Matrices and Determinants JEE 2026: Double Adjoint

What is the answer to this matrices and determinants JEE 2026 question?

D) 16 is correct for this matrices and determinants JEE 2026 question. Two square matrices of order three have corresponding entries related by:

P = [p_{i j}], Q = [q_{i j}], q_{i j} = 2^{i + j - 1} p_{i j}, det (Q) = 2^{10} .

Find the determinant after taking the adjoint of the first matrix twice:

det (adj (adj P)) .

The supplied choices are:

The supplied record identifies this as JEE Main, January 2026, Slot 1, under Properties of Determinants and Adjoint of Matrix. Hard is the question bank’s difficulty classification; 180 seconds is the question bank’s expected solve time, not a measured student average.

Why is the determinant multiplier a power of two with exponent nine?

The determinant multiplier is 512, because you extract one factor from each row and then one from each column. Do not multiply the scaling factors of all nine entries. Separate the row-dependent and column-dependent powers first.

Write the second matrix explicitly:

Q = (\begin{matrix} 2 p_{11} & 4 p_{12} & 8 p_{13} \\ 4 p_{21} & 8 p_{22} & 16 p_{23} \\ 8 p_{31} & 16 p_{32} & 32 p_{33} \end{matrix}) .

Split the entry multiplier into two parts:

2^{i + j - 1} = 2^{i - 1} \times 2^{j} .

The first part depends only on the row; the second depends only on the column. Extract factors 1, 2 and 4 from rows 1, 2 and 3 respectively:

det (Q) = (1 \times 2 \times 4) | \begin{matrix} 2 p_{11} & 4 p_{12} & 8 p_{13} \\ 2 p_{21} & 4 p_{22} & 8 p_{23} \\ 2 p_{31} & 4 p_{32} & 8 p_{33} \end{matrix} | .

Every entry in the first column now contains a factor of 2. The second and third columns contain factors of 4 and 8 respectively. Extract those factors, leaving the original matrix:

det (Q) = (1 \times 2 \times 4) (2 \times 4 \times 8) | \begin{matrix} p_{11} & p_{12} & p_{13} \\ p_{21} & p_{22} & p_{23} \\ p_{31} & p_{32} & p_{33} \end{matrix} | .

The total multiplier is:

(1 \times 2 \times 4) (2 \times 4 \times 8) = 8 \times 64 = 512 = 2^{9} .

Using the given determinant:

2^{9} det (P) = 2^{10} ⟹ det (P) = 2 .

The supplied working contains an inconsistent all-entry multiplication passage. Multiplying all nine entry factors does not give the determinant multiplier; retain the valid factor-extraction step and final answer.

Each term of a third-order determinant contains three entries, one from every row and every column, not all nine entries. Each row factor and each column factor therefore appears exactly once in every term. Their product scales the whole determinant.

How do you apply the adjoint determinant identity twice?

Both applications use exponent two, giving intermediate determinant 4 and final determinant 16. Apply the identity once to the original matrix, then once to its adjoint. Taking an adjoint changes the entries, but not the number of rows or columns.

For a square matrix, the identity is:

A \in ℝ^{n \times n} ⟹ det (adj A) = (det A)^{n - 1} .

The original matrix and its adjoint both have order three. The first application gives:

det (adj P) = (det P)^{2} = 2^{2} = 4 .

Apply the same identity to the intermediate adjoint:

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

The combined exponent comes from raising a power to another power:

(n - 1)^{2} = (3 - 1)^{2} = 4 .

These exponents multiply; they are not two unrelated contributions to add. Use this compact checking chain:

det (Q) = 2^{9} det (P) ⟹ det (P) = 2 ⟹ det (adj P) = 4 ⟹ det (adj (adj P)) = 16 .

Answer: D) 16.

How can a matrix-identity mix-up produce C) 32?

One possible erroneous route produces C) 32 by putting the wrong matrix into the double-adjoint identity. This is an algebra error, not a claim about observed student responses. The determinant scaling afterward can be correct while the answer is wrong.

For a matrix of order three, the correct supplementary identity is:

adj (adj P) = (det P) P .

False substitution, do not use:

adj (adj P) \overset{false}{=} (det P) adj P .

Using the established determinant values, that false substitution leads to:

det (2 adj P) = 2^{3} det (adj P) = 8 \times 4 = 32 .

The error is the final matrix: it must be the original matrix, not its adjoint. The scalar determinant rule is not the problem.

The correct identity instead gives:

det (2 P) = 2^{3} det (P) = 8 \times 2 = 16 .

Keep this comparison as an error check. For the main solution, I would use the two successive determinant-identity applications because each intermediate value is easy to verify.

Can you solve three related questions without expanding cofactors?

Factor extraction and the adjoint determinant identity are enough for all three questions below. These are original related practice questions, not verified past-year questions. Each tests a different change:

Question 1: Changed entry factors.
Question 2: Changed matrix order.
Question 3: Sign information lost through squaring.

What changes when the entry factors use powers of three?

The determinant multiplier changes, but the double-adjoint exponent stays four for matrices of order three. Question 1: Two such matrices satisfy:

A = [a_{i j}], B = [b_{i j}], b_{i j} = 3^{i + j - 2} a_{i j}, det (B) = 3^{7} .

Find:

det (adj (adj A)) .

Answer: 81. Split the multiplier into row and column factors:

3^{i + j - 2} = 3^{i - 1} 3^{j - 1} .

Both sets of factors are 1, 3 and 9, giving:

det (B) = (1 \times 3 \times 9)^{2} det (A) = 3^{6} det (A) .

det (A) = 3, det (adj (adj A)) = 3^{4} = 81 .

What changes when the matrix has order four?

Each adjoint application now uses exponent three, not two. Question 2: A matrix of order four has determinant 2. Find its double-adjoint determinant.

A \in ℝ^{4 \times 4}, det (A) = 2 .

Answer: 512. Both adjoints remain matrices of order four:

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

det (adj (adj A)) = 8^{3} = 512 .

Equivalently, the combined exponent is:

(4 - 1)^{2} = 9, 2^{9} = 512 .

Can the adjoint determinant tell you the original determinant’s sign?

For a matrix of order three, the original sign is not determined, because the determinant is squared. Question 3: Such a matrix satisfies:

A \in ℝ^{3 \times 3}, det (adj A) = 9 .

Find every possible original determinant and then the double-adjoint determinant. The possible values follow from:

(det A)^{2} = 9 ⟹ det (A) = \pm 3 .

The double-adjoint determinant is 81 in either case:

det (adj (adj A)) = 9^{2} = 81 .

Squaring loses the original sign, but the given adjoint determinant is enough for the second application. Before checking your work, write the row factors, column factors and adjoint exponent separately for each drill.

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

Why is the determinant multiplier 2^9 in this JEE 2026 question?

For the third-order matrices with q_ij = 2^(i+j-1)p_ij, split each multiplier as 2^(i-1) × 2^j. Extract row factors 1, 2 and 4, then column factors 2, 4 and 8; their product is 512 = 2^9. Do not multiply all nine entry factors, because each determinant term contains only one entry from each row and column.

How do you find the determinant of a double adjoint?

For a square matrix A of order n, apply det(adj A) = (det A)^(n-1) twice. This gives det(adj(adj A)) = (det A)^((n-1)^2). In the article's third-order problem, det P = 2, so the result is 2^4 = 16.

Why is 32 the wrong answer to the double-adjoint question?

For a third-order matrix P, the correct identity is adj(adj P) = (det P)P, not (det P)adj P. With det P = 2, the correct calculation is det(2P) = 2^3 × 2 = 16. Substituting adj P in place of P incorrectly gives 32.

If det(adj A) is 9, is det A always 3?

For a real third-order matrix, det(adj A) = (det A)^2, so det A can be either 3 or -3. The original sign cannot be recovered from the square. The double-adjoint determinant is 9^2 = 81 in either case.