MATRIX INVERSE METHOD
for solving a system of equations
See the similar discussion in our text, Rolf, on Pages 165167 ;
the following continues an earlier discussion.
We start with a system of equations (right),
to be solved. The number of variables must equal
the number of equations, both equal to 3 here: 

x 
 
y 
+ 
3z 
= 
2 
2x 
+ 
y 
+ 
2z 
= 
2 
2x 
 
2y 
+ 
z 
= 
3 


On Prof. M^{c}Farland's tests, you would be asked
to solve the above problem
"by the matrix inverse method" in three distinct steps, as follows :
[1] 
Write the given system (above) as a single matrix
equation: 


Capital letter variables represent the matrices
(not numbers) which sit directly above them. Hence, the
above equation is a matrix equation.
Notice how the matrix entries correspond to
numbers in the original system of 3 equations 




[2] 
Solve the matrix equation obtained in
step [1] above; i.e., find X. 

In the MATRIX INVERSE METHOD
(unlike Gauss/Jordan),
we solve for the matrix variable X by
leftmultiplying both sides of the above matrix
equation (AX=B) by A^{1}.
Typically, A^{1} is calculated as
a separate exercize ;
otherwise, we must pause here to calculate A^{1}. 

The left side (above) is easy to calculate
because an identity matrix
I
_{3} appears.
Note: A^{1} is on the LEFT
side of both products. 

Step [2] is now completed by finding
A^{1.} B ===> 




[3] 
Using [2] above, write the solution to the
original system: 

original system 
x 
 
y 
+ 
3z 
= 
2 
2x 
+ 
y 
+ 
2z 
= 
2 
2x 
 
2y 
+ 
z 
= 
3 




solution to original system 



