How do you solve using gaussian elimination or gauss-jordan elimination, #x-2y+z=1#, #2x-3y+z=5#, #-x-2y+3z=-13#?

1 Answer
Oct 11, 2017

Answer:

#x = 7+t#, #y=3+t#, #z=t#

Explanation:

Begin by creating the augmented matrix, or a matrix with the #x#, #y#, and #z# coefficients on the left side of a vertical bar, and the constant values on the right side:

# [ [1,-2,1,|,1], [2,-3,1,|,5] , [-1,-2,3,|,-13] ] #

We now perform row operations on the 3 rows in order to reduce the left portion of the augmented matrix to the identity form:

# [ [1,0,0], [0,1,0], [0,0,1] ]#

We can do any of the following:

  • Swap any two rows
  • Multiply a single row by a non-zero constant value
  • Add/Subtract a constant multiple of a row to another row

Using the notation #R_n# to refer to row #n#, here's the steps I'd use:

# [ [1,-2,1,|,1], [2,-3,1,|,5] , [-1,-2,3,|,-13] ] {:(),(-2R_1 rightarrow),(+R_1):} #

# [ [1,-2,1,|,1], [0,1,-1,|,3] , [0,-4,4,|,-12] ] {:(),( rightarrow),(+4R_2):} #

# [ [1,-2,1,|,1], [0,1,-1,|,3] , [0,0,0,|,0] ] {:(+2R_2),( rightarrow),():} #

# [ [1,0,-1,|,7], [0,1,-1,|,3] , [0,0,0,|,0] ] #

We can do nothing further [useful]. Notice the bottom line has all zero values. This indicates that this system of equations does not result in a unique #z# value. In fact, we can translate this reduced row matrix back to a system of three unknowns in two equations:

#{(x,,-z,=,7),(,y,-z,=,3):} #

Any value of #z# chosen (out of infinitely many) will result in a unique solution. For instance, if #z=1#, we can find #x=8# and #y=4#.

It's commonly the case for problems like this with infinite solutions that we define everything in terms of a template variable #t# like so:

#x = 7+t#, #y=3+t#, #z=t#