How do you solve #x+2y+3w+4z = 10#, #2x+y+w-z = 1#, #3x+y+4w+3z=22#, and #-2x+6y+4w+20z = 18# using matrices?

1 Answer
Apr 7, 2016

Answer:

See explanation...

Explanation:

Write these equations as a #4 xx 5# matrix, then perform a sequence of row operations until the left hand #4 xx 4# matrix is the identity matrix. Then the right hand column will be the solution.

Start with:

#((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (3, 1, 4, 3, 22), (-2, 6, 4, 20, 18))#

Add row #2# to row #4# to get:

#((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (3, 1, 4, 3, 22), (0, 7, 5, 19, 19))#

Subtract rows #1# and #2# from row #3# to get:

#((1, 2, 3, 4, 10), (2, 1, 1, -1, 1), (0, -2, 0, -2, 11), (0, 7, 5, 19, 19))#

Subtract twice row #1# from row #2# to get:

#((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -2, 0, -2, 11), (0, 7, 5, 19, 19))#

Add row #2# plus twice row #3# to row #4# to get:

#((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -2, 0, -2, 11), (0, 0, 0, 6, 22))#

Multiply row #3# by #3# to get:

#((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, -6, 0, -6, 33), (0, 0, 0, 6, 22))#

Subtract twice row #2# from row #3# to get:

#((1, 2, 3, 4, 10), (0, -3, -5, -9, -19), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))#

Multiply row #2# by #-1/3# to get:

#((1, 2, 3, 4, 10), (0, 1, 5/3, 3, 19/3), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))#

Subtract twice row #2# from row #1# to get:

#((1, 0, -1/3, -2, -8/3), (0, 1, 5/3, 3, 19/3), (0, 0, 10, 12, 71), (0, 0, 0, 6, 22))#

Divide row #3# by #10# to get:

#((1, 0, -1/3, -2, -8/3), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 6, 22))#

Add #1/3# row #3# to row #1# to get:

#((1, 0, 0, -8/5, -3/10), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 6, 22))#

Divide row #4# by #6# to get:

#((1, 0, 0, -8/5, -3/10), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))#

Add #8/5# row #4# to row #1# to get:

#((1, 0, 0, 0, 167/30), (0, 1, 5/3, 3, 19/3), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))#

Subtract #5/3# row #3# from row #2# to get:

#((1, 0, 0, 0, 167/30), (0, 1, 0, 1, -11/2), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))#

Subtract row #4# from row #2# to get:

#((1, 0, 0, 0, 167/30), (0, 1, 0, 0, -55/6), (0, 0, 1, 6/5, 71/10), (0, 0, 0, 1, 11/3))#

Subtract #6/5# row #4# from row #3# to get:

#((1, 0, 0, 0, 167/30), (0, 1, 0, 0, -55/6), (0, 0, 1, 0, 27/10), (0, 0, 0, 1, 11/3))#

Provided I have made no arithmetic errors:

#{ (x = 167/30), (y = -55/6), (w = 27/10), (z = 11/3) :}#