How do you solve the system #-3a-b-3c=-8#, #-5a+3b+6c=-4#, and #-6a-4b+c=-20#?

The matrix of coefficients is:

#[ (-3,-1,-3), (-5,3,6), (-6,-4,1) ]#

The determinant is:

#| (-3,-1,-3), (-5,3,6), (-6,-4,1) | = -164#

The determinant is not zero, therefore, the system has a unique solution.

Append the constants to the matrix of coefficients, to create an augmented matrix :

#[ (-3,-1,-3,|,-8), (-5,3,6,|,-4), (-6,-4,1,|,-20) ]#

Peform elementary row operations on the augmented matrix, until you obtain an identity matrix on the left:

#R_2 - R_3 rarr R_2#

#[ (-3,-1,-3,|,-8), (1,7,7,|,16), (-6,-4,1,|,-20) ]#

#R_1 harr R_2#

#[ (1,7,7,|,16), (-3,-1,-3,|,-8), (-6,-4,1,|,-20) ]#

#R_3 - 2R_2 rarr R_3#

#[ (1,7,7,|,16), (-3,-1,-3,|,-8), (0,-2,7,|,-4) ]#

#R_2 + 3R_1 to R_2#

#[ (1,7,7,|,16), (0,20,18,|,40), (0,-2,7,|,-4) ]#

#R_2 + 10R_3 to R_3#

#[ (1,7,7,|,16), (0,20,18,|,40), (0,0,88,|,0) ]#

#(1/88)R_3 to R_3#

#[ (1,7,7,|,16), (0,20,18,|,40), (0,0,1,|,0) ]#

#R_2 - 18R_2 to R_2#

#[ (1,7,7,|,16), (0,20,0,|,40), (0,0,1,|,0) ]#

#R_1 - 7R_3 to R_1#

#[ (1,7,0,|,16), (0,20,0,|,40), (0,0,1,|,0) ]#

#(1/20)R_2 to R_2#

#[ (1,7,0,|,16), (0,1,0,|,2), (0,0,1,|,0) ]#

#R_1 - 7R_2 to R_1#

#[ (1,0,0,|,2), (0,1,0,|,2), (0,0,1,|,0) ]#

We have an identity matrix on the left, therefore, we can read the values of the variables on the right:

#a = 2, b = 2, and c = 0#

Check:

#-3a-1b-3c =-8#
#-5a+3b+6c=-4#
#-6a-4b+1c=-20#

#-3(2)-1(2)-3(0) =-8#
#-5(2)+3(2)+6(0)=-4#
#-6(2)-4(2)+1(0)=-20#

#-8 =-8#
#-4=-4#
#-20=-20#

This checks.

The solution is: #a = 2, b = 2, and c = 0#