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

1 Answer
Jan 5, 2017

Answer:

Verify that the determinant of the matrix of coefficients is not zero. If so, then append the constants to the matrix to form an augmented matrix and perform elementary row operations to solve it.

Explanation:

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#