How do you solve #((2, 2, 0), (4, -3, 2), (0, -3, 5))x=((-10), (2), (-9))#?

1 Answer
Mar 1, 2016

Answer:

#x_1 = -17/29; x_2=-128/29; x_3 = -129/29#

Explanation:

Firs and foremost correct the question the way it is written is meaningless. What the question should read is:
#((2, 2, 0), (4, -3, 2), (0, -3, 5)) ((x_1), (x_2), (x_3))=((-10), (2), (-9))#
x is a vector #vecx = ((x_1), (x_2), (x_3)) # OK?
Now Either use a determinant or process of elimination to:
#((2, 2, 0, |, -10), (4, -3, 2, |, 2), (0, -3, 5, |, -9))#
#color(blue)(R_2 => R_2 -2*R_1)#
#((2, 2, 0, |, -10), (color(blue)0, color(blue)(-7), color(blue)2, |, color(blue)22), (0, -3, 5, |, -9))#
#color(red)(R_3 => R_3 -3/7*R_2)#
#((2, 2, 0, |, -10), (0, -7, 2, |, 22), (0, 0, color(red)(29/7), |, color(red)(-129/7)))#
you're almost there:
#color(red)(29/7 x_3 = -129/7; x_3 = -129/29 #
Now work your way back to determine #x_2#
#color(blue) (-7x_2 -2*129/29 = 22; x_2=-128/29#
and last equation say
#2x_1 - 2*128/29 = -10; x_1 = -17/29#