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

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

q=Ap(1)

for column vectors **p** and **q** and N×N matrix **A** .
We know **A** and **q**. We are trying to figure out what **p** must be.

What if there were a matrix **B**_{N×N} such that

p=Bq(2)

If there were such a matrix, then we could calculate what we want ( **p** ) from **q**.

Substitute (2) into (1):

q=A(Bq) (3)

q= (AB)q(4)

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}.

So now you can solve

q=Ap(1)

for **p** by multiplying each side by **A**^{-1}:

A^{-1}q=A^{-1}Ap

A^{-1}q=p

Of course, the above assumes that you somehow managed to
calculate **A**^{-1}.

For any square matrix **A**, is there always going to be
an inverse **A**^{-1} ?
Hint: consider the zero matrix.