Suppose A is any $$3 \times 3$$ non-singular matrix and $$(A - 3I)(A - 5I) = O$$, where $$I = I_3$$ and $$O = O_3$$. If $$\alpha A + \beta A^{-1} = 4I$$, then $$\alpha + \beta$$ is equal to:

We have a non-singular $$3 \times 3$$ matrix $$A$$ which satisfies the quadratic matrix equation

$$ (A-3I)(A-5I)=O. $$

First we expand the left side exactly as we would expand two algebraic brackets:

$$ (A-3I)(A-5I)=A^2-5A-3A+15I =A^2-8A+15I. $$

Since the product is the zero matrix $$O$$, we obtain

$$ A^2-8A+15I=O \;\;\Longrightarrow\;\; A^2=8A-15I. $$

Because $$A$$ is non-singular it has an inverse $$A^{-1}$$, so we may right-multiply every term by $$A^{-1}$$ (this is legal for matrices of the same size):

$$ A^2A^{-1}=8AA^{-1}-15IA^{-1}. $$

Now we simplify each product:

$$ A^2A^{-1}=A,\qquad AA^{-1}=I,\qquad IA^{-1}=A^{-1}. $$

Substituting these three simplifications gives

$$ A = 8I - 15A^{-1}. $$

We want an explicit expression for $$A^{-1}$$, so we move the term containing $$A^{-1}$$ to the left:

$$ 15A^{-1}=8I-A \;\;\Longrightarrow\;\; A^{-1}=\dfrac{8I-A}{15}. $$

The statement of the problem tells us that real numbers $$\alpha$$ and $$\beta$$ satisfy

$$ \alpha A+\beta A^{-1}=4I. $$

We now replace $$A^{-1}$$ by the expression just found:

$$ \alpha A+\beta\left(\dfrac{8I-A}{15}\right)=4I. $$

Next we distribute the factor $$\beta/15$$ inside the parentheses:

$$ \alpha A+\dfrac{\beta}{15}\,8I-\dfrac{\beta}{15}\,A =4I. $$

Collecting like terms, we separate the coefficients of $$A$$ and $$I$$:

$$\bigl(\alpha-\dfrac{\beta}{15}\bigr)A+\dfrac{8\beta}{15}I =4I.$$

For two matrices to be equal, corresponding coefficients must be equal. Hence

$$ \alpha-\dfrac{\beta}{15}=0 \quad\text{and}\quad \dfrac{8\beta}{15}=4. $$

From the second equation we solve for $$\beta$$:

$$ \dfrac{8\beta}{15}=4 \;\;\Longrightarrow\;\; 8\beta = 60 \;\;\Longrightarrow\;\; \beta = 7.5 =\dfrac{15}{2}. $$

Using $$\alpha=\dfrac{\beta}{15}$$ from the first equation we get

$$ \alpha = \dfrac{\,7.5\,}{15}=\dfrac12. $$

Finally, we add the two numbers:

$$ \alpha+\beta = \dfrac12 + 7.5 = 8. $$

Hence, the correct answer is Option A.