MCQMediumJEE 2025Inverse & Adjoint of a Matrix

JEE Mathematics 2025 Question with Solution

Let AA be a 3×33 \times 3 matrix such that adj(adjA)=81.| \operatorname{adj}(\operatorname{adj} A) | = 81. If S={nZ:adj(adjA)(n1)22=A(3n25n4)},S = \left\{ n \in \mathbb{Z}: \left| \operatorname{adj}(\operatorname{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, then the value of nSA(n2+n)\sum_{n \in S} |A| (n^2 + n) is:

  • A

    866866

  • B

    750750

  • C

    820820

  • D

    732732

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: adj(adj(adjA))=81|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))| = 81 is used in the extracted working, and S={nZ:adj(adjA)(n1)22=A3n25n4}.S = \left\{ n \in \mathbb{Z}: \left| \operatorname{adj}(\operatorname{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{3n^2 - 5n - 4} \right\}. Find: nSA(n2+n).\sum_{n \in S} |A| (n^2+n).

Using the determinant property for a 3×33 \times 3 matrix,

adjA=A2|\operatorname{adj} A| = |A|^2

and hence

adj(adjA)=A4.|\operatorname{adj}(\operatorname{adj} A)| = |A|^4.

The second extracted approach gives

adj(adj(adjA))=81|\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))| = 81

which leads to

adjA4=81,|\operatorname{adj} A|^4 = 81,

so

adjA=3,|\operatorname{adj} A| = 3,

and therefore

A2=3.|A|^2 = 3.

Now use the given equation in SS:

(A4)(n1)22=A3n25n4.\left(|A|^4\right)^{\frac{(n-1)^2}{2}} = |A|^{3n^2-5n-4}.

Equating exponents,

2(n1)2=3n25n4.2(n-1)^2 = 3n^2-5n-4.

So,

2n24n+2=3n25n42n^2-4n+2 = 3n^2-5n-4

which gives

n2n6=0.n^2-n-6 = 0.

Factorising,

(n3)(n+2)=0.(n-3)(n+2)=0.

Thus,

n=3,2.n=3,-2.

Therefore,

nSA(n2+n)=A(12)+A(2)=14A.\sum_{n\in S} |A| (n^2+n) = |A|(12)+|A|(2)=14|A|.

The extracted the solution concludes the final value as 732732 and marks the correct option as D. There is a discrepancy between the intermediate expressions shown and the printed question expression, but the source solution concludes that the correct option is D.

Therefore, the correct option is D.

Extracted Working and Discrepancy Note

Given: the solution contains two inconsistent approaches. Find: The option supported by the solution.

In Approach Solution - 1, the working states

adj(adjA)=A3,|\operatorname{adj}(\operatorname{adj} A)| = |A|^3,

then uses A=3|A|=3 and finally states the answer is 732732. However, for a 3×33 \times 3 matrix, the standard determinant identity is

adjA=A2,|\operatorname{adj} A| = |A|^{2},

so applying it twice gives

adj(adjA)=A4,|\operatorname{adj}(\operatorname{adj} A)| = |A|^{4},

not A3|A|^3.

Approach Solution - 2 also begins from a different statement,

adj(adj(adjA))=81,|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))| = 81,

which does not match the printed question. It then solves

2(n1)2=3n25n42(n-1)^2 = 3n^2-5n-4

and obtains

n=3,2.n=3,-2.

After that, it concludes with

732.\boxed{732}.

Since the solution explicitly states The Correct Option is D and both extracted approaches conclude 732732, the answer recorded from the source is D.

Therefore, based on the source solution authority, the correct option is D.

Common mistakes

  • Using the wrong determinant identity for the adjugate. For an n×nn \times n matrix, adjA=An1|\operatorname{adj} A| = |A|^{n-1}, so for a 3×33 \times 3 matrix it is A2|A|^2, not A3|A|^3. Apply the dimension-specific formula before substituting.

  • Equating powers before rewriting both sides with the same base. The exponent comparison is valid only after expressing both sides as powers of the same quantity. First convert adj(adjA)|\operatorname{adj}(\operatorname{adj} A)| in terms of A|A|.

  • Ignoring the condition nZn \in \mathbb{Z}. Even if an algebraic equation has real roots, only integer roots belong to SS. Always filter the solutions using the set definition.

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