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ResoluciΓ³n de ecuaciones de 2 incognitas 2x2 por determinantes
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matrices y determinantes 2x2.pptx
1.
2 3 5 2 2
3 5 2 = 2 π₯ 2 β 4 β 15 β11 MATRICES Y DETERMINANTES 5 π₯ 3
2.
2 8 1 β2 2
8 1 β2 = 2 π₯ (β2) β β4 β 8 β12 MATRICES Y DETERMINANTES 8 π₯ 1
3.
3 β1 β2 β3 1 6 3
β1 β2 β3 = 3 π₯ (β3) β β9 β 2 = β9 β 2 β= β11 MATRICES Y DETERMINANTES 3π₯ β 1π¦ = 1 β2π₯ β 3π¦ = 6 Encontrar valor general Encontrar valor para x 1 β1 6 β3 = 1 π₯ (β3) β β3 β β6 = β3 + 6 β1 = 3 Encontrar valor para y 3 1 β2 6 = 3 π₯ 6 β 18 β β2 = 18 + 2 β2 =20 β1 π₯(β2) β1 π₯ 6 1 π₯(β2)
4.
3 β1 β2 β3 1 6 3
β1 β2 β3 = β MATRICES Y DETERMINANTES 3π₯ β 1π¦ = 1 β2π₯ β 3π¦ = 6 1 β1 6 β3 = β1 3 1 β2 6 = β2 π₯ = β1 β = 3 β11 π¦ = β2 β = 20 β11
5.
MATRICES Y DETERMINANTES 3π₯
β 1π¦ = 1 (1) β2π₯ β 3π¦ = 6 (2) Despejando y en (1) 3π₯ β 1π¦ = 1 3π₯ β 1π¦ = 1 β 1 β3π₯ + 1π¦ = β1 π¦ = β1 + 3π₯ Sustituyendo y= -1 + 3x en (2) β2π₯ β 3π¦ = 6 β2π₯ β 3(β1 + 3π₯) = 6 β2π₯ + 3 β 9π₯ = 6 β2π₯ β 9π₯ = 6 β 3 β11π₯ = 3 π₯ = 3 β11 Sustituyendo x en y= -1 + 3x π¦ = β1 + 3π₯ π¦ = β1 + (3) 3 β11 π¦ = β1 + 9 β11 π¦ = β1 β 9 11 π₯ = β 3 11 π¦ = β 11 11 β 9 11 = β20 11 π¦ = β 20 11
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