How do you solve #[(1,-2,3), (2,-4,6), (3,-6,9)][(x), (y), (z)]=[(0),(0),(0)]#?

1 Answer
Oct 5, 2017

Answer:

# x - 2y+3z = 0 #

Explanation:

We can solve this system of linear equations by using Gaussian Elimination by setting up an augmented matrix of the equation coefficients.

#( (1, -2, 3, |, 0), (2, -4, 6, |, 0), (3, -6, 9, |, 0) )#

We can now perform elementary row operations:

#( (1, -2, 3, |, 0), (2, -4, 6, |, 0), (3, -6, 9, |, 0) ) stackrel(R_2-2R_1 rarr R_2)(rarr) ( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (3, -6, 9, |, 0) )#

#( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (3, -6, 9, |, 0) ) stackrel(R_3-3R_1 rarr R_3)(rarr) ( (1, -2, 3, |, 0), (0, 0, 0, |, 0), (0, 0, 0, |, 0) )#

We can now use back substitution to get the values of #x#, #y#, and #z#:

From Row #1# we have:

# x - 2y+3z = 0 #

Thus we have infinitely many solutions.

To interpret this geometrically we have the insertion of three identical planes with equation:

# x - 2y+3z = 0 #

whose points of intersection are the entire plane.