for solving a system of equations

See the similar discussion in our text, Rolf, on Pages 165-167 ;
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. McFarland's tests, you would be asked to solve the above problem

"by the matrix inverse method" in three distinct steps, as follows :

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

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 left-multiplying 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 ===>
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
x = -3
y = 14
z = 13