SOLUTION: Solve the system using the inverse method. x + 2y + 3z = -2 x + y + z = -7 -x + y + 2z = -10

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Question 620071: Solve the system using the inverse method.
x + 2y + 3z = -2
x + y + z = -7
-x + y + 2z = -10

Answer by Edwin McCravy(20060)   (Show Source): You can put this solution on YOUR website!
x + 2y + 3z = -2
x + y + z = -7
-x + y + 2z = -10



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 = . 
 
2. Matrix X is the 3x1 matrix of variables  X = 
 
3. Matrix B is the 3x1 matrix, whose only column is the
column of constants:  B = 
 
Next we form the matrix equation:
 
                               AX = B
 
or
 

 
To solve the equation
 
                               AX = B
 
we left-multiply both sides by A-1, the inverse of A.
 

                          A-1(AX) = A-1B
 
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-1A)X = A-1B
 
Now since A-1A = I, where I is the identity matrix, the
above becomes:
 
                              IX = A-1B
 
and by the identity property:
 
                               X = A-1B
 
Performing these operations with the actual matrices we have
the equation AX = B
 

 
Next we find the inverse of A, which is written A-1.
 
A-1 = 
 
Then we indicate the left multiplication of both sides by
 to get the equation A-1(A*X) = A-1B:
 

 
Next we use the associative principle to move the parentheses so that
they are around the first two factors to get the equation 

                (A-1A)X=A-1B:
 
 to get the equation (A-1A)X = A-1B:
 

 
When we perform the matrix multiplication we get:



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-1B
 


or x=15, y=-49, z=27
 
Edwin

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