JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 5)
Let A be a 3 $$ \times $$ 3 matrix such that A2 $$-$$ 5A + 7I = 0
Statement - I :
A$$-$$1 = $${1 \over 7}$$ (5I $$-$$ A).
Statement - II :
The polynomial A3 $$-$$ 2A2 $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).
Then :
Statement - I :
A$$-$$1 = $${1 \over 7}$$ (5I $$-$$ A).
Statement - II :
The polynomial A3 $$-$$ 2A2 $$-$$ 3A + I can be reduced to 5(A $$-$$ 4I).
Then :
Statement-I is true, but Statement-II is false.
Statement-I is false, but Statement-II is true.
Both the statements are true.
Both the statements are false
Explanation
Given,
A2 $$-$$ 5A + 7I = 0
$$ \Rightarrow $$ A2 $$-$$ 5A = $$-$$ 7I
$$ \Rightarrow $$ AAA$$-$$1 $$-$$ 5AA$$-$$1 = $$-$$ 7IA$$-$$1
$$ \Rightarrow $$ AI $$-$$ 5I = $$-$$ 7A$$-$$1
$$ \Rightarrow $$ A $$-$$ 5I = $$-$$ 7A$$-$$1
$$ \Rightarrow $$ A$$-$$1 = $${1 \over 7}$$(5I $$-$$ A)
Hence, statement 1 is true.
Now A3 $$-$$ 2A2 $$-$$ 3A + I
= A(A2) $$-$$ 2A2 $$-$$ 3A + I
= A(5A $$-$$ 7I) $$-$$ 2A2 $$-$$ 3A + I
= 5A2 $$-$$ 7A $$-$$ 2A2 $$-$$ 3A + I
= 3A2 $$-$$ 10A + I
= 3(5A $$-$$ 7I) $$-$$ 10A + I
= 15A $$-$$ 21A $$-$$ 10A + I
= 5A $$-$$ 20I
= 5(A $$-$$ 4I)
So, statement 2 is also correct.
A2 $$-$$ 5A + 7I = 0
$$ \Rightarrow $$ A2 $$-$$ 5A = $$-$$ 7I
$$ \Rightarrow $$ AAA$$-$$1 $$-$$ 5AA$$-$$1 = $$-$$ 7IA$$-$$1
$$ \Rightarrow $$ AI $$-$$ 5I = $$-$$ 7A$$-$$1
$$ \Rightarrow $$ A $$-$$ 5I = $$-$$ 7A$$-$$1
$$ \Rightarrow $$ A$$-$$1 = $${1 \over 7}$$(5I $$-$$ A)
Hence, statement 1 is true.
Now A3 $$-$$ 2A2 $$-$$ 3A + I
= A(A2) $$-$$ 2A2 $$-$$ 3A + I
= A(5A $$-$$ 7I) $$-$$ 2A2 $$-$$ 3A + I
= 5A2 $$-$$ 7A $$-$$ 2A2 $$-$$ 3A + I
= 3A2 $$-$$ 10A + I
= 3(5A $$-$$ 7I) $$-$$ 10A + I
= 15A $$-$$ 21A $$-$$ 10A + I
= 5A $$-$$ 20I
= 5(A $$-$$ 4I)
So, statement 2 is also correct.
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