Question 629331
<pre><font size = 4 color = "indigo"><b>
Below is a problem just like yours with different numbers.  
You can use it as a guide to solve yours.  Yours has this
solution:

(x,y,z) = ({{{-25/61}}}, {{{-88/61}}}, {{{-117/61}}})

This one has solution (x,y,z) = (15, -49, 27)

 x + 2y + 3z =  -2
 x +  y +  z =  -7
-x +  y + 2z = -10


<pre><font size = 4 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(3,3,1,2,3,1,1,1,-1,1,2))}}}. 
 
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:  B = {{{(matrix(3,1,-2,-7,-10))}}}
 
Next we form the matrix equation:
 
                               AX = B
 
or
 
{{{(matrix(3,3,1,2,3,1,1,1,-1,1,2))(matrix(3,1,x,y,z))=(matrix(3,1,-2,-7,-10))}}}
 
To solve the equation
 
                               AX = B
 
we left-multiply both sides by A<sup>-1</sup>, the inverse of A.
 

                          A<sup>-1</sup>(AX) = A<sup>-1</sup>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<sup>-1</sup>A)X = A<sup>-1</sup>B
 
Now since A<sup>-1</sup>A = I, where I is the identity matrix, the
above becomes:
 
                              IX = A<sup>-1</sup>B
 
and by the identity property:
 
                               X = A<sup>-1</sup>B
 
Performing these operations with the actual matrices we have
the equation AX = B
 
{{{(matrix(3,3,1,2,3,1,1,1,-1,1,2))(matrix(3,1,x,y,z))=(matrix(3,1,-2,-7,-10))}}}
 
Next we find the inverse of A, which is written A<sup>-1</sup>.
 
A<sup>-1</sup> = {{{(matrix(3,3,1, -1, -1, -3, 5, 2, 2, -3, -1))
}}}
 
Then we indicate the left multiplication of both sides by
{{{A^(-1)}}} to get the equation A<sup>-1</sup>(A*X) = A<sup>-1</sup>B:
 
{{{
 
(matrix(3,3,1, -1, -1, -3, 5, 2, 2, -3,-1))
 

((matrix(3,3,2,-5,7,3,1,-12,5,2,-4))(matrix(3,1,x,y,z)))=
 
(matrix(3,3,1, -1, -1, -3, 5, 2, 2, -3, -1))
 
(matrix(3,1,-2,-7,-10))}}}
 
Next we use the associative principle to move the parentheses so that
they are around the first two factors to get the equation 

                (A<sup>-1</sup>A)X=A<sup>-1</sup>B:
 
{{{A^(-1)}}} to get the equation (A<sup>-1</sup>A)X = A<sup>-1</sup>B:
 
{{{
 
((matrix(3,3,1, -1, -1, -3, 5, 2, 2, -3, -1))
 

(matrix(3,3,2,-5,7,3,1,-12,5,2,-4)))(matrix(3,1,x,y,z))=
 
(matrix(3,3,1, -2, -1, -3, 5, 2, 2, -3,-1))
 
(matrix(3,1,-2,-7,-10))}}}
 
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(3,1,15,-49,27))}}}

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<sup>-1</sup>B
 
{{{(matrix(3,1,x,y,z))=(matrix(3,1,15,-49,27))}}}

or x=15, y=-49, z=27
 
Edwin</pre>