Question

How do you solve the simultaneous equations `x+y+z=-2`, `2x+5y+2z=-10`, `-x+6y-3z=-16` ?

Answer 1

`(x, y, z) = (-2, -2, 2)`

Explanation:

Given:

`{ (x+y+z=-2),(2x+5y+2z=-10),(-x+6y-3z=-16) :}`

Subtracting twice the first equation from the second, we get:

`3y = -6`

Dividing both sides by `3` we find:

`y = -2`

Adding the first and third equation together, we get:

`7y-2z = -18`

Substituting `y=-2` into this equation, we get:

`-14-2z = -18`

Add `14` to both sides to get:

`-2z = -4`

Divide both sides by `-2` to get:

`z = 2`

Then putting `y=-2` and `z = 2` in the first equation, we find:

`x-color(red)(cancel(color(black)(2)))+color(red)(cancel(color(black)(2)))=-2`

Hence:

`x = -2`

Answer 2

Use the 3 equations to write an Augmented Matrix and then perform elementary row operations until you obtain an identity matrix.

Explanation:

Write the augmented matrix:

#[ (1,1,1,|,-2), (2,5,2,|,-10), (-1,6,-3,|,-16) ]#

Perform elementary row operations.

`-2R_1+R_2toR_2`

#[ (1,1,1,|,-2), (0,3,0,|,-6), (-1,6,-3,|,-16) ]#

`R_2/3`

#[ (1,1,1,|,-2), (0,1,0,|,-2), (-1,6,-3,|,-16) ]#

`R_1+R_3toR_3`

#[ (1,1,1,|,-2), (0,1,0,|,-2), (0,7,-2,|,-18) ]#

`-7R_2+R_3toR_3`

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

`R_3/-2`

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

`R_1-R_2toR_1`

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

`R_1-R_3toR_1`

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

We have obtained an identity matrix and the right column contains the solution set:

`x = -2, y = -2, and z = 2`