Question #bb623

1 Answer
Feb 1, 2018

Given an invertible #nxxn# matrix A, a #1xxn# matrix of variables X, and a #1xxn# matrix of constants C such that:
#A*X=C#
The solution set can be found by multiplying both sides by #A^-1#:
#X=A^-1*C#

Explanation:

We can find #A^-1# by starting with the matrix of coefficients A:

#A = [(1,1,3), (0,1,2), (3,5,-1)]#

Augment A with an identity matrix:

#A|I = [(1,1,3,|,1,0,0), (0,1,2,|,0,1,0), (3,5,-1,|,0,0,1)]#

Perform row operations until the left side becomes an identity matrix, then the inverse will be the matrix on the right,

Multiply Row 1 by -3, add to Row 3 and put the result in Row 3:

Shorthand: #-3R_1+R_3toR_3#:

#[(1,1,3,|,1,0,0), (0,1,2,|,0,1,0), (0,2,-10,|,-3,0,1)]#

#-2R_2+R_3toR_3#:

#[(1,1,3,|,1,0,0), (0,1,2,|,0,1,0), (0,0,-14,|,-3,-2,1)]#

#-1R_2+R_1toR_1#:

#[(1,0,1,|,1,-1,0), (0,1,2,|,0,1,0), (0,0,-14,|,-3,-2,1)]#

#-1/14R_3toR_3#

#[(1,0,1,|,1,-1,0), (0,1,2,|,0,1,0), (0,0,1,|,3/14,1/7,-1/14)]#

#-2R_3+R_2toR_2#

#[(1,0,1,|,1,-1,0), (0,1,0,|,-3/7,5/7,1/7), (0,0,1,|,3/14,1/7,-1/14)]#

#-1R_3+R_1toR_1#

#[(1,0,0,|,11/14,-8/7,1/14), (0,1,0,|,-3/7,5/7,1/7), (0,0,1,|,3/14,1/7,-1/14)]= I|A^-1#

We have an identity matrix on the left, therefore, #A^-1# is the matrix on the right:

#A^-1= [(11/14,-8/7,1/14), (-3/7,5/7,1/7), (3/14,1/7,-1/14)]#

We are given that #A*X=C# is:

# [(1,1,3), (0,1,2), (3,5,-1)] [(x), (y), (z)] = [(a),(b),(c) ]#

Multiply both sides by #A^-1#:

#[(11/14,-8/7,1/14), (-3/7,5/7,1/7), (3/14,1/7,-1/14)] [(1,1,3), (0,1,2), (3,5,-1)] [(x), (y), (z)] = [(11/14,-8/7,1/14), (-3/7,5/7,1/7), (3/14,1/7,-1/14)] [(a),(b),(c) ]#

The left side collapses to the column of variables:

# [(x), (y), (z)] = [(11/14,-8/7,1/14), (-3/7,5/7,1/7), (3/14,1/7,-1/14)] [(a),(b),(c) ]#

NOTE: Actually, I have eliminated a step where the inverse and the matrix become the identity matrix, which collapses to X as follows,
#A^-1*A*X= I*X = X#, but this level of formality is not needed here.

Perform the matrix multiplication:

# [(x), (y), (z)] = [(11/14a-8/7b+1/14c), (-3/7a+5/7b+1/7c), (3/14a+1/7b-1/14c)]#