Ìåòîä Êðàìåðà

Îòûñêàíèå ðåøåíèÿ ñèñòåìû ïî òåîðåìå Êðàìåðà íàçûâàþò ìåòîäîì Êðàìåðà ðåøåíèÿ ñèñòåìû ëèíåéíûõ óðàâíåíèé.

Òåîðåìà Êðàìåðà. Ïóñòü D – îïðåäåëèòåëü ìàòðèöû ñèñòåìû À; Dj – îïðåäåëèòåëü ìàòðèöû, ïîëó÷àåìîé èç ìàòðèöû À çàìåíîé j- ãî ñòîëáöà ñòîëáöîì ñâîáîäíûõ ÷ëåíîâ. Òîãäà, åñëè D ¹ 0, òî ñèñòåìà èìååò åäèíñòâåííîå ðåøåíèå, îïðåäåëÿåìîå ïî ôîðìóëàì:

, j = 1, 2, …, n

Ýòè ôîðìóëû íàçûâàþòñÿ ôîðìóëàìè Êðàìåðà.

Ïðèìåð 2. Ðåøèòü ñèñòåìó ëèíåéíûõ óðàâíåíèé ìåòîäîì Êðàìåðà.

Ðåøåíèå.

Íàéäåì îáùèé îïðåäåëèòåëü ñèñòåìû:

∆ = = 3∙(–2) ∙(–5) + 1∙(–3) ∙2 + 1∙2∙(–2) – (–2) ∙(–2) ∙2 –

– 2∙(–3) ∙3 – 1∙1∙(–5) = 30 – 4 – 6 – 8 + 18 + 5 = 35.

Âû÷èñëèì îïðåäåëèòåëè ïåðåìåííûõ ∆1, ∆2, ∆3, êîòîðûå ïîëó÷àþòñÿ èç îáùåãî îïðåäåëèòåëÿ çàìåíîé ñîîòâåòñòâóþùåãî ñòîëáöà ñòîëáöîì ñâîáîäíûõ ÷ëåíîâ (ïðàâûõ ÷àñòåé êàæäîãî óðàâíåíèÿ):

1 = = 5∙(–2) ∙(–5) + 1∙(–3) ∙(–3) + 4∙2∙(–2) – (–2) ∙(–2) ∙(–3) –

– 2∙(–3) ∙5 – 1∙4∙(–5) = 50 + 9 – 16 + 12 + 30 + 20 = 105.

2 = = 3∙4∙(–5) + 1∙(–3)∙(–2) + 5∙(–3)∙2 – (–2)∙4∙2 – (–3)∙(–3)∙3 –

– 1∙5∙(–5) = – 60 + 6 – 30 + 16 – 27 + 25 = – 70.

3 = = 3∙(–2) ∙(–3) + 1∙2∙5 + 1∙4∙2– 5∙(–2)∙2 – 2∙4∙3 – 1∙1∙(–3) =
= 18 + 10 + 8 + 20 – 24 + 3 = 35.

Ðåøåíèå ñèñòåìû ëèíåéíûõ óðàâíåíèé íàéäåì ïî ôîðìóëàì Êðàìåðà:

x1 = ; x2 = ; x3 = .