Marsia,

Think about solving a simple algebraic equation such as

5x = 10

I expect that you would divide both sides by 5 to obtain

x = 2

But 1/5 is 5^-1 which is the inverse of 5 so you can think about solving

5x = 10

as multiplying both sides by the inverse of 5 to obtain

5x = 10 thus

5^-1 × 5 x = 5^-1 × 10

but 5^-1 × 5 = 1 so

1 × x = 5^-1 × 10 or
x = 5^-1 × 10

The same technique works with a linear system if you can write it as

A X = B

and A has an inverse A^-1. Then A^-1 ×A = I and hence

A X = B thus

A^-1 × A × X = A^-1 × B so

I × X = A^-1 × B or

X = A^-1 × B

Penny