Question

How do you solve using gaussian elimination or gauss-jordan elimination, `X + 2Y- 2Z=1`, `2X + 3Y + Z=14`, `4Y + 5Z=27`?

Answer 1

`implies ((x),(y),(z)) = ((1),(3),(3))`

Explanation:

We will use Gaussian and back-substitution.

We kick off with the starting augmented matrix (see the NB at bottom):

`((1,2,-2),(2,3,1),(0,4,5))((1),(14),(27))`

`R2 -> R2 - 2 R1`

`((1,2,-2),(0,-1,5),(0,4,5))((1),(12),(27))`

`R3 -> R3 + 4 R2`

`((1,2,-2),(0,-1,5),(0,0,25))((1),(12),(75))`

That was really quick, actually; and this is now in row echelon form.

We can see that:

`((1,2,-2),(0,-1,5),(0,0,25))((x),(y),(z)) = ((1),(12),(75))`

We back substitute from:

`25z = 75 implies z = 3`

So:

`-y + 5 z = 12 implies y = 5z - 12 = 3`

And:

`x + 2y - 2z = 1 implies x = 1 - 2(3) + 2(3) = 1`

`implies ((x),(y),(z)) = ((1),(3),(3))`

NB

The augmented matrix will not necessarily be represented as it might be in your book as I can't find the functionality in the typesetter here. (So I've got a 3x3 and a column vector.)