix} adjuncta matricii A .

In conditiile notatiilor de mai sus demonstram egalitatea:

det\left ( A^{*} \right )=\left [ det(A) \right ]^{n-1}      (1).

Pentru demonstratie consideram urmatoarele cazuri:

\mathit{Cazul\: 1:}det(A)\neq 0. In acest caz matricea A este inversabila si inversa sa este A^{-1}=\frac{1}{det(A)}A^{*}. Se obtine

1=det\left ( I_{n} \right )=det(A)\cdot det\left ( A^{-1} \right )=det(A)\cdot \left (\frac{1}{det(A)} \right )^{n}\cdot det\left ( A^{*} \right ) ,

de unde rezulta egalitatea (1).

\mathit{Cazul\: 2:}A=O_{n}. Avem evident A^{*}=O_{n} , deci det\left ( A^{*} \right )=0.
\mathit{Cazul\: 3:}A\neq O_{n}\: si\: det(A)=0. Pentru a demonstra egalitatea (1) vom considera sistemul omogen:

^{t}\left ( A^{*} \right )X=O_{n,1}\Leftrightarrow \begin{pmatrix} A_{11} &.... &A_{1j} &.... &A_{1n} \\ .... &.... &.... &.... &.... \\ A_{i1} &.... &A_{ij} &.... &A_{in} \\ .... &.... &.... &.... &... \\ A_{n1} &.... &A_{nj} &.... &A_{nn} \end{pmatrix}\cdot \begin{pmatrix} x_{1}\\ ....\\ x_{i}\\ ....\\ x_{n}\end{pmatrix}=\begin{pmatrix} 0\\ ....\\ 0\\ ....\\ 0\end{pmatrix}\; \;(S)

Cum A\neq O_{n} , exista i,j∈{1,2,….,n } astfel incat a_{ij}\neq 0 si atunci sistemul omogen (S ) va avea solutia nebanala x_{1}=a_{i1},....,x_{j}=a_{ij},....,x_{n}=a_{in}, ceea ce conduce la  det\left ( A^{*} \right )=0.

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Calculul unui determinant folosind metoda lui Laplace

In cele ce urmeaza vom considera A=\left ( a_{_{ij}} \right )_{_{1\leqslant i\leq n,1\leqslant j\leq \leqslant n}} o matrice patratica de ordinul n, n ≥ 2,al carui determinant il notam cu D.

Pentru p  numar natural, 1≤p<n, vom considera numerele naturale i{_{1}},i{_{2}},....,i{_{p}},i_{p+1},.....,i{_{n}} si j{_{1}},j{_{2}},....,j{_{p}},j_{p+1},.....,j{_{n}}, cu proprietatile

1\leq i{_{1}}<i{_{2}}<....<i{_{p}}\leqslant n,1\leqslant i_{p+1}<.....<i{_{n}}\leqslant n
\left \{ i_{1},i_{2},....,i_{n} \right \}=\left \{ 1,2,....,n \right \}

1\leq j{_{1}}<j{_{2}}<....<j{_{p}}\leqslant n,1\leqslant j_{p+1}<.....<j{_{n}}\leqslant n
\left \{ j_{1},j_{2},....,j_{n} \right \}=\left \{ 1,2,....,n \right \}

si vom defini:

  • minor de ordinul p  al matricei A , determinantul matricei de ordinul p , formata cu elementele situate la intersectia a linii si p  coloane din matricea A. Pentru un minor de ordinul p  vom folosi notatia:

D\begin{bmatrix} i_{1} &i_{2} &.... &i_{p} \\ j{_{1}} &j_{2} &.... &j{_{p}} \end{bmatrix}=\begin{vmatrix} a{_{i{_{1}j{_{1}}}}} &a{_{i{_{1}j{_{2}}}}} &.... &a{_{i{_{1}j{_{p}}}}} \\ a{_{i{_{2}j{_{1}}}}} &a{_{i{_{2}j{_{2}}}}} &.... &a{_{i{_{2}j{_{p}}}}} \\ .... &.... &.... &.... \\ a{_{i{_{p}j{_{1}}}}} &a{_{i{_{p}j{_{2}}}}} &.... &a{_{i{_{p}j{_{p}}}}} \end{vmatrix} .

  • minor complementar al minorului de ordinul p , D\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ]  al matricei A, determinantul matricei cu n-p linii si n-p coloane, obtinuta din matricea A  prin eliminarea celor linii si p  coloane corespunzatoare minorului D\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ] , adica minorul D\left [ \begin{matrix} i{_{p+1}} &i_{p+2} &.... &i_{n} \\ j_{p+1} &j_{p+2} &.... &j_{n} \end{matrix} \right ] .
  • complement algebric sau cofactor al minorului de ordinul p , D\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ]  al matricei A, numarul \overline{D}\left [ \begin{matrix} i{_{1}} &i{_{2}} &.... &i{_{p}} \\ j_{1} &j{_{2}} &.... &j{_{p}} \end{matrix} \right ]=(-1)^{{i_{1}+i_{2}+....+i_{p}+j_{1}+j_{2}+....+j_{p}}}\cdot D\left [ \begin{matrix} i{_{p+1}} &i_{p+2} &.... &i_{n} \\ j_{p+1} &j_{p+2} &.... &j_{n} \end{matrix} \right] .

**

Pentru  1≤p<n, fixand liniile 1\leq i{_{1}}<i_{2}<....<i{_{p}}\leqslant n , determinantul D al matricei A se poate calcula cu formula urmatoare, numita si dezvoltarea dupa liniile i{_{1}},i_{2},....,i{_{p}} (teorema lui Laplace) .

D=\sum_{1\leqslant j{_{1}<j_{2}<....<j{_{p}\leqslant n}}}^{ }D\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ]\cdot \overline{D}\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ]

Observatii:

1.O dezvoltarea asemanatoare se poate face dupa coloanele fixate 1\leq j{_{1}}<j_{2}<....<j{_{p}}\leqslant n :

D=\sum_{1\leqslant i{_{1}<i_{2}<....<i{_{p}\leqslant n}}}^{ }D\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ]\cdot \overline{D}\left [ \begin{matrix} i{_{1}} &i_{2} &.... &i_{p} \\ j_{1} &j_{2} &.... &j_{p} \end{matrix} \right ] .

2.In cazul particular cand p=1, se obtine ceea ce se numeste dezvoltarea determinantului dupa o linie sau dupa o coloana.

Aplicatie.

Daca a{_{i}},b{_{i}},c{_{i}},d{_{i}}\epsilon \, C,i=\overline{1,2} , sa se demonstreze egalitatea:

\left | \begin{matrix} a{_{1}}c_{1} &a{_{2}}d{_{1}} &a_{1}c_{2} &a_{2}d_{2} \\ a_{3}c_{1} &a_{4}d_{1} &a_{3}c_{2} &a_{4}d_{2} \\ b_{1}c_{3} &b_{2}d_{3} &b{_{1}}c_{4} &b_{2}d_{4} \\ b_{3}c_{3} &b_{4}d_{3} &b_{3}c_{4} &b_{4}c_{4} \end{matrix} \right |=\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |\cdot \left | \begin{matrix} c_{1} &c_{2} \\ c_{3} &c_{4} \end{matrix} \right |\cdot \left | \begin{matrix} d_{1} &d_{2} \\ d_{3} &d_{4} \end{matrix} \right | .

Solutie.

Dezvoltand determinantul de ordinul patru dupa primele doua linii, conform formulei lui Laplace, obtinem:

\left | \begin{matrix} a{_{1}}c_{1} &a{_{2}}d{_{1}} &a_{1}c_{2} &a_{2}d_{2} \\ a_{3}c_{1} &a_{4}d_{1} &a_{3}c_{2} &a_{4}d_{2} \\ b_{1}c_{3} &b_{2}d_{3} &b{_{1}}c_{4} &b_{2}d_{4} \\ b_{3}c_{3} &b_{4}d_{3} &b_{3}c_{4} &b_{4}c_{4} \end{matrix} \right |=

(-1)^{1+2+1+2}\left | \begin{matrix} a_{1}c_{1} &a{_{2}}d_{1} \\ a_{3}c_{1} &a_{4}d_{1} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1}c_{4} &b_{2}d_{4} \\ b_{3}c_{4} &b_{4}c_{4} \end{matrix} \right |+(-1)^{1+2+1+3}\left | \begin{matrix} a_{1}c_{1} &a_{1}c_{2} \\ a_{3}c_{1} &a_{3}c{_{2}} \end{matrix} \right |\cdot \left | \begin{matrix} b_{2}d_{3} &b_{2}d_{4} \\ b_{4}d_{3} &b_{4}c_{4} \end{matrix} \right |+

(-1)^{1+2+1+4}\left | \begin{matrix} a_{1}c_{1} &a{_{2}}d_{2} \\ a_{3}c_{1} &a_{4}d_{2} \end{matrix} \right |\cdot \left | \begin{matrix} b_{2}d_{3} &b_{1}c_{4} \\ b_{4}d_{3} &b_{3}c_{4} \end{matrix} \right |+(-1)^{1+2+2+3}\left | \begin{matrix} a_{2}d_{1} &a_{1}c_{2} \\ a_{4}d_{1} &a_{3}c{_{2}} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1}c_{3} &b_{2}d_{4} \\ b_{3}c_{3} &b_{4}c_{4} \end{matrix} \right |+

(-1)^{1+2+2+4}\left | \begin{matrix} a_{2}d_{1} &a{_{2}}d_{2} \\ a_{4}d_{1} &a_{4}d_{2} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1}c_{3} &b_{1}c_{4} \\ b_{3}c_{3} &b_{3}c_{4} \end{matrix} \right |+(-1)^{1+2+3+4}\left | \begin{matrix} a_{1}c_{2} &a_{2}d_{2} \\ a_{3}c_{2} &a_{4}d{_{2}} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1}c_{3} &b_{2}d_{3} \\ b_{3}c_{3} &b_{4}d_{3} \end{matrix} \right |=

c_{1}c{_{4}}d_{1}d_{4}\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |-0+c_{1}c_{4}d_{2}d_{3}\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{2} &b_{1} \\ b_{4} &b_{3} \end{matrix} \right |+

c_{2}c_{3}d_{1}d_{4}\left | \begin{matrix} a_{2} &a_{1} \\ a_{4} &a_{3} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |-0+c_{2}c_{3}d_{2}d_{3}\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |=

\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |\left (c_{1}c_{4}d_{1}d_{4} -c{_{1}c_{4}d_{2}d_{3}-c_{2}c_{3}d{_{1}d_{4}}}+c_{2}c_{3}d_{2}d{_{3}} \right )=

\left | \begin{matrix} a_{1} &a_{2} \\ a_{3} &a_{4} \end{matrix} \right |\cdot \left | \begin{matrix} b_{1} &b_{2} \\ b_{3} &b_{4} \end{matrix} \right |\cdot \left | \begin{matrix} c_{1} &c_{2} \\ c_{3} &d_{4} \end{matrix} \right |\cdot \left | \begin{matrix} d_{1} &d_{2} \\ d_{3} &d_{4} \end{matrix} \right | .

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