To determine whether the given statements hold for two nonsingular (invertible) matrices $A$ and $B$, we will analyze both statements individually.

Given:

• $A = (a_{ij})_{n,n}$ and $B = (b_{ij})_{n,n}$ are nonsingular (invertible) matrices of size $n \times n$.

(a) Prove or disprove: $(AB)^{-1} = B^{-1} A^{-1}$

Solution for (a):

To find the inverse of the product of two matrices, we can use the property: $(AB)^{-1} = B^{-1}A^{-1}$

Proof:

1. Start with the expression $AB$.
2. Suppose $C = B^{-1} A^{-1}$.
3. Then, we want to verify if $C$ satisfies the property of an inverse for $AB$, i.e., $(AB) C = I$ and $C (AB) = I$, where $I$ is the identity matrix.

Let's calculate $(AB)C$: $(AB)(B^{-1}A^{-1}) = A(B B^{-1})A^{-1} = A I A^{-1} = A A^{-1} = I$

Now, let's calculate $C(AB)$: $(B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1} B = I$

Since both $(AB)C = I$ and $C(AB) = I$, we conclude that $C = (AB)^{-1}$ and therefore: $(AB)^{-1} = B^{-1} A^{-1}$

Conclusion for (a): The statement $(AB)^{-1} = B^{-1}A^{-1}$ is true.

(b) Prove or disprove: $(AB)^{\top} = B^{\top} A^{\top}$

Solution for (b):

To determine the transpose of the product of two matrices, we use the property: $(AB)^{\top} = B^{\top} A^{\top}$

Proof:

The transpose of a product of matrices satisfies: $(AB)^{\top} = B^{\top} A^{\top}$

This property states that the transpose of the product of two matrices $A$ and $B$ is equal to the product of their transposes in reverse order. This can be shown by examining the elements of $(AB)^{\top}$ and $B^{\top} A^{\top}$:

1. The element in the $i,j$-position of $AB$ is: $(AB)_{ij} = \sum_k a_{ik} b_{kj}$

2. Taking the transpose, the element in the $j,i$-position of $(AB)^{\top}$ is: $(AB)^{\top}_{ji} = (AB)_{ij} = \sum_k a_{ik} b_{kj}$

3. Now consider $B^{\top} A^{\top}$:

   • The element in the $j,i$-position of $B^{\top} A^{\top}$ is: $(B^{\top} A^{\top})_{ji} = \sum_k (B^{\top})_{jk} (A^{\top})_{ki} = \sum_k b_{kj} a_{ik}$

Since both expressions are equal, we have: $(AB)^{\top} = B^{\top} A^{\top}$

Conclusion for (b): The statement $(AB)^{\top} = B^{\top}A^{\top}$ is true.

Final Summary:

Both statements (a) and (b) are true for any two nonsingular matrices $A$ and $B$.

Would you like further clarification on any part of this solution, or do you have questions about specific matrix properties?

Here are some related questions to explore:

1. What are the properties of determinants when applied to matrix products?
2. How does matrix inversion relate to linear transformations?
3. Can these properties be extended to block matrices?
4. What happens if either $A$ or $B$ is singular in this context?
5. How does the transpose property work in higher-dimensional tensor products?

Tip: Always remember the order of operations in matrix multiplication, especially when dealing with inverses and transposes, as it often affects the outcome.