CAPITULO XXV
Hallar el valor de una determinante de tercer orden
Hallar el valor de una determinante de tercer orden
- Ejercicio 187
Hallar el valor de las siguientes determinantes:
- | 1 2 1 1 3 4 1 0 2 | | 1 2 1 1 3 4 1 0 2 1 2 1 1 3 4 | = [ ( 1 Γ 3 Γ 2 ) +( 1 Γ 0 Γ 1 ) +( 1 Γ 2 Γ 4 ) ] β[ ( 1 Γ 3 Γ 1 ) +( 4 Γ 0 Γ 1 ) +( 2 Γ 2 Γ 1 ) ] = [ 6+0+8 ] β[ 3+0+4 ] = 14β7 = 7
- | 1 2 β 2 1 β 3 3 β 1 4 5 | | 1 2 β 2 1 β 3 3 β 1 4 5 1 2 β 2 1 β 3 3 | = [ ( 1 Γ β3 Γ 5 ) +( 1 Γ 4 Γ β2 ) +( β1 Γ 2 Γ 3 ) ] β[ ( β2 Γ β3 Γ β1 ) +( 3 Γ 4 Γ 1 ) +( 5 Γ 2 Γ 1 ) ] = [ β15β8β6 ] β[ β6+12+10 ] = β29β16 = β45
- | β 3 4 1 2 β 3 0 1 2 7 | | β 3 4 1 2 β 3 0 1 2 7 β 3 4 1 2 β 3 0 | = [ ( β3 Γ β3 Γ 7 ) +( 2 Γ 2 Γ 1 ) +( 1 Γ 4 Γ 0 ) ] β[ ( 1 Γ β3 Γ β1 ) +( β3 Γ 2 Γ 0 ) +( 2 Γ 4 Γ 7 ) ] = [ 63+4+0 ] β[ β3+0+56 ] = 67β53 = 14
- | 2 5 β 1 3 β 4 3 6 2 4 | | 2 5 β 1 3 β 4 3 6 2 4 2 5 β 1 3 β 4 3 | = [ ( 2 Γ β4 Γ 4 ) +( 3 Γ 2 Γ β1 ) +( 6 Γ 5 Γ 3 ) ] β[ ( β1 Γ β4 Γ 6 ) +( 2 Γ 2 Γ 3 ) +( 3 Γ 5 Γ 4 ) ] = [ β32β6+90 ] β[ 24+12+60 ] = 52β96 = β44
- | 5 β 1 β 6 β 2 5 3 3 4 2 | | 5 β 1 β 6 β 2 5 3 3 4 2 5 β 1 β 6 β 2 5 3 | = [ ( 5 Γ 5 Γ 2 ) +( β2 Γ 4 Γ β6 ) +( 3 Γ β1 Γ 3 ) ] β[ ( 3 Γ 5 Γ β6 ) +( 5 Γ 4 Γ 3 ) +( β2 Γ β1 Γ 2 ) ] = [ 50+48β9 ] β[ β90+60+4 ] = 89+26 = 115
- | 4 1 5 3 2 β 6 12 3 2 | | 4 1 5 3 2 β 6 12 3 2 4 1 5 3 2 β 6 | = [ ( 4 Γ 2 Γ 2 ) +( 3 Γ 3 Γ 5 ) +( 12 Γ 1 Γ β6 ) ] β[ ( 12 Γ 2 Γ 5 ) +( 4 Γ 3 Γ β6 ) +( 3 Γ 1 Γ 2 ) ] = [ 16+45β72 ] β[ 120β72+6 ] = β11β54 = β65
- | 5 2 β 8 β 3 β 7 3 4 0 β 1 | | 5 2 β 8 β 3 β 7 3 4 0 β 1 5 2 β 8 β 3 β 7 3 | = [ ( 5 Γ β7 Γ β1 ) +( β3 Γ 0 Γ β8 ) +( 4 Γ 2 Γ 3 ) ] β[ ( 4 Γ β7 Γ β8 ) +( 5 Γ 0 Γ 3 ) +( β3 Γ 2 Γ β1 ) ] = [ 35+0+24 ] β[ 224+0+6 ] = 59β230 = β171
- | 3 2 5 β 1 β 3 4 3 2 5 | | 3 2 5 β 1 β 3 4 3 2 5 3 2 5 β 1 β 3 4 | = [ ( 3 Γ β3 Γ 5 ) +( β1 Γ 2 Γ 5 ) +( 3 Γ 2 Γ 4 ) ] β[ ( 3 Γ β3 Γ 5 ) +( 3 Γ 2 Γ 4 ) +( β1 Γ 2 Γ 5 ) ] = [ β45β10+24 ] β[ β45+24β10 ] = β31+31 = 0
- | 5 2 3 6 1 2 3 4 5 | | 5 2 3 6 1 2 3 4 5 5 2 3 6 1 2 | = [ ( 5 Γ 1 Γ 5 ) +( 6 Γ 4 Γ 3 ) +( 3 Γ 2 Γ 2 ) ] β[ ( 3 Γ 1 Γ 3 ) +( 5 Γ 4 Γ 2 ) +( 6 Γ 2 Γ 5 ) ] = [ 25+72+12 ] β[ 9+40+60 ] = 109β109 = 0
- | 12 5 10 8 β 6 9 7 4 β 2 | | 12 5 10 8 β 6 9 7 4 β 2 12 5 10 8 β 6 9 | = [ ( 12 Γ β6 Γ β2 ) +( 8 Γ 4 Γ 10 ) +( 7 Γ 5 Γ 9 ) ] β[ ( 7 Γ β6 Γ 10 ) +( 12 Γ 4 Γ 9 ) +( 8 Γ 5 Γ β2 ) ] = [ 144+320+315 ] β[ β420+432β80 ] = 779+68 = 847
- | β 9 3 β 4 7 β 5 β 3 4 6 1 | | β 9 3 β 4 7 β 5 β 3 4 6 1 β 9 3 β 4 7 β 5 β 3 | = [ ( β9 Γ β5 Γ 1 ) +( 7 Γ 6 Γ β4 ) +( 4 Γ 3 Γ β3 ) ] β[ ( 4 Γ β5 Γ β4 ) +( β9 Γ 6 Γ β3 ) +( 7 Γ 3 Γ 1 ) ] = [ 45β168β36 ] β[ 80+162+21 ] = β159β263 = β422
- | 11 β 5 7 β 12 3 8 β 13 1 9 | | 11 β 5 7 β 12 3 8 β 13 1 9 11 β 5 7 β 12 3 8 | = [ ( 11 Γ 3 Γ 9 ) +( β12 Γ 1 Γ 7 ) +( β13 Γ β5 Γ 8 ) ] β[ ( β13 Γ 3 Γ 7 ) +( 11 Γ 1 Γ 8 ) +( β12 Γ β5 Γ 9 ) ] = [ 297β84+520 ] β[ β273+88+540 ] = 733β355 = 378