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

##### 1 Answer

#(x,y,z) = (2,1/2,-2)#

Try setting up a matrix:

#[(3,-2,-1,|,7),(1,2,-1,|,5),(-1,4,2,|,-4)]#

Do note that your second equation has

To row-reduce this using gaussian elimination, simply row-reduce this using elementary row operations, i.e. any of the following in any combination:

- Scale a row, i.e.
#cR_i# , where#c# is a constant. - Add row
#i# onto row#j# and store in row#j# , i.e.#R_i + R_j#

Hence, if I write

Our process in general will be to get the matrix down to a solvable form. Gauss-Jordan elimination would have gotten it down to reduced-row-echelon form, though that is not necessary to solve this system.

#stackrel(R_2 + R_3" ")(->)[(3,-2,-1,|,7),(1,2,-1,|,5),(0,6,1,|,1)]#

#stackrel(-1/3R_1 + R_2" ")(->)[(3,-2,-1,|,7),(0,8/3,-2/3,|,8/3),(0,6,1,|,1)]#

#stackrel(3/2R_2" ")(->)[(3,-2,-1,|,7),(0,4,-1,|,4),(0,6,1,|,1)]#

#stackrel(R_3 + R_2; 1/5R_2" ")(->)[(3,-2,-1,|,7),(0,2,0,|,1),(0,6,1,|,1)]#

Good enough; we can solve backwards now. The current system of equations looks like:

#3x - 2y - z = 7#

#0x + 2y + 0 = 1#

#0x + 6y + 1 = 1#

Thus, from the second equation we have:

#2y = 1 => color(green)(y = 1/2)#

#6y + z = 1#

#=> 6(1/2) + z = 1 => color(green)(z = -2)#

#3x - 2y - z = 7#

#=>3x - 2(1/2) - (-2) = 7#

#=> 3x = 6 => color(green)(x = 2)#

Hence, we got

Let's check...

#3x - 2y - z#

#= 3(2) - 2(1/2) - (-2) = 6 - 1 + 2 = 7 color(blue)(sqrt"")#

#x + 2y - 5#

#= 2 + 2(1/2) - 5 = 2 + 1 - 5 = -2 color(blue)(sqrt"")#

#-x + 4y + 2z#

#= -(2) + 4(1/2) + 2(-2) = -2 + 2 - 4 = -4 color(blue)(sqrt"")#