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Provide step-by-step solutions for the problems in the screenshot; show proof for solutions

To solve this problem, let’s verify each of the statements individually using properties of matrix transposition and inverses.

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### Part (a): \((A^T)^{-1} = (A^{-1})^T\)

To prove this, let’s consider properties of transposes and inverses of matrices:

1. **Property**: If \( A \) is an invertible (nonsingular) matrix, then \((A^{-1})^T\) is also invertible, and \((A^{-1})^T = (A^T)^{-1}\).

   This property holds because:
   \[
   (A \cdot A^{-1})^T = I^T = I
   \]
   Expanding this, we get:
   \[
   A^T \cdot (A^{-1})^T = I
   \]
   This equation implies that \((A^{-1})^T\) is the inverse of \(A^T\), which means:
   \[
   (A^T)^{-1} = (A^{-1})^T
   \]

Therefore, **Statement (a) is true**.

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### Part (b): \(((A \cdot B)^T)^{-1} = ((A \cdot B)^{-1})^T\)

To prove this, we use the properties of transposes and inverses for matrix products:

1. **Property of Transposes**: \((A \cdot B)^T = B^T \cdot A^T\).

   So,
   \[
   ((A \cdot B)^T)^{-1} = (B^T \cdot A^T)^{-1}
   \]

2. **Property of Inverses for Products**: For any two invertible matrices \(X\) and \(Y\), we have \((X \cdot Y)^{-1} = Y^{-1} \cdot X^{-1}\).

   Applying this to \(B^T \cdot A^T\), we get:
   \[
   (B^T \cdot A^T)^{-1} = (A^T)^{-1} \cdot (B^T)^{-1}
   \]

3. From part (a), we know that \((A^T)^{-1} = (A^{-1})^T\) and \((B^T)^{-1} = (B^{-1})^T\).

   Substituting these, we get:
   \[
   (A^T)^{-1} \cdot (B^T)^{-1} = (A^{-1})^T \cdot (B^{-1})^T
   \]

4. Using the property of transpose for products again, we have:
   \[
   (A^{-1} \cdot B^{-1})^T = ((A \cdot B)^{-1})^T
   \]

Therefore, **Statement (b) is also true**.

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### Summary

Both statements are true:
1. \((A^T)^{-1} = (A^{-1})^T\)
2. \(((A \cdot B)^T)^{-1} = ((A \cdot B)^{-1})^T\)

These follow from the properties of matrix transposition and inversion, specifically that the inverse of a transpose is the transpose of the inverse, and the inverse of a product is the product of the inverses in reverse order.

Would you like more details or have any questions?

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Here are some related questions to explore further:

1. What other properties of transposes and inverses apply to block matrices?
2. How would these properties extend if \( A \) and \( B \) were not square matrices?
3. Can you show a concrete example using specific \(2 \times 2\) matrices to verify these properties?
4. What is the relationship between the determinant of a matrix and its transpose?
5. How would these proofs change if \( A \) and \( B \) were complex matrices?

**Tip:** When working with matrix transposes and inverses, remember that operations on products are often non-commutative, so reversing the order of matrices in expressions like \((A \cdot B)^{-1}\) is essential.

Let A and B be any two nonsingular matrices. Determine whether (a) (A^T)^(-1) = (A^(-1))^T and (b) ((A · B)^T)^(-1) = ((A · B)^(-1))^T.

Learn how to prove that for two nonsingular matrices A and B, (a) (A^T)^(-1) = (A^(-1))^T and (b) ((A · B)^T)^(-1) = ((A · B)^(-1))^T. This problem explores key linear algebra concepts, including matrix inversion and transposition properties.

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