Question

How do you solve the system of equations `5a + 3b + c = - 5`, `a - 3b + 2c = - 1`, and `14a - 2b + 3c = 6`?

Answer 1

Please see the explanation for steps leading to: `a = 1, b = -2, and c = -4`

Explanation:

Put the coefficients in a augmented matrix:
#a - 3b + 2c = -1 to [ (1, -3,2,|-1) ]#

#5a + 3b + c = -5 to [ (1, -3,2,|-1), (5,3,1,|-5) ]#

#14a - 2b + 3c = 6 to [ (1, -3,2,|,-1), (5,3,1,|,-5), (14, -2,3,|,6) ]#

Perform row operations, until you get an identity matrix:

#[ (1, -3,2,|,-1), (5,3,1,|,-5), (14, -2,3,|,6) ]#

Multiply row 1 by -5 and add to row 2:

#[ (1, -3,2,|,-1), (0,18,-9,|,0), (14, -2,3,|,6) ]#

Multiply row 1 by -14 and add to row 2:

#[ (1, -3,2,|,-1), (0,18,-9,|,0), (0, 40,-25,|,20) ]#

Divide row 2 by 9 and row 3 by 5:

#[ (1, -3,2,|,-1), (0,2,-1,|,0), (0, 8,-5,|,4) ]#

Multiply row 2 by -4 and add to row 3:

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

Multiply row 3 by -1 and leave it that way after adding to row 2:

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

Divide row 3 by 2:

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

Multiply row 3 by -2 and add to row 1:

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

Multiply row 2 by 3 and add to row 1:

#[ (1, 0,0,|,1), (0,1,0,|,-2), (0, 0,1,|,-4) ]#

The identity matrix says, `a = 1, b = -2, and c = -4`

Check:

`1 - 3(-2) + 2(-4) = -1`

`5(1) + 3(-2) + -4 = -5`

`14(1) - 2(-2) + 3(-4) = 6`

`-1 = -1`

`-5 = -5`

`6 = 6`

This checks.