Let A−1 be a 3×3 matrix such that A2−5A+7I=0. Statement-I :-A−1=17(5I−A). Statement-II :-The polynomial A3−2A2−3A+I can be reduced to 5(A−4I).Then
A
Both the statement are true.
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B
Statement-I is true, but Statement-II is false.
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C
Statement-I is false, but Statement-II is true.
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D
Both the statement are false.
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Solution
The correct option is A Both the statement are true. Multiply the equation with A−1. Now we have
A−1A2−5A−1A+7IA−1=0.A−5I+7A−1=0 A−1=17(5I−A).
Now for second statement
The polynomial A3−2A2−3A+I can be reduced to 5(A−4I) A3–2A2–3A+I=A(A2)–2A2–3A+I=A(5A–7I)–2A2–3A+I=3A2–10A+I=3(5A–7I)–10A+I=5A–20I=5(A–4I)
So both statements are true