We are given that $A$ is a square matrix and $A^{-1}$ is its inverse matrix. The equation provided is:

$A \cdot A - 3A - I = 0$

Where $I$ is the identity matrix of the same order as $A$. We are asked to find $A^{-1}$, the inverse of $A$, and show the steps involved in the process.

Step-by-Step Solution

Step 1: Simplify the given equation

The given equation is:

$A \cdot A - 3A - I = 0$

We can rewrite this as:

$A^2 - 3A - I = 0$

Rearranging this equation, we get:

$A^2 = 3A + I$

This will be useful in deriving $A^{-1}$.

Step 2: Multiply both sides by $A^{-1}$

Since we want to find $A^{-1}$, one approach is to multiply both sides of the equation $A^2 = 3A + I$ by $A^{-1}$ from the left. This gives:

$A^{-1} \cdot A^2 = A^{-1} \cdot (3A + I)$

Using the properties of the inverse matrix, where $A^{-1} \cdot A = I$, we can simplify the left side:

$I \cdot A = A^{-1} \cdot (3A + I)$

This simplifies to:

$A = 3I + A^{-1}$

Step 3: Solve for $A^{-1}$

Now, we solve for $A^{-1}$. Rearranging the equation $A = 3I + A^{-1}$, we get:

$A^{-1} = A - 3I$

Final Answer

The inverse of $A$ is:

$A^{-1} = A - 3I$

Summary of the Steps:

1. Start with the equation $A^2 - 3A - I = 0$.
2. Rearrange it to $A^2 = 3A + I$.
3. Multiply both sides by $A^{-1}$ to isolate $A^{-1}$.
4. Solve for $A^{-1}$ to get $A^{-1} = A - 3I$.

Let me know if you need any more details or have any questions about the solution!

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Here are some related questions to further explore this topic:

1. How can we prove that $A^{-1}$ is the unique solution to the equation $A \cdot A^{-1} = I$?
2. What happens if $A$ is a diagonal matrix? How does it affect the inverse?
3. Can this approach be applied to non-square matrices? Why or why not?
4. How does the equation $A^2 - 3A - I = 0$ relate to finding eigenvalues of $A$?
5. What are the geometric interpretations of the matrix inverse in terms of transformations?

Tip: When working with matrix equations, always remember to utilize the properties of the identity matrix and the inverse matrix to simplify expressions effectively.