Let A be a 3 [ ] 3 matrix such that A2 - 5A + 7I = 0 Statement - I : A-1 = [1 7] (5I - A).

Question

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 :

Accepted Answer

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.

Solutions · 1