maths-matrix

Larry

2011-05-07 5:38 am

Solve thesystem of equations, where p≠-1, p≠0 and p≠4.px+2y-2z=0……….(1)-px-(p+1)y+(1-2p)z=p(p+2)……….(2)2px+6y+(p-2)z=-p^2……….(3)

回答 (1)

匿名

2011-05-08 1:59 am

✔ 最佳答案

[ p 2 -2 0 ]
[ -p -p-1 -2p+1 p^2+2p ]
[ 2p 6 p-2 -p^2 ]

(2)=(1)+(2), (3)=(3)-2*(1)
[ p 2 -2 0 ]
[ 0 -p+1 -2p-1 p^2+2p ]
[ 0 2 p+2 -p^2 ]

(1)=(1)-(3), (2)=(3)-2*(2)
[ p 0 -p-4 p^2 ]
[ 0 2p 5p+4 -3p^2-4p ]
[ 0 2 p+2 -p^2 ]

(3)=p*(3)-(2)
[ p 0 -p-4 p^2 ]
[ 0 2p 5p+4 -3p^2-4p ]
[ 0 0 p^2-3p-4 -p^3+3p^2+4p ]

(3)=(3)/(p^2-3p-4)
[ p 0 -p-4 p^2 ]
[ 0 2p 5p+4 -3p^2-4p ]
[ 0 0 1 -p ]

(1)=(1)+(p+4)*(3), (2)=(2)-(5p+4)*(3)
[ p 0 0 -4p ]
[ 0 2p 0 2p^2 ]
[ 0 0 1 -p ]

(1)=(1)/p, (2)=(2)/(2p)
[ 1 0 0 -4 ]
[ 0 1 0 p ]
[ 0 0 1 -p ]

So, x = -4, y = p, and z = -p

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