Verified Solution • 2025–26 Session moderate

QJEEM16Maae47: JEE Main 2016 Mathematics - Class 12 JEEM16Maae47 Solution: Let A be a 3 3 matrix such that A 2 5A + 7I = 0 Statement - I : …

Let A be a 3 ×\times 3 matrix such that A2 −- 5A + 7I = 0

Statement - I :  

A −- 1 = 17{1 \over 7} (5I −- A).

Statement - II :

The polynomial A3 −- 2A2 −- 3A + I can be reduced to 5(A −- 4I).

Then :

Multiple Choice Options

  • A Statement-I is true, but Statement-II is false.
  • B Statement-I is false, but Statement-II is true.
  • C Both the statements are true.
  • D Both the statements are false

Step-by-Step Verified Solution

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 = 17{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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