Key Concepts and Principles

This problem primarily relies on the fundamental properties of matrix algebra, especially concerning the identity matrix ($I$), the zero matrix ($O$), and the inverse of a matrix ($A^{-1}$).

1. Matrix Multiplication and Distributive Property: Matrix multiplication is distributive over addition, similar to scalar algebra. For matrices $A, B, C$, we have $A(B+C) = AB + AC$ and $(A+B)C = AC + BC$.
2. Identity Matrix Properties: For any $n \times n$ matrix $A$, $AI = IA = A$. Also, $I \cdot I = I$.
3. Zero Matrix Properties: For any matrix $A$, $AO = OA = O$.
4. Inverse of a Matrix: A square matrix $A$ is called non-singular (or invertible) if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I$. If a matrix is non-singular, its inverse is unique.
5. Using a Polynomial Equation to Find the Inverse: If a matrix $A$ satisfies a polynomial equation $P(A) = O$, and if $A$ is non-singular, we can often rearrange this equation to express $A^{-1}$ in terms of $A$ and $I$.

---

Step-by-Step Derivation

Step 1: Expand the given matrix equation. We are given the matrix equation: $(A - 3I)(A - 5I) = O$ Here, $A$ is a $3 \times 3$ non-singular matrix, $I$ is the $3 \times 3$ identity matrix, and $O$ is the $3 \times 3$ zero matrix.

To expand this expression, we apply the distributive property of matrix multiplication, similar to expanding a product of two binomials in scalar algebra. Remember that $AI = A$ and $I \cdot I = I$. $(A - 3I)(A - 5I) = A \cdot A - A \cdot 5I - 3I \cdot A + 3I \cdot 5I$ $= A^2 - 5A - 3A + 15I$ $= A^2 - 8A + 15I$ So, the given equation becomes: $A^2 - 8A + 15I = O$ This equation is a matrix polynomial, often related to the characteristic polynomial or minimal polynomial of the matrix $A$.

Step 2: Utilize the non-singularity of A to find its inverse. We are given that $A$ is a non-singular matrix. This is a crucial piece of information because it guarantees that its inverse, $A^{-1}$, exists. Our goal is to derive an expression involving $A$ and $A^{-1}$ that matches the form $\alpha A + \beta A^{-1} = 4I$.

To introduce $A^{-1}$ into our equation $A^2 - 8A + 15I = O$, we can multiply both sides of the equation by $A^{-1}$. Since $A$ commutes with $I$ and $A^{-1}$ also commutes with $I$, multiplying by $A^{-1}$ from the left or right will yield the same result. Let's multiply by $A^{-1}$ from the right: $(A^2 - 8A + 15I) A^{-1} = O \cdot A^{-1}$ Now, we distribute $A^{-1}$ to each term: $A^2 A^{-1} - 8A A^{-1} + 15I A^{-1} = O$ Let's simplify each term:

• $A^2 A^{-1} = (A \cdot A) A^{-1} = A (A A^{-1}) = A \cdot I = A$
• $8A A^{-1} = 8I$
• $15I A^{-1} = 15A^{-1}$
• $O \cdot A^{-1} = O$ (The product of the zero matrix and any matrix is the zero matrix)

Substituting these simplified terms back into the equation, we get: $A - 8I + 15A^{-1} = O$

Step 3: Rearrange the equation to match a desired form. We have the equation $A - 8I + 15A^{-1} = O$. Our target equation is $\alpha A + \beta A^{-1} = 4I$. Notice that the identity matrix term is on the right-hand side in the target equation. Let's move the $8I$ term to the right side of our current equation: $A + 15A^{-1} = 8I$

Step 4: Scale the equation to match the coefficient of I. The target equation is $\alpha A + \beta A^{-1} = 4I$. Our derived equation is $A + 15A^{-1} = 8I$. To make the right-hand side $4I$, we need to divide the entire equation by 2. This is equivalent to multiplying by the scalar $\frac{1}{2}$: $\frac{1}{2} (A + 15A^{-1}) = \frac{1}{2} (8I)$ $\frac{1}{2} A + \frac{15}{2} A^{-1} = 4I$

Step 5: Compare and solve for $\alpha$ and $\beta$. Now we have two equations:

1. $\alpha A + \beta A^{-1} = 4I$ (Given)
2. $\frac{1}{2} A + \frac{15}{2} A^{-1} = 4I$ (Derived)

By comparing the coefficients of $A$ and $A^{-1}$ on both sides, we can determine the values of $\alpha$ and $\beta$: Comparing the coefficients of $A$: $\alpha = \frac{1}{2}$ Comparing the coefficients of $A^{-1}$: $\beta = \frac{15}{2}$

Step 6: Calculate $\alpha + \beta$. Finally, we need to find the value of $\alpha + \beta$: $\alpha + \beta = \frac{1}{2} + \frac{15}{2}$ $\alpha + \beta = \frac{1+15}{2}$ $\alpha + \beta = \frac{16}{2}$ $\alpha + \beta = 8$

---

Tips and Common Mistakes to Avoid

• Matrix Commutativity: Remember that matrix multiplication is generally NOT commutative ($AB \neq BA$). However, matrices always commute with scalar multiples of the identity matrix (e.g., $AI = IA = A$, $A^{-1}I = I A^{-1} = A^{-1}$). In this problem, because we are dealing with $A$, $I$, and $A^{-1}$ only, the order of multiplication by $A^{-1}$ (left or right) does not affect the outcome.
• Existence of Inverse: Always check or be given that the matrix is non-singular before attempting to multiply by its inverse. The problem explicitly states $A$ is non-singular, so $A^{-1}$ exists.
• Scalar vs. Matrix Multiplication: Be careful when multiplying a matrix by a scalar. For example, $k(A+B) = kA + kB$.
• Zero Matrix: Remember that $O \cdot M = O$ for any matrix $M$.
• Identity Matrix: $I^n = I$ for any positive integer $n$.

---

Summary and Key Takeaway

This problem demonstrates a common technique in matrix algebra where a polynomial equation satisfied by a matrix $A$ can be used to find an expression for its inverse, $A^{-1}$. By judiciously multiplying the polynomial equation by $A^{-1}$ (after ensuring $A$ is non-singular) and rearranging terms, we can relate $A$ and $A^{-1}$ to the identity matrix. This technique is very useful in simplifying matrix expressions and solving matrix equations, particularly in the context of the Cayley-Hamilton Theorem, which states that every square matrix satisfies its own characteristic equation. The ability to manipulate matrix expressions algebraically is fundamental to solving problems in linear algebra.

The final answer is $\boxed{\text{8}}$.