Adjoint and Inverse JEE 2024: Why the Expression Gives 14

What does the adjoint and inverse JEE 2024 question actually ask?

The stated expression in this adjoint and inverse JEE 2024 entry gives 14, while the supplied solution reports 56 through invalid determinant transformations. Publication as a verified PYQ solution requires source correction.

The source labels the question 2024 but supplies the date 2023-01-24. Verify the original paper through jeemain.nta.nic.in before asserting its year, date or shift. It is labelled numerical, supplies no options, and records C in the answer field. That does not establish an option-to-value mapping.

For an invertible real matrix of order three, define its non-zero determinant:

A \in ℝ^{3 \times 3}, d = det (A) \neq 0 .

Given

det (3 adj (2 adj (d A))) = 3^{- 13} \times 2^{- 10},

find the integer prime exponents in

det (3 adj (2 A)) = 2^{m} \times 3^{n},

then evaluate $| 3 m + 2 n | .$

Which two determinant rules should I use?

Cube a scalar multiplying an order-three matrix, and square the determinant when taking its adjoint. These are the valid foundations of the supplied solution’s determinant-identity method. Every matrix in the nested expression remains of order three: taking an adjoint does not change the matrix order.

For every matrix of order three:

det (k B) = k^{3} det (B), det (adj (B)) = [det (B)]^{2} .

The scalar rule takes one factor from each of the three rows. A scalar multiplying the matrix multiplies every row, not just one.

For an invertible matrix, derive the adjoint rule by taking determinants in its defining identity:

B adj (B) = det (B) I_{3} .

det (B) det (adj (B)) = [det (B)]^{3} .

Divide by the non-zero determinant to obtain the square rule. No cofactor expansion or individual matrix entries are needed.

How do I evaluate the nested adjoints without losing an exponent?

Work from the innermost matrix outward, recording its determinant after each operation. First cube the determinant factor multiplying the original matrix, then apply the adjoint’s square rule. Keep scalar multiplication and taking an adjoint as separate steps.

Start with the scalar multiple of the original matrix:

det (d A) = d^{3} det (A) = d^{4} .

The first adjoint squares that entire determinant:

det (adj (d A)) = (d^{4})^{2} = d^{8} .

Multiplication by two contributes its cube:

det (2 adj (d A)) = 2^{3} d^{8} .

The second adjoint squares both factors:

det (adj (2 adj (d A))) = (2^{3} d^{8})^{2} = 2^{6} d^{16} .

The outer multiplication by three contributes its cube:

det (3 adj (2 adj (d A))) = 3^{3} \times 2^{6} \times d^{16} .

Equate this ledger’s last line with the given value:

d^{16} = \frac{3^{- 13} \times 2^{- 10}}{3^{3} \times 2^{6}} = 3^{- 16} \times 2^{- 16} = 6^{- 16} .

For the real-matrix interpretation used here, the squared determinant is positive. Taking the positive eighth root gives:

(d^{2})^{8} = {(\frac{1}{36})}^{8}, d^{2} > 0 ⟹ d^{2} = \frac{1}{36} .

There is no need to choose the sign of the original determinant. The second expression needs only its square.

What does the second determinant give, and how can I check it?

The second determinant equals 48, giving the requested value 14 under the intended integer-exponent reading. Calculate it separately rather than copying the powers in the given first determinant. Use the same sequence: scalar multiplication, adjoint, then outer scalar multiplication.

First, apply the scalar rule: $det (2 A) = 2^{3} d .$

Next, apply the adjoint rule:

det (adj (2 A)) = (2^{3} d)^{2} = 2^{6} d^{2} .

Finally, include the outer scalar:

det (3 adj (2 A)) & = 3^{3} \times 2^{6} \times d^{2} & = \frac{3^{3} \times 2^{6}}{2^{2} \times 3^{2}} & = 2^{4} \times 3 = 48 .

Reading the intended integer prime exponents gives:

m = 4, n = 1,

| 3 m + 2 n | = | 12 + 2 | = 14 .

Check the first expression using the determinant power already found:

3^{3} \times 2^{6} \times 6^{- 16} = 3^{3 - 16} \times 2^{6 - 16} = 3^{- 13} \times 2^{- 10} .

This reproduces the given value exactly. 14 is the result of the expression as supplied, not a verified replacement official answer. The reported 56 cannot honestly be certified from these premises. Do not manufacture intermediate steps to force agreement.

Why does the supplied solution report 56?

It transfers the exponents of the first determinant to a different second determinant without deriving a relationship between them. The supplied solution assigns:

m = - 10, n = - 13,

and obtains:

| 3 m + 2 n | = | - 30 - 26 | = 56 .

Those exponents belong to the given first determinant. They do not describe the second determinant, which the question asks us to express in prime powers.

The supplied solution also handles the outer scalar incorrectly. The valid transformation is:

det (3 adj (2 adj (d A))) = 3^{3} {[det (2 adj (d A))]}^{2} .

The square acts only on the determinant inside the brackets. The outer factor of three contributes its cube once, outside that square. Turning this contribution into the sixth power is incorrect.

Call 56 a supplied wrong-result pathway, not a listed distractor. No wrong options were supplied; attach an option label only after verifying the original options.

Which two practice questions test whether I understand the rules?

Compare a repeated adjoint with an inverse. For order three, an adjoint squares the determinant, while an inverse takes its reciprocal. Both questions below are original practice, not verified PYQs.

What is the determinant after taking the adjoint twice?

Question 1 answer: 16. For a matrix of order three, the original practice task is:

det (B) = - 2, find det (adj (adj (B))) .

Apply the square rule once for each adjoint. Keep the intermediate determinant visible:

det (adj (B)) = (- 2)^{2} = 4,

det (adj (adj (B))) = 4^{2} = 16 .

What is the determinant of three times the inverse?

Question 2 answer: negative twenty-seven halves. For the same matrix, the original practice task is to evaluate: $det (3 B^{- 1}) .$

Take the reciprocal first. Then cube the scalar multiplying the order-three matrix:

det (B^{- 1}) = \frac{1}{det (B)} = - \frac{1}{2},

det (3 B^{- 1}) = 3^{3} (- \frac{1}{2}) = - \frac{27}{2} .

The inverse preserves the negative sign here; the adjoint’s square removes it. Before checking any nested-expression answer, write one determinant line per operation and keep every outer scalar outside any square it does not belong to.

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).

Frequently asked questions

Which determinant formulas are used for adjoint and inverse questions?

For a matrix B of order three, det(kB) = k³det(B) and det(adj(B)) = [det(B)]². If B is invertible, det(B⁻¹) = 1/det(B). Apply each rule separately, working from the innermost matrix outward.

Why does the stated adjoint question give 14 instead of 56?

Writing d = det(A), the given nested determinant yields d¹⁶ = 6⁻¹⁶, so d² = 1/36. The requested second determinant is then 48 = 2⁴ × 3¹, giving m = 4, n = 1 and |3m + 2n| = 14. The supplied result 56 incorrectly reuses the first determinant's exponents for the second determinant.

Do I need to find the sign of det(A) in this question?

No: the requested determinant depends only on [det(A)]². The given expression establishes [det(A)]² = 1/36 for the real, invertible matrix. Either possible sign of det(A) therefore gives the same result.

Is this a verified JEE 2024 question with an official answer?

No: the supplied entry is labelled 2024 but records the date 2023-01-24, so its year, date and shift require verification against the original paper. It also records C in the answer field despite supplying no options. The stated expression gives 14, but that is not a verified replacement official answer.