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Step-by-Step Solution
Step 1: Understand the Given Matrix Equation
We are given that
$$(A - 3I)(A - 5I) = O,$$
where \(A\) is a \(3 \times 3\) nonsingular matrix, \(I\) is the \(3 \times 3\) identity matrix, and \(O\) is the \(3 \times 3\) zero matrix.
Step 2: Expand the Product
Expanding \((A - 3I)(A - 5I)\) yields:
$$(A - 3I)(A - 5I) = A^2 - 5A - 3A + 15I = A^2 - 8A + 15I.$$
Since the product is equal to the zero matrix, we get
$$A^2 - 8A + 15I = O.$$
This implies
$$A^2 - 8A + 15I = 0.$$
Step 3: Multiply Both Sides by \(A^{-1}\)
Because \(A\) is nonsingular, it has an inverse \(A^{-1}\). Multiply both sides of
$$A^2 - 8A + 15I = 0$$
by \(A^{-1}\):
\[
A^{-1} \left(A^2 - 8A + 15I\right) = A^{-1} \cdot 0 \quad \Longrightarrow \quad
A^{-1} A^2 - 8 A^{-1} A + 15 A^{-1}I = 0.
\]
Note that \(A^{-1} A = I\). Thus, we get:
\[
A - 8I + 15A^{-1} = 0.
\]
Step 4: Rearrange to Isolate \(A + 15A^{-1}\)
Rewriting the above equation:
\[
A + 15A^{-1} = 8I.
\]
Step 5: Compare with \(\alpha A + \beta A^{-1}\)
We want to find \(\alpha\) and \(\beta\) in the expression:
\[
\alpha A + \beta A^{-1} = 4I.
\]
From
\[
A + 15A^{-1} = 8I,
\]
we can divide everything by 2 on the left-hand side to match the form \(4I\):
\[
\frac{A}{2} + \frac{15A^{-1}}{2} = 4I.
\]
Hence,
\[
\alpha = \frac{1}{2}, \quad \beta = \frac{15}{2}.
\]
Step 6: Find \(\alpha + \beta\)
Finally,
\[
\alpha + \beta
= \frac{1}{2} + \frac{15}{2}
= \frac{16}{2}
= 8.
\]
Therefore, \(\alpha + \beta = 8\).
