Eliminacja Gaussa i macierze odwrotne
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Eliminacja Gaussa i macierze odwrotne
Eliminacja Gaussa i macierze odwrotne 1. (a) (c) 2. Stosuj¡c eliminacj¦ Gaussa, rozwi¡» nast¦puj¡ce ukªady równa« x1 +x2 +x3 2x1 +x2 −4x3 x +2x +7x 1 2 3 2x1 +x2 −x3 x1 +2x3 3x1 +x2 +x3 [ 2 −1 4 −3 (b) 4x1 +5x2 +6x3 x1 +2x2 +3x3 0 −2 −1 5 ] 1 2 (h) −1 3 0 1 2 1 [ (c) 2 −1 0 1 −1 (e) 0 −1 −1 2 = 6 = 4 x1 +x2 +x3 −x −x2 +2x3 1 x1 −2x2 −x3 3x1 −x2 +2x3 3 −2 −1 −4 3 2 (g) 8 −6 −3 2 −1 1 3 1 0 2 −1 2 1 5 3 = 0 = 3 = −2 = −1 ] 1 0 0 (f ) 2 −1 0 −2 1 1 6 3 2 4 3 1 (i) 1 −1 2 3 2 1 1 3 . 5 3 Korzystaj¡c z podanej macierzy odwrotnej do macierzy ukªadu, rozwi¡» poni»sze równania. (c) (d) [ 1 1 1 (d) 1 2 3 1 3 4 (b) = 1 = 2 = 3 ] (a) (b) Znajd¹ macierze odwrotne do poni»szych (o ile istniej¡). (a) 3. { = 0 = 1 = −1 [ ] 2 1 5 3 { 1 2 1 1 4 2 1 3 2 = −3 2x1 −x2 +x2 −x3 = 1 −x1 −x2 +2x3 = 1 x1 −24x2 −5x3 +32x4 −x +25x +5x −33x 1 2 3 4 −x1 +37x2 +8x3 −50x4 −5x2 −x3 +7x4 6 3 2 4 3 1 1 −1 2 3 2 1 1 3 5 3 3x1 −x2 −5x1 +2x2 = 0 = 1 = −9 = 15 = −2 = 0 4. Rozwi¡» równania macierzowe: [ (a) [ (c) 2 1 −5 −2 −1 0 −1 2 ] [ X= ] [ X 2 3 1 5 3 2 −2 −1 [ ] (b) X ] [ = 0 2 1 4 ] [ (d) 2 3 1 5 3 −2 4 2 [ [ ] = ] [ X ] [ ]2 [ 3 2 1 −2 5 (e) X = −2 −1 1 0 −3 [ ] [ ] [ 1 −1 −1 −2 3 (f ) − X= 2 0 0 1 0 ] [ [ ] [ 6 2 1 2 −1 = + (g) X −8 0 −2 0 3 −1 0 1 1 −2 1 −1 X = 0 −2 (h) 0 1 1 1 −2 0 −1 2 0 1 2 (j) 1 1 −1 X = 6 7 2 5 1 1 1 4 −4 2 −9 ] 3 1 −2 −2 −26 14 ] 1 3 ] 2 8 ] [ = 2 0 −1 3 ] ] [ 0 1 2 8 7 −6 (i) X 1 1 −1 = 12 14 2 1 1 1 ]