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X - FIIT STU

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2. krok. Vykonáme vynulovanie prvku pod diagonálou v druhom stĺpci<br />

⎛2 −3 1 0 ⎞ ⎛2 −3 1 0 ⎞<br />

⎜ ⎟ ⎜ ⎟<br />

0 −7 3 −6 ∼ 0 −7 3 −6<br />

⎜0 4 0 12⎟ ⎜0 7 0 21⎟<br />

⎝ ⎠ ⎝ ⎠<br />

K tretiemu riadku pripočítame druhý riadok<br />

⎛2 −3 1 0 ⎞ ⎛2 −3 1 0 ⎞<br />

⎜ ⎟ ⎜ ⎟<br />

0 −7 3 −6 ∼ 0 −7 3 −6<br />

⎜0 7 0 21⎟ ⎜0 0 3 15⎟<br />

⎝ ⎠ ⎝ ⎠<br />

Posledná matica znamená, že pôvodný systém rovníc bol pretransformovaný do<br />

tvaru<br />

2x1− 3x2 + x3<br />

= 0<br />

− 7x2 + 3x3<br />

=−6<br />

3x<br />

= 15<br />

x<br />

T<br />

=<br />

3<br />

( 135 , , )<br />

Priesvitka 18

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