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Analiza 2

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1.8. IZREK O IMPLICITNI FUNKCIJI 45<br />

(ii) Oglejmo si sistem n2 linearnih enačb z n1 +n2 neznankami (∗).<br />

⎧<br />

⎪⎨<br />

(∗)<br />

⎪⎩<br />

α11x1 +...+α1n1xn1 +β11y1 +...+β1n2yn2 = 0<br />

α21x1 +...+α2n1xn1 +β21y1 +...+β2n2yn2 = 0<br />

...<br />

αn21x1 +...+αn2n1xn1 +βn21y1 +...+βn2n2yn2 = 0<br />

Kdaj lahko ta sistem enolično razreˇsimo na y1,y2,...,yn2, t.j. kdaj za vsak<br />

x1,x2,...,xn1 obstajanatankoenan2-terica(y1,y2,...,yn2), kiizpolnjujezgornji<br />

sistem enačb. Odgovor: Natanko tedaj, ko je matrika<br />

⎡<br />

⎤<br />

β11 ⎢ β21<br />

[β] = ⎢<br />

.<br />

⎣<br />

β12<br />

β22<br />

··· β1n2 ⎥<br />

β2n2<br />

⎥<br />

.<br />

⎥<br />

.. ⎥<br />

⎦<br />

βn21 βn22<br />

βn2n2<br />

nesingularna, t.j. det([β]) = 0. V matričnem zapisu lahko tedaj zapiˇsemo<br />

[α]X +[β]Y = 0<br />

[β]Y = −[α]X<br />

Y = −[β] −1 [α]X.<br />

Zgornji sistem enačb lahko dopolnimo do ekvivalentnega sistema n1+n2 enačb<br />

(∗∗),<br />

⎧<br />

⎪⎩<br />

α11x1 +...+α1n1xn1 +β11y1 +...+β1n2yn2 = 0<br />

α21x1 +...+α2n1xn1 +β21y1 +...+β2n2yn2 = 0<br />

...<br />

⎪⎨ αn21x1 +...+αn2n1xn1 +βn21y1 +...+βn2n2yn2 = 0<br />

(∗∗)<br />

Matrika sistema enačb (∗∗)<br />

x1 = x1<br />

x2 = x2<br />

...<br />

xn1 = xn1<br />

⎡ ⎤<br />

α β<br />

A = ⎣ ⎦,<br />

I 0

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