Question 891936
<pre>
 x +  y +  z = 2
3x + 2y – 2z = 8
2x – 3y – 4z = 0

By matrix inversion.

I will assume you already know how to find the inverse
of a matrix, and how to multiply two matrices. If you don't, 
post again asking how.

First we form three matrices, A, X, and B.

1. Matrix A is the 3x3 coefficient matrix A, which consists 
of just the three columns of x, y, and z coefficients. in 
that order, but does not contain the column of constants.

{{{A=(matrix(3,3,1,1,1,3,2,-2,2,-3,-4))}}}. 

2. Matrix X is the 3x1 matrix of variables {{{X=(matrix(3,1,x,y,z))}}}

3. Matrix B is the 3x1 matrix, whose only column is the
column of constants: {{{(matrix(3,1,2,8,0))}}}

Next we form the matrix equation:

       {{{A*X = B}}}

or

{{{(matrix(3,3,1,1,1,3,2,-2,2,-3,-4))(matrix(3,1,x,y,z))=(matrix(3,1,2,8,0))}}}

To solve the equation

       {{{A*X = B}}}

we left-multiply both sides by {{{A^(-1)}}}, the inverse of {{{A}}}.

 {{{A^(-1)*(A*X) = A^(-1)*B}}}

Then since the associatitive principle holds for matrix multiplication,
(even though the commutative principle DOES NOT!!!), we can move
the parentheses on the left around the first two matrix factors:

{{{(A^(-1)*A)*X = A^(-1)*B}}}

Now since {{{(A^(-1)*A)=I}}}, where I is the identity matrix, the
above becomes:

{{{I*X = A^(-1)*B}}}

and by the identity property:

{{{X=A^(-1)*B}}}

Performing these operations with the actual matrices we have
the equation {{{AX=B}}}

{{{(matrix(3,3,1,1,1,3,2,-2,2,-3,-4))(matrix(3,1,x,y,z))=(matrix(3,1,2,8,0))}}}

Next we form the inverse of A, which is written A<sup>-1</sup>.

{{{A^(-1)=(matrix(3,3,14/19, -1/19, 4/19, -8/19, 6/19, -5/19, 13/19, -5/19, 1/19))}}}

Then we indicate the left multiplication of both sides by
{{{A^(-1)}}} to get the equation {{{A^(-1)(A*X)=A^(-1)B}}}:

{{{

(matrix(3,3,14/19, -1/19, 4/19, -8/19, 6/19, -5/19, 13/19, -5/19, 1/19))


((matrix(3,3,1,1,1,3,2,-2,2,-3,-4))(matrix(3,1,x,y,z)))=

(matrix(3,3,14/19, -1/19, 4/19, -8/19, 6/19, -5/19, 13/19, -5/19, 1/19))

(matrix(3,1,2,8,0))}}}

Next we use the associative principle to move the parentheses so that
they are around the first two factors to get the equation {{{(A^(-1)*A)*X=A^(-1)*B}}}:

{{{

((matrix(3,3,14/19, -1/19, 4/19, -8/19, 6/19, -5/19, 13/19, -5/19, 1/19))


(matrix(3,3,1,1,1,3,2,-2,2,-3,-4)))(matrix(3,1,x,y,z))=

(matrix(3,3,14/19, -1/19, 4/19, -8/19, 6/19, -5/19, 13/19, -5/19, 1/19))

(matrix(3,1,2,8,0))}}}


Now we perform the actual multiplications and we get the equation {{{IX=A^(-1)*B}}}:

{{{(matrix(3,3,1,0,0,0,1,0,0,0,1))*(matrix(3,1,x,y,z))=(matrix(3,1,20/19,32/19,-14/19))}}}

Then when we multiply the identity matrix {{{I}}} by the column matrix of
variables, we just get the matrix of variables, or the 
equation {{{X=A^(-1)B}}}

{{{(matrix(3,1,x,y,z))=(matrix(3,1,20/19,32/19,-14/19))}}}

------------------

Checking:

x + y + z = 2
{{{20/19+32/19-14/19=2}}}
{{{38/19=2}}}

That checks.

3x + 2y – 2z = 8
{{{3(20/19)+2(32/19)-2(-14/19)=8}}}
{{{60/19+64/19+28/19=8}}}
{{{152/19=8}}}
{{{8=8}}}

That checks.

2x – 3y – 4z = 0
{{{2(20/19)-3(32/19)-4(-14/19)=0}}}
{{{40/19-96/19+56/19=0}}}
{{{0/19=0}}}
{{{0=0}}}

That checks.

Terrible answers, but they're correct.  Did you copy the problem
correctly?  It's the correct solution for the system you posted.

Edwin</pre>