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Question

Let A1 be a 3×3 matrix such that A25A+7I=0.
Statement-I :-A1=17(5IA).
Statement-II :-The polynomial A32A23A+I can be reduced to 5(A4I).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 A1. Now we have

A1A25A1A+7IA1=0.A5I+7A1=0
A1=17(5IA).
Now for second statement
The polynomial A32A23A+I can be reduced to 5(A4I)
A32A23A+I=A(A2)2A23A+I=A(5A7I)2A23A+I=3A210A+I=3(5A7I)10A+I=5A20I=5(A4I)
So both statements are true

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