Question 389772
2x-5y+7z=4
3x+y-12z=-8
5x+2y-4z=3
<pre><font size = 4 face = "batangche" color = "indigo"><b>

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(3,3,2,-5,7,3,1,-12,5,2,-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(3,1,4,-8,3))}}}
 
Next we form the matrix equation:
 
       {{{A*X = B}}}
 
or
 
{{{(matrix(3,3,2,-5,7,3,1,-12,5,2,-4))(matrix(3,1,x,y,z))=(matrix(3,1,4,-8,3))}}}
 
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 associative 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(3,3,2,-5,7,3,1,-12,5,2,-4))(matrix(3,1,x,y,z))=(matrix(3,1,4,-8,3))}}}
 
Next we find the inverse of A, which is written A<sup>-1</sup>.
 
{{{A^(-1)=(matrix(3,3,20/297, -6/297, 33/287, -48/29, -43/287, 45/287, 1/287, -29/287, 17/287))
}}}
 
Then we indicate the left multiplication of both sides by
{{{A^(-1)}}} to get the equation {{{A^(-1)(A*X)=A^(-1)B}}}:
 
{{{
 
(matrix(3,3,20/297, -6/297, 33/287, -48/29, -43/287, 45/287, 1/287, -29/287, 17/287))
 

((matrix(3,3,2,-5,7,3,1,-12,5,2,-4))(matrix(3,1,x,y,z)))=
 
(matrix(3,3,20/297, -6/297, 33/287, -48/29, -43/287, 45/287, 1/287, -29/287, 17/287))
 
(matrix(3,1,4,-8,3))}}}
 
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}}}:
 
{{{A^(-1)}}} to get the equation {{{A^(-1)(A*X)=A^(-1)B}}}:
 
{{{
 
((matrix(3,3,20/297, -6/297, 33/287, -48/29, -43/287, 45/287, 1/287, -29/287, 17/287))
 

(matrix(3,3,2,-5,7,3,1,-12,5,2,-4)))(matrix(3,1,x,y,z))=
 
(matrix(3,3,20/297, -6/297, 33/287, -48/29, -43/287, 45/287, 1/287, -29/287, 17/287))
 
(matrix(3,1,4,-8,3))}}}
 
When we perform the matrix multiplication we get:

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

The matrix on the left is the identity matrix

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(3,1,1,1,1))}}}

or x=1, y=1, z=1
 
Edwin</pre>