Lineáris egyenletrendszer megoldása? X-2y+3z=1 2x-4y+az=3

1x-2y+3z=1

2x-4y+az=3

1x-2y=1-3z

2x-4y=3-az

1x-2y=1-3z

2(1x-2y)=3-az

1x-2y=1-3z

1x-2y=(3-az)/2

1 - 3z = (3 - az)/2

1 = (3 - az)/2 + 3z

1 = 3/2 - az/2 + 3z

1 - 3/2 = - az/2 + 3z

2*1/2 - 3/2 = - az/2 + 2*3z/2

2/2 - 3/2 = - az/2 + 6z/2

(2 - 3)/2 = (- az + 6z)/2

2 - 3 = - az + 6z

-1 = 6z - az

-1 = z(6 - a)

-1/z = 6 - a

-1/z - 6 = - a

1/z + 6 = a

a = 1/z + 6

Mivel "a" eleme R-nek, ezért "z" csak 1, 0 vagy -1 lehet.

z = 1

a = 1/z + 6

a = 1/1 + 6

a = 1 + 6

a = 7

z = 0

a = 1/z + 6

a = 1/0 + 6

a = 0 + 6

a = 6

z = -1

a = 1/z + 6

a = 1/-1 + 6

a = -1 + 6

a = 5