A2=5A−7I.....(i)
AAA−1−5AA−1−7IA−1=O
AI=5I−7A−1
A−1=71(5I−A)
Hence, the statement I is true.
Now A3−2A2−3A+I
=A(5A−7I)−2A2−3A+I {from(i)}
=5A2−7A−2A2−3A+I
=3A2−10A+I
=3(5A−7I)−10A+I {from(i)}
=5A−20I
=5(A−4I)
Hence, Statement II is true.
JEE Main 2016 — Mathematics Algebra
Let A, be a 3×3 matrix, such that A2−5A+7I=O.
Statement - I : A−1=71(5I−A).
Statement - II : The polynomial A3−2A2−3A+I,can be reduced to 5(A−4I). Then :
Held on 10 Apr 2016 · Verified 6 Jul 2026.
Both the statements are true
Both the statements are false
Statement - I is true, but Statement - II is false
Statement - I is false, but Statement - II is true
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