We have (A−3I)(A−5I)=O⇒A2−8A+15I=O Multiplying both sides by A−1, we get; A−1A⋅A−8A−1A+15A−1I=A−1O⇒∴A−8I+15A−1=OA+15A−1=8I2A+215A−1=4Iα+β=21+215=216=8
Suppose A is any 3×3 non-singular matrix and (A−3I)(A−5I)=O, where I=I3 and O=O3. If αA+ βA−1=4I, then α+β is equal to
Held on 15 Apr 2018 · Verified 6 Jul 2026.
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