https://file.aiquickdraw.com/mathbot/file/1730125634301m8ttpzup.jpg

Let's analyze and solve the two parts of the problem given in the image.

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**Problem Statement:**

Let \( A \) be an \( n \times n \) matrix. Show that:

### Part (a)
If \( \lambda \) is an eigenvalue of \( A \), then \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \), and vice versa.

### Solution for Part (a):
1. By definition, if \( \lambda \) is an eigenvalue of \( A \), there exists a nonzero vector \( \mathbf{v} \) such that:
   \[
   A \mathbf{v} = \lambda \mathbf{v}.
   \]

2. Assuming \( \lambda \neq 0 \), we can multiply both sides by \( A^{-1} \) (since \( A \) is invertible):
   \[
   A^{-1} A \mathbf{v} = A^{-1} (\lambda \mathbf{v}).
   \]
   This simplifies to:
   \[
   \mathbf{v} = \lambda A^{-1} \mathbf{v}.
   \]

3. Rearranging this equation gives:
   \[
   A^{-1} \mathbf{v} = \frac{1}{\lambda} \mathbf{v}.
   \]

4. This shows that \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \) corresponding to the same eigenvector \( \mathbf{v} \).

5. Conversely, if \( \mu \) is an eigenvalue of \( A^{-1} \), then there exists a nonzero vector \( \mathbf{w} \) such that:
   \[
   A^{-1} \mathbf{w} = \mu \mathbf{w}.
   \]
   Multiplying both sides by \( A \) gives:
   \[
   A A^{-1} \mathbf{w} = A (\mu \mathbf{w}),
   \]
   which simplifies to:
   \[
   \mathbf{w} = \mu A \mathbf{w}.
   \]
   Rearranging, we get:
   \[
   A \mathbf{w} = \frac{1}{\mu} \mathbf{w}.
   \]
   Hence, \( \frac{1}{\mu} \) is an eigenvalue of \( A \).

Thus, we have shown that if \( \lambda \) is an eigenvalue of \( A \), then \( \frac{1}{\lambda} \) is an eigenvalue of \( A^{-1} \), and vice versa.

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### Part (b)
Show that the product of the eigenvalues of \( A \) is equal to \( \det(A) \).

### Solution for Part (b):
1. The eigenvalues \( \lambda_1, \lambda_2, \ldots, \lambda_n \) of \( A \) are the roots of the characteristic polynomial of \( A \), given by:
   \[
   p(\lambda) = \det(A - \lambda I).
   \]

2. For an \( n \times n \) matrix \( A \), the characteristic polynomial \( p(\lambda) \) is a polynomial of degree \( n \) with the form:
   \[
   p(\lambda) = (-1)^n \lambda^n + c_{n-1} \lambda^{n-1} + \cdots + c_1 \lambda + c_0.
   \]
   where \( c_0 = \det(A) \) is the constant term of the polynomial, which is also the product of the eigenvalues of \( A \) (assuming they are non-zero).

3. By the Fundamental Theorem of Algebra, the characteristic polynomial can be factored as:
   \[
   p(\lambda) = (-1)^n (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n).
   \]

4. Setting \( \lambda = 0 \) in this polynomial gives:
   \[
   p(0) = (-1)^n (-\lambda_1)(-\lambda_2) \cdots (-\lambda_n) = \lambda_1 \lambda_2 \cdots \lambda_n = \det(A).
   \]

Therefore, the product of the eigenvalues of \( A \) is equal to \( \det(A) \).

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Let me know if you would like further details on any step, or if you have additional questions!

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### Further Related Questions

1. How does the eigenvalue relationship between \( A \) and \( A^{-1} \) change if \( A \) is not invertible?
2. Can we use the trace of \( A \) to find any specific sum of its eigenvalues?
3. How would the results change if \( A \) is a non-square matrix?
4. What is the relationship between the eigenvalues of \( A \) and those of \( A^T \) (the transpose of \( A \))?
5. How does the product of the eigenvalues relate to the determinant for other special types of matrices, like diagonal matrices?

### Tip:
The determinant of a matrix is an important property that provides insights into its invertibility, eigenvalues, and overall behavior in transformations.

Let A be a n × n matrix. Show that:

(a) λ is an eigenvalue of A if and only if 1/λ is an eigenvalue of A^{-1}.

(b) The product of the eigenvalues of A is equal to det(A).

Explore the relationship between the eigenvalues of an n × n matrix A and its inverse, A^-1. This problem also demonstrates that the product of A's eigenvalues equals its determinant. Ideal for college-level linear algebra studies.

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