Let's analyze each statement to determine which ones are true for all invertible $n \times n$ matrices $A$ and $B$:

A. $(A + A^{-1})^3 = A^3 + A^{-3}$

This statement is false. Matrix addition and multiplication do not follow simple distribution rules like for scalars. Expanding $(A + A^{-1})^3$ would not result in $A^3 + A^{-3}$. There are cross terms, so this is not true in general.

B. $(A + B)(A - B) = A^2 - B^2$

This statement is false. Matrix multiplication is not commutative, meaning $AB \neq BA$ in general. The left-hand side would expand to $A^2 - AB + BA - B^2$, and $AB$ may not equal $BA$, so the expression does not simplify to $A^2 - B^2$.

C. $A B A^{-1} = B$

This statement is false in general. The result of $A B A^{-1}$ is not necessarily $B$. In fact, it's a similarity transformation, and it transforms $B$ into a conjugate form. Only when $B$ commutes with $A$ (i.e., $AB = BA$) would this hold true.

D. $A + I_n$ is invertible

This statement is false in general. The matrix $A + I_n$ is not necessarily invertible for all invertible matrices $A$. For example, if $A = -I_n$, then $A + I_n = 0$, which is not invertible.

E. $(I_n + A)(I_n + A^{-1}) = 2I_n + A + A^{-1}$

This statement is true. Expanding both sides, we get: $(I_n + A)(I_n + A^{-1}) = I_n + A + A^{-1} + I_n = 2I_n + A + A^{-1}$ Thus, this holds for all invertible matrices $A$.

F. $A^4$ is invertible

This statement is true. If $A$ is invertible, then $A^{-1}$ exists. Therefore, $A^4 = A \cdot A \cdot A \cdot A$ is also invertible, since the product of invertible matrices is invertible.

True Statements:

• E. $(I_n + A)(I_n + A^{-1}) = 2I_n + A + A^{-1}$
• F. $A^4$ is invertible

Would you like more details on any of the explanations above?

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Follow-up questions:

1. Why doesn't matrix multiplication follow commutative rules?
2. Can you provide an example where $A + I_n$ is not invertible?
3. What are similarity transformations in matrices?
4. What does it mean for a matrix to be invertible?
5. How do you prove that a matrix power $A^n$ is invertible?

Tip:

Remember that matrix multiplication is generally not commutative—i.e., $AB \neq BA$ in most cases.