Desarrollo punto 2 Wilmer Pupo.docx

# Desarrollo punto 2 Wilmer Pupo.docx - 2 Dadas las matrices...

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2. Dadas las matrices: A = | 3 5 1 4 | B = | 4 2 5 0 4 2 | C = | 1 4 3 4 1 2 6 2 0 | Calcule si es posible: a) C .B. A Para esto primero asociamos y resolvemos: ( C .B ) . A ( | 1 4 3 4 1 2 6 2 0 | | 4 2 5 0 4 2 | ) | 3 5 1 4 | Multiplicamos la primera parte: | 1 4 3 4 1 2 6 2 0 | | 4 2 5 0 4 2 | Al ser C una matriz 3X3 y B una matriz 3X2, observamos: 3 X 3 = 3 X 2 vemos que sus elementos centrales son 3 y si se puede hacer el producto y su resultado serán sus externos, es decir, 3X2: C.B = | 1 4 3 4 1 2 6 2 0 | | 4 2 5 0 4 2 | = | 1 ( 4 ) + 4 ( 5 ) + 3 ( 4 ) 1 ( 2 ) + 4 ( 0 ) + 3 ( 2 ) 4 ( 4 ) + 1 ( 5 ) +(− 2 )( 4 ) 4 ( 2 ) + 1 ( 0 ) +(− 2 )( 2 ) 6 ( 4 ) +(− 2 ) ( 5 ) +( 0 )( 4 ) 6 ( 2 ) +(− 2 ) ( 0 ) +( 0 )( 2 ) | C.B = | 4 + 20 + 12 2 + 0 + 6 16 + 5 8 8 + 0 4 24 10 + 0 12 + 0 + 0 | C.B = | 36 4 13 12 14 12 |

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Ahora procedemos a multiplicar el C.B por A, revisando que esta matriz es 3X2 y la matriz A es 2X2, sus interiores son 2 por lo que si se pueden multiplicar y su resultado será una matriz 3X2: ( C .B ) . A = | 36 4 13 12 14 12 | | 3 5 1 4 | = | 36 ( 3 ) + 4 ( 1 ) 36 ( 5 ) + 4 ( 4 ) 13 ( 3 ) +(− 12 )( 1 ) 13 ( 5 ) +(− 12 )( 4 ) 14 ( 3 ) +(− 12 )( 1 ) 14 ( 5 ) +(− 12 )( 4 ) | ( C .B ) . A = | 108 + 4 180 + 16 39 12 65 48
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• Spring '17
• jerson gonzalez

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