We know that
∣AdjA∣=∣A∣n−1
Here n is 3×3 matrix
so ∣AdjA∣=∣A∣3−1=∣A∣2
Now find ∣A∣=∣1−1−a010a01∣
=1+a2
Now ∣AdjA∣=∣A∣2=(1+a2)2
Now
a∈SΣdet(adjA)=Σ(1+a2)2=100λ
⇒22+42...+482+502=100λ
⇒4(12+22...+252)=100λ
⇒4×625×26×51=100λ
⇒221=λ
JEE Main 2022 — Mathematics Algebra
Let S={n:1⩽n⩽50 and n is odd}. Let a∈S and A=[1−1−a010a01]. If a∈SΣdet(adjA)=100λ, then λ is equal to
Held on 24 Jun 2022 · Verified 6 Jul 2026.
218
221
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1717
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