Prove that if A and B are both invertible then AB must also be invertible.
The correct answer and explanation is:
To prove that if matrices A and B are both invertible, then the matrix product AB is also invertible, we proceed as follows:
Given:
- A is an invertible matrix.
- B is an invertible matrix.
Goal:
We want to show that AB is also invertible and find its inverse.
Proof:
Since A is invertible, there exists a matrix A−1A^{-1} such that: A−1A=IA^{-1}A = I
where II is the identity matrix. Similarly, since B is invertible, there exists a matrix B−1B^{-1} such that: B−1B=IB^{-1}B = I
Now, consider the product ABAB. We aim to find a matrix XX such that: (AB)X=IandX(AB)=I(AB)X = I \quad \text{and} \quad X(AB) = I
To find such a matrix XX, let’s test with X=B−1A−1X = B^{-1}A^{-1}, the product of the inverses of B and A. First, compute: (AB)(B−1A−1)=A(BB−1)A−1=AIA−1=AA−1=I(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I
Thus, (AB)(B−1A−1)=I(AB)(B^{-1}A^{-1}) = I.
Next, compute the other product: (B−1A−1)(AB)=B−1(A−1A)B=B−1IB=B−1B=I(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I
Thus, (B−1A−1)(AB)=I(B^{-1}A^{-1})(AB) = I.
Since both products result in the identity matrix, B−1A−1B^{-1}A^{-1} is the inverse of ABAB. Therefore, ABAB is invertible, and its inverse is: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
Conclusion:
If both A and B are invertible, then the product ABAB is also invertible, and its inverse is B−1A−1B^{-1}A^{-1}.