Question 354999
{{{system(x+3y+z = 4,
2x+2y+z = -1,
2x+3y+z = 3)}}} 
<pre>
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.  Each of those is is a separate topic.

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(3,3,1,3,1,2,2,1,2,3,1))}}}. 

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(3,1,4,-1,3))}}}

Next we form the matrix equation:

       {{{A*X = B}}}

or

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

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

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

{{{A^(-1)=(matrix(3,3,-1, 0, 1, 0, -1, 1, 2, 3, -4))}}}

Remember I assume you know where I got this inverse.  It is a whole separate
problem on how to find it.  If you don't know how, post again asking how.

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,-1, 0, 1, 0, -1, 1, 2, 3, -4))


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

(matrix(3,3,-1, 0, 1, 0, -1, 1, 2, 3, -4))

(matrix(3,1,4,-1,3))}}}

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,-1, 0, 1, 0, -1, 1, 2, 3, -4))


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

(matrix(3,3,-1, 0, 1, 0, -1, 1, 2, 3, -4))

(matrix(3,1,4,-1,3))}}}

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,-1,3,-3))}}}

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,-1,4,-7))}}}

So {{{x=-1}}}, {{{y=4}}}, and {{{z=-7}}}

Edwin</pre>