**p** = (1, 2)^{T}

It was tedious to figure that out
(and would be much worse if **A** were, say, 5 × 5).
It would be nice to have a better way.
Say that

(1)q=Ap

for column vectors **p** and **q** and n×n matrix **A** .
We know **A** and **q** and are trying to figure out what **p** is.
Is it possible to find a matrix **B**_{n×n} so that

(2)p=Bq

If it is, then using (1) and (2)

(3)q=A(Bq)

(4)q= (AB)q

If (4) is true, then (**A****B**) = **I**.
(Remember that **I** is unique).
**B**, if it exists, is the **inverse** of **A**, written **A**^{-1},
and the following is true:

AA^{-1}=A^{-1}A=I