If A and B are both square n x n matrices,
If AB = I,
prove BA = I
Presumably you have to do this without using the usual properties of the inverse of matrices. But we do need to use that if there exists a matrix B such that
A B = I
then the equation A X = 0 has the unique solution:
X = 0
So, let's start with:
AB = I
Multiply both sides by A on the right:
(AB)A = A
Now you use that (AB)A = A(BA) and you can rewrite the above equation as:
A(BA - I) = 0
And it follows that
BA = I
well if u decmial the 7 to a 6 good luck with that ..........
To prove that BA = I, given AB = I, without using the usual properties of the inverse of matrices, we can follow these steps:
1. Start with the equation AB = I.
2. Multiply both sides of the equation by A on the right: (AB)A = A.
3. Use the property that (AB)A = A(BA) to rewrite the equation as: A(BA - I) = 0.
4. Since A(BA - I) = 0, it means the equation A X = 0 has a unique solution: X = 0.
5. This implies that BA - I = 0, since any other value for X would result in A X ≠ 0.
6. Therefore, BA = I, completing the proof.
It's important to note that the proof relies on the fact that the equation A X = 0 has a unique solution of X = 0 when AB = I. If this property is not assumed or proved separately, the proof would not be complete.