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Mathematics 18 Online
OpenStudy (anonymous):

Use the Gauss jordan elimination to find the solution to the following system of equations. -x +y+z = -1 -x +5y-19z = -13 3x-2y-8z= 0 Select the correct choice below and fill in any answer boxes within your choice. a. there is one solution. the solution is x=__, y=__ and z=___ b. there are infinitely many solutions. if z is any real number, x=__ and y=__. c. there is no solution.

OpenStudy (anonymous):

@jim_thompson5910

jimthompson5910 (jim_thompson5910):

-x +y+z = -1 -x +5y-19z = -13 3x-2y-8z= 0 -1 1 1 | -1 -1 5 -19 | -13 3 -2 -8 | 0 1 -1 -1 | 1 -1*R1 -1 5 -19 | -13 3 -2 -8 | 0 1 -1 -1 | 1 0 4 -20 | -12 R2 + 1*R1 3 -2 -8 | 0 1 -1 -1 | 1 0 4 -20 | -12 0 1 -5 | 3 R3 - 3*R1 1 -1 -1 | 1 0 1 -5 | -3 (1/4)*R2 0 1 -5 | 3 1 -1 -1 | 1 0 1 -5 | -3 0 0 0 | 0 R3 - 1*R2 1 0 -6 | -2 R1 + 1*R2 0 1 -5 | -3 0 0 0 | 0 Since the last row is 0 0 0 | 0 which means 0x + 0y + 0z = 0 or just 0 = 0

jimthompson5910 (jim_thompson5910):

since 0 = 0 is always true, there are infinitely many solutions

jimthompson5910 (jim_thompson5910):

I'll let you fill out the rest

OpenStudy (anonymous):

okay so z= 0 and x= -2 and y= -3

OpenStudy (anonymous):

@jim_thompson5910 which one should i circle? B.?? if its infintely many solutions and what should i type in the boxes?

jimthompson5910 (jim_thompson5910):

1 0 -6 | -2 turns into x - 6z = -2 so x = -2 + 6z see how I got this?

jimthompson5910 (jim_thompson5910):

that is your first solution do the same thing to find y

OpenStudy (anonymous):

can u help me find the y i found the x

jimthompson5910 (jim_thompson5910):

0 1 -5 | -3 turns into what?

OpenStudy (anonymous):

y=-5-3

jimthompson5910 (jim_thompson5910):

y - 5z = -3 solve for y

OpenStudy (anonymous):

y= -3+5z

jimthompson5910 (jim_thompson5910):

yep

jimthompson5910 (jim_thompson5910):

so x = -2 + 6z and y = -3 + 5z

jimthompson5910 (jim_thompson5910):

z is the free variable

OpenStudy (anonymous):

yes cuz its 0 :)

jimthompson5910 (jim_thompson5910):

well z can be any number you want it to be x and y will depend on whatever value of z you pick

OpenStudy (anonymous):

oh okay that makes sense!

jimthompson5910 (jim_thompson5910):

that adds further confirmation that there are infinitely many solutions

OpenStudy (anonymous):

ohh okay thank you! :)

jimthompson5910 (jim_thompson5910):

np

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