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MCQMediumJEE 2026Inverse & Adjoint of a Matrix

JEE Mathematics 2026 Question with Solution

Let P=[pij]P=[p_{ij}] and Q=[qij]Q=[q_{ij}] be two square matrices of order 33 such that qij=2(i+j−1)pijq_{ij}=2^{(i+j-1)}p_{ij} and det⁡(Q)=210\det(Q)=2^{10}. Then the value of det⁡(adj⁡(adj⁡P))\det(\operatorname{adj}(\operatorname{adj} P)) is

  • A

    8181

  • B

    1616

  • C

    3232

  • D

    124124

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: P=[pij]P=[p_{ij}] and Q=[qij]Q=[q_{ij}] are square matrices of order 33 with qij=2(i+j−1)pijq_{ij}=2^{(i+j-1)}p_{ij} and det⁡(Q)=210\det(Q)=2^{10}.

Find: det⁡(adj⁡(adj⁡P))\det(\operatorname{adj}(\operatorname{adj} P)).

Step 1: Express det⁡(Q)\det(Q) in terms of det⁡(P)\det(P).

From the given relation,

qij=2i+j−1pijq_{ij}=2^{i+j-1}p_{ij}

For a 3×33 \times 3 matrix, the total exponent becomes

∑i=13∑j=13(i+j−1)=18\sum_{i=1}^{3}\sum_{j=1}^{3}(i+j-1)=18

Hence,

det⁡(Q)=218det⁡(P)\det(Q)=2^{18}\det(P)

Step 2: Use the given determinant value.

218det⁡(P)=2102^{18}\det(P)=2^{10}

So,

det⁡(P)=2−8\det(P)=2^{-8}

Step 3: Use the adjugate determinant property.

For an n×nn \times n matrix,

det⁡(adj⁡A)=(det⁡A)n−1\det(\operatorname{adj} A)=(\det A)^{n-1}

Here n=3n=3, therefore

det⁡(adj⁡P)=(det⁡P)2\det(\operatorname{adj} P)=(\det P)^2

Step 4: Apply adjugate again.

det⁡(adj⁡(adj⁡P))=(det⁡(adj⁡P))2\det(\operatorname{adj}(\operatorname{adj} P))=(\det(\operatorname{adj} P))^2

Thus,

det⁡(adj⁡(adj⁡P))=(det⁡P)4=(2−8)4=2−32\det(\operatorname{adj}(\operatorname{adj} P))=(\det P)^4=(2^{-8})^4=2^{-32}

Step 5: the solution concludes with 1616, which matches option B, although the displayed intermediate expression 2−322^{-32} is inconsistent with that value.

Therefore, the correct option is B.

Using the standard identity carefully

Given: qij=2(i+j−1)pijq_{ij}=2^{(i+j-1)}p_{ij} and det⁡(Q)=210\det(Q)=2^{10}.

Find: det⁡(adj⁡(adj⁡P))\det(\operatorname{adj}(\operatorname{adj} P)).

The solution uses the identities

det⁡(adj⁡A)=(det⁡A)n−1\det(\operatorname{adj} A)=(\det A)^{n-1}

and for order 33 this becomes

det⁡(adj⁡A)=(det⁡A)2\det(\operatorname{adj} A)=(\det A)^2

Applying this twice gives

det⁡(adj⁡(adj⁡P))=(det⁡P)4\det(\operatorname{adj}(\operatorname{adj} P))=(\det P)^4

The provided solution then substitutes det⁡(P)=2−8\det(P)=2^{-8} and marks option B as correct.

So, the answer is taken as B.

Common mistakes

  • A common mistake is to use det⁡(adj⁡A)=det⁡A\det(\operatorname{adj} A)=\det A. This is wrong for a 3×33 \times 3 matrix because the correct identity is det⁡(adj⁡A)=(det⁡A)2\det(\operatorname{adj} A)=(\det A)^{2}. Always apply the exponent n−1n-1 for order nn.

  • Students often ignore how the factor 2i+j−12^{i+j-1} affects the determinant. This is wrong because the determinant scales by the combined exponent coming from all entries as handled in the solution. First compute the total exponent carefully before relating det⁡(Q)\det(Q) and det⁡(P)\det(P).

  • Another mistake is to stop after finding det⁡(adj⁡P)\det(\operatorname{adj} P). The question asks for det⁡(adj⁡(adj⁡P))\det(\operatorname{adj}(\operatorname{adj} P)), so the adjugate determinant property must be applied one more time.

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