Adjoint and Inverse JEE 2025: Why the Answer Is 280

What is the correct answer to the adjoint and inverse JEE 2025 question?

Option [4], 280, is correct for the adjoint and inverse JEE 2025 question discussed here. The supplied “Correct answer: 4” is an option index, not the numerical answer.

The given matrix has three rows and three columns, with determinant and diagonal sum:

Take the adjoint twice of the doubled matrix, then find the determinant-plus-trace sum:

Trace means the sum of diagonal entries. Adjoint here means the classical adjugate, not the conjugate transpose.

The bank labels this JEE Main question 2025, Hard, but supplies 2023-01-24, Morning as its date and slot. Its exact year and shift need checking against the original paper.

Which three identities should I write before substituting?

Use the double-adjoint identity, determinant scaling and trace scaling. The first two depend on matrix order; trace scaling does not. Write them before substituting so that you keep the determinant’s cube separate from the trace’s first power.

For a matrix with three rows and three columns:

Multiplying the matrix by a scalar multiplies each of its three rows by that scalar. Therefore: $det(cM)=c^{3}det(M).$

Each diagonal entry acquires the scalar once, so for any square matrix:

The determinant uses the cube; the trace uses the first power. The double-adjoint identity also changes with matrix order. Its general form is:

How does the double adjoint simplify to eight times the original matrix?

The doubled matrix has determinant 4, so the double-adjoint identity gives four times the doubled matrix: eight times the original. Use the determinant here, not the trace.

Follow the supplied solution by setting: $M=2A.$

First calculate its determinant, cubing the scalar because the matrix has three rows:

Substitute into the double-adjoint identity:

By contrast, the trace of the doubled matrix is: $trace(2A)=2\times 3=6.$

That diagonal sum does not belong in the double-adjoint identity. No entries of the original matrix, cofactor expansion or explicit inverse are needed.

How do I calculate the final determinant and trace?

Use eight as the scalar in both calculations. The determinant is 256 and the trace is 24, giving 280. Write the calculations separately: the same multiplier appears in both, but only the determinant cubes it.

For the determinant:

For the trace:

Hence:

Select option [4]. The scalar audit is simple: once the reduced matrix is eight times the original, neither calculation may switch to a different multiplier.

The bank gives an 180-second expected solve time. Treat that supplied estimate as a practice benchmark, not as observed student performance.

Why does the previously keyed answer 132 fail?

The supplied previously keyed option [2], 132, can be reconstructed by using six for the determinant calculation while retaining eight for the trace calculation. This is not an alternative method: it is an inconsistent calculation that cannot follow from one expression for the resulting matrix.

The incorrect determinant arithmetic is: $6^3\times \frac{1}{2}=108.$

Keeping the correct trace then produces the previously keyed value: $108+24=132.$

A possible source of six is the trace of the doubled original matrix: $trace(2A)=6.$

This reconstruction explains how the number can arise; it does not establish how the earlier key was produced. The diagonal sum has been confused with a determinant-related multiplier.

If the multiplier really were six, consistency would require:

not 24. The correct identity instead establishes: $B=8A.$

132 mixes two different scalar assumptions. It fails both the double-adjoint identity and the scalar audit.

Which three related questions should I solve next?

Use these original related practice questions, not verified PYQs, to test:

• Scalar consistency with a fractional multiplier.
• Matrix-order dependence.
• The distinction between adjoint and inverse.

For each drill, keep the reduced matrix expression above the determinant and trace calculations. Check both scalars against it.

How does removing the initial factor of two change the answer?

The double-adjoint identity now gives half the original matrix. Use that fractional multiplier consistently.

Question 1: For the same original matrix, define:

Find: $det(C)+trace(C).$

The reduced matrix is:

Therefore:

Thus:

Checks: scalar consistency, including a fractional multiplier.

What changes when the matrix has two rows and two columns?

Taking the adjugate twice returns the matrix itself for this order. Do not reuse the three-row identity unchanged.

Question 2: Let:

Find the determinant-plus-trace sum. The double-adjoint identity gives: $Q=2P.$

Therefore:

Checks: matrix-order dependence.

Are the determinants of the adjoint and inverse equal?

They are different for the original matrix. Its nonzero determinant also guarantees that the inverse exists.

Question 3: Find:

Since: $det(A)=\frac{1}{2}\ne 0,$ the required determinants are:

Checks: adjoint and inverse are not interchangeable. On your next attempt, write the matrix order first, reduce the double adjoint second, and audit the scalar before selecting an answer.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

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