Question 66250
<pre><font size = 4 color = "darkgreen"><b>Solve the linear system using the given
 inverse of the coefficent matrix. 

 w + 6x + 3y - 3z = 2
2w + 7x +  y + 2z = 5
 w + 5x + 3y - 3z = 3
   - 6x - 2y + 3z = 6

A<sup>-1</sup> = {{{(matrix(4,4,40,-3,-33,9,1,0,-1,0,-39,3,33,-8,-24,2,20,-5))}}}

Do I plug in the number from the inverse coefficent matrix
into the linear system, and then possibly use Cramer's Rule?
<font color = "blue">
No, no. This has nothing whatsoever to do with Cramer's rule.
<font size = 4 color = "darkgreen">
If not, how do I even begin this problem??
<font size = 4 color = "darkblue">
First of all you have to know how to multiply two matrices
before you can do this problem.  If you don't know how,
then post again asking how to do it.  Or maybe your teacher
will let you do it on the calculator.

A = {{{(matrix( 4, 4, 1,6,3,-3,2,7,1,2,1,5,3,-3,0,-6,-2,3))}}} = the coefficient matrix
X = {{{(matrix(4,1,w,x,y,z))}}} = the column vector of unknowns
B = {{{(matrix(4,1,2,5,3,6))}}} = the column vector of constants

The steps are:
                              AX = B

Left-multiply both sides by the inverse matrix of matrix A,
which is denoted by A<sup>-1</sup>:

                           A<sup>-1</sup>AX = A<sup>-1</sup>B

When you multiply a matrix by its inverse you get the identity matrix I

                              IX = A<sup>-1</sup>B

When you multiply the identity matrix I by a matrix X you just get X
(It's just like multiplying x by 1 and getting x in ordinary algebra)

                               X = A<sup>-1</sup>B


You start with the matrix equation   

                              AX = B
which is        

                 {{{(matrix( 4, 4, 1,6,3,-3,2,7,1,2,1,5,3,-3,0,-6,-2,3))}}}{{{(matrix(4,1,w,x,y,z))}}} = {{{(matrix(4,1,2,5,3,6))}}}

Then you left-multiply both sides by the inverse matrix:

                           A<sup>-1</sup>AX = A<sup>-1</sup>B

which is

   {{{(matrix(4,4,40,-3,-33,9,1,0,-1,0,-39,3,33,-8,-24,2,20,-5))}}}{{{(matrix( 4, 4, 1,6,3,-3,2,7,1,2,1,5,3,-3,0,-6,-2,3))}}}{{{(matrix(4,1,w,x,y,z))}}} = {{{(matrix(4,4,40,-3,-33,9,1,0,-1,0,-39,3,33,-8,-24,2,20,-5))}}} {{{(matrix(4,1,2,5,3,6))}}}

Then you PERFORM the multiplication of the two 4×4 matrices A and 
A<sup>-1</sup> on the left, which will give I, the identity matrix.  Also 
PERFORM the multiplication of the 4×4 matrix A<sup>-1</sup> by the 4×1 matrix
B on the right:

                              IX = A<sup>-1</sup>B

                    {{{(matrix( 4, 4, 1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1))}}}{{{(matrix(4,1,w,x,y,z))}}} = {{{(matrix(4,1,20,-1,-12,-8))}}}

It always turns out that when you multiply a matrix by its inverse,
you get the identity matrix, I, which has 1's down the main diagonal 
and zeros elsewhere.  Now when you multiply that identity matrix
by the 4×1 matrix which has only the four letters w,x,y,z, in it
you just get that same 4×1 matrix, so you have:

                               X = A<sup>-1</sup>B

                            {{{(matrix(4,1,w,x,y,z))}}} = {{{(matrix(4,1,20,-1,-12,-8))}}}

That means of course that w = 20, x = -1, y = -12, z = -8  

Edwin</pre>