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Let's work through the problems one by one:

**Exercise 2.4.9**: In each case, either prove the assertion or give an example showing that it is false.

### a. If \( A \neq 0 \) is a square matrix, then \( A \) is invertible.

**False.** 
A square matrix being non-zero does not guarantee invertibility. For example, consider the matrix:
\[
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
\]
This is a square matrix, and it is not the zero matrix, but it is not invertible because its determinant is 0 (a requirement for invertibility is that the determinant must be non-zero).

### b. If \( A \) and \( B \) are both invertible, then \( A + B \) is invertible.

**False.**
The sum of two invertible matrices is not always invertible. Consider the following counterexample:
\[
A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}
\]
Both \( A \) and \( B \) are invertible, but:
\[
A + B = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}
\]
The resulting matrix is the zero matrix, which is not invertible.

### c. If \( A \) and \( B \) are both invertible, then \( (A^{-1}B)^T \) is invertible.

**True.**
If both \( A \) and \( B \) are invertible, then their product is also invertible. We know that:
\[
(A^{-1}B)^T = B^T (A^{-1})^T
\]
Both \( B^T \) and \( (A^{-1})^T \) are invertible since the transpose of an invertible matrix is also invertible. Therefore, the transpose \( (A^{-1}B)^T \) is invertible.

### d. If \( A^4 = 3I \), then \( A \) is invertible.

**True.**
If \( A^4 = 3I \), then:
\[
A^4 = 3I \implies \frac{1}{3} A^4 = I
\]
So, \( A^4 \) is a scalar multiple of the identity matrix, meaning that \( A \) must be invertible (since we can multiply by \( \frac{1}{\sqrt[4]{3}} \) and still have a valid inverse matrix).

### e. If \( A^2 = A \) and \( A \neq 0 \), then \( A \) is invertible.

**False.**
This statement is generally false. Consider the matrix:
\[
A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
\]
Clearly:
\[
A^2 = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} = A
\]
and \( A \neq 0 \). However, \( A \) is not invertible because its determinant is 0.

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Do you want more details or have any further questions?

Here are five related questions to expand your understanding:

1. What are the conditions for a matrix to be invertible?
2. Can the sum of more than two invertible matrices be invertible?
3. How does the property of the transpose affect invertibility?
4. Can a matrix with non-zero determinant always be invertible?
5. How does the rank of a matrix relate to its invertibility?

**Tip:** Always check the determinant of a matrix when determining if it is invertible!

In each case either prove the assertion or give an example showing that it is false.

a. If A ≠ 0 is a square matrix, then A is invertible.
b. If A and B are both invertible, then A + B is invertible.
c. If A and B are both invertible, then (A⁻¹B)ᵀ is invertible.
d. If A⁴ = 3I, then A is invertible.
e. If A² = A and A ≠ 0, then A is invertible.

Explore a series of matrix invertibility problems, including proofs and counterexamples. Learn about conditions for matrix inversion, matrix transposition, and how the determinant plays a key role. Ideal for university-level students studying Linear Algebra.

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