How do you solve using gaussian elimination or gauss-jordan elimination, 2x + 5y - 2z = 14, 5x -6y + 2z = 0, 4x - y + 3z = -7?

2 Answers
Jan 6, 2018

P={(2,0,-5)}

Explanation:

([2,5,-2,|,14],[5,-6,2,|,0],[4,-1,3,|,-7])~~([10,25,-10,|,70],[10,-12,4,|,0],[0,-11,7,|,-35])
R_3=R_3-2xxR_1
R_2=R_2xx2
R_1=R_1xx5
R_2=R_2-R_1

([10,25,-10,|,70],[0,-37,14,|,-70],[0,-11,7,|,-35])~~([10,25,-10,|,70],[0,-407,154,|,-770],[0,-407,259,|,-1295])
R_2=R_2xx11
R_3=R_3xx37
R_3=R_3-R_2

([10,25,-10,|,70],[0,-407,154,|,-770],[0,0,105,|,-525])~~([10,25,-10,|,70],[0,-37,14,|,-70],[0,0,1,|,-5])
R_3=R_3xx1/105
R_2=R_2xx1/11
R_2=R_2-14xxR_3
R_1=R_1+10xxR_3

([10,25,0,|,20],[0,-37,0,|,0],[0,0,1,|,-5])
R_2=R_2xx-1/37
R_1=R_1-25xxR_2
R_1=R_1xx1/10

([1,0,0,|,2],[0,1,0,|,0],[0,0,1,|,-5])

P={(2,0,-5)}

Jan 7, 2018

x=2, y=0 and z=-5

Explanation:

From third equation, y=4x+3z+7

Hence,

2x+5*(4x+3z+7)-2z=14 or 22x+13z=-21 (1) and,

5x-6*(4x+3z+7)+2z=0 or -19x-16z=42 (2)

From (1), z=(-22x-21)/13

Consequently,

-19x-16*(-22x-21)/13=42

(352x+336)/13-19x=42

(352x+336)/13=19x+42

352x+336=13*(19x+42)

352x+336=247x+546

105x=210, so x=2

Thus,

z=((-22)*2-21)/13=-5 and,

y=4*2+3*(-5)+7=0