Answer:
p = (1, 2)T
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 BN×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 (AB) = 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-1q = A-1Ap
A-1q = 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.