Question 519029
<pre>
This system

{{{system(x+4y-z=7,
x+y+z=2,
-2x-2y+2z=-8)}}}

becomes

{{{(matrix(3,3,1,4,-1,1,1,1,-2,-2,2))(matrix(3,1,x,y,z))=(matrix(3,1,7,2,-8))}}}

You didn't say to solve it, but I will anyway, because no doubt
you'll be given some later to solve, and you can use this as a
model to go by:

To solve it find the inverse of the coefficient matrix, which is

{{{red((matrix(3,3, -1/3,1/2,-5/12,1/3,0,1/6,0,1/2,1/4)))}}}

Then left-multiply both sides of the matrix equation by this inverse:

{{{red((matrix(3,3, -1/3,1/2,-5/12,1/3,0,1/6,0,1/2,1/4)))(matrix(3,3,1,4,-1,1,1,1,-2,-2,2))(matrix(3,1,x,y,z))=red((matrix(3,3, -1/3,1/2,-5/12,1/3,0,1/6,0,1/2,1/4)))(matrix(3,1,7,2,-8))}}}

Multiply the first two matrices on the left, and multiply the
matrices on the right:

{{{(matrix(3,3,1,0,0,0,1,0,0,0,1))(matrix(3,1,x,y,z))=(matrix(3,1,2,1,-1))}}}

Multiplying the matrices on the left (the one on the left is the identity
matrix)

{{{(matrix(3,1,x,y,z))=(matrix(3,1,2,1,-1))}}}

So the solution is x=2, y=1, z=-1.

Edwin</pre>