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)]#