The given problem is as follows (translated from Vietnamese):

"Let $A$ be a square matrix of size $n$ that satisfies the equation:

$A^2 - A = I_n$

Prove that $A$ is invertible and find the inverse of $A$."

Step 1: Show that $A$ is invertible

The given equation is:

$A^2 - A = I_n$

Rewriting this:

$A(A - I_n) = I_n$

This equation tells us that the product of $A$ and $(A - I_n)$ is the identity matrix $I_n$. Therefore, $A$ has a right inverse, which is $A - I_n$.

Next, if we multiply both sides of the equation on the left by $A^{-1}$ (assuming $A$ is invertible), we would get:

$A^{-1} A(A - I_n) = A^{-1} I_n$

This simplifies to:

$(A - I_n) = A^{-1}$

Thus, we can conclude that $A$ is invertible and its inverse is given by:

$A^{-1} = A - I_n$

Conclusion:

• $A$ is invertible.
• The inverse of $A$ is $A^{-1} = A - I_n$.

Would you like more details or have any other questions?

Here are 5 related questions to explore:

1. How can we prove that a matrix is invertible in general?
2. What properties of square matrices ensure they are invertible?
3. Can the equation $A(A - I_n) = I_n$ have other types of solutions?
4. What are the implications of matrix inverses in solving systems of linear equations?
5. How does the identity matrix function in matrix multiplication?

Tip: When working with matrix equations, always check if the matrix is square before attempting to find its inverse!