B=A−1
Let A∗ be adjoint of A,
B∗ be adjoint of B.
1)
A−1=detAA∗ and B−1=detBB∗
B=A−1=detAA∗ (multyply by B−1 )
BB−1=detAA∗B−1
E=detAA∗detBB∗
E=detAA∗det(A−1)B∗
E=A∗B∗
B∗=(A∗)−1
2)
A=B−1
AT=(B−1)T
AT=(BT)−1
Another proof:
AB=E
(AB)T=ET
BTAT=E
(BT)−1BTAT=(BT)−1E
AT=(BT)−1
Comments