Matrices and Determinants JEE 2024: Why the Answer Is 64

What is the nested-product question, and can its answer key be trusted?

The displayed matrices give determinant 64, not the source’s reported 729. This matrices and determinants JEE 2024 solution follows the supplied multiplication-first method while separating the calculation from the inconsistent answer key.

The supplied record labels the question JEE Main 2024 but gives the date 2023-01-24. That conflict prevents a verified year-and-shift attribution. It reports option B, 729; publication as a verified PYQ solution requires source correction.

Using the following matrices and definitions, find the determinant of the final matrix:

A = [\begin{matrix} \sqrt{2} & 1 \\ - 1 & 1 / \sqrt{2} \end{matrix}], B = [\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}] .

C = A B A^{T}, X = A^{T} C^{2} A .

The other options are absent from the supplied DATA block, despite the editorial description. There is no basis for reconstructing them or assigning the calculated result an option label.

How do I transpose the first matrix and calculate the first product?

Transpose by interchanging row and column positions, then multiply each row of the first matrix by each column of the second. Transposition does not independently change signs: the negative entry moves with its sign unchanged.

A^{T} = [\begin{matrix} \sqrt{2} & - 1 \\ 1 & 1 / \sqrt{2} \end{matrix}] .

Following the supplied solution’s first two steps, calculate all four entries of the initial product. Keep radicals exact throughout; decimal approximations add rounding error without simplifying the later products.

(A B)_{11} & = \sqrt{2} \cdot 1 + 1 \cdot 1 = \sqrt{2} + 1, (A B)_{12} & = \sqrt{2} \cdot 0 + 1 \cdot 1 = 1, (A B)_{21} & = (- 1) \cdot 1 + \frac{1}{\sqrt{2}} \cdot 1 = - 1 + \frac{1}{\sqrt{2}}, (A B)_{22} & = (- 1) \cdot 0 + \frac{1}{\sqrt{2}} \cdot 1 = \frac{1}{\sqrt{2}} .

A B = [\begin{matrix} \sqrt{2} + 1 & 1 \\ - 1 + 1 / \sqrt{2} & 1 / \sqrt{2} \end{matrix}] .

How do I calculate the middle matrix, its square and the final product?

Multiply the first product by the transpose, square the resulting matrix by matrix multiplication, then include both outer factors. The supplied solution omits these calculations; working entry by entry makes the signs and factor order checkable. $C = (A B) A^{T} .$

C_{11} & = (\sqrt{2} + 1) \sqrt{2} + 1 \cdot 1 = 3 + \sqrt{2}, C_{12} & = - (\sqrt{2} + 1) + \frac{1}{\sqrt{2}} = - 1 - \frac{\sqrt{2}}{2}, C_{21} & = (- 1 + \frac{1}{\sqrt{2}}) \sqrt{2} + \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2}, C_{22} & = - (- 1 + \frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{3 - \sqrt{2}}{2} .

C = [\begin{matrix} 3 + \sqrt{2} & - 1 - \sqrt{2} / 2 \\ 1 - \sqrt{2} / 2 & (3 - \sqrt{2}) / 2 \end{matrix}] .

For the square, each entry is a row-column dot product. Squaring individual entries is not matrix squaring.

(C^{2})_{11} & = (3 + \sqrt{2})^{2} + (- 1 - \frac{\sqrt{2}}{2}) (1 - \frac{\sqrt{2}}{2}) = \frac{21}{2} + 6 \sqrt{2}, (C^{2})_{12} & = (3 + \sqrt{2}) (- 1 - \frac{\sqrt{2}}{2}) + (- 1 - \frac{\sqrt{2}}{2}) \frac{3 - \sqrt{2}}{2} = - 5 - \frac{11 \sqrt{2}}{4}, (C^{2})_{21} & = (1 - \frac{\sqrt{2}}{2}) (3 + \sqrt{2}) + \frac{3 - \sqrt{2}}{2} (1 - \frac{\sqrt{2}}{2}) = 4 - \frac{7 \sqrt{2}}{4}, (C^{2})_{22} & = (1 - \frac{\sqrt{2}}{2}) (- 1 - \frac{\sqrt{2}}{2}) + {(\frac{3 - \sqrt{2}}{2})}^{2} = \frac{9}{4} - \frac{3 \sqrt{2}}{2} .

Name the resulting entries to keep the next multiplication readable. This postpones substitution, not any matrix operation.

C^{2} = [\begin{matrix} 21 / 2 + 6 \sqrt{2} & - 5 - 11 \sqrt{2} / 4 \\ 4 - 7 \sqrt{2} / 4 & 9 / 4 - 3 \sqrt{2} / 2 \end{matrix}] = [\begin{matrix} p & q \\ r & s \end{matrix}] .

Multiply on the right first. Each row of the squared matrix is multiplied by each column of the original matrix.

C^{2} A = [\begin{matrix} \sqrt{2} p - q & p + q / \sqrt{2} \\ \sqrt{2} r - s & r + s / \sqrt{2} \end{matrix}] .

Now multiply on the left by the transpose. The four row-by-column expressions, before substitution, are:

X_{11} & = \sqrt{2} (\sqrt{2} p - q) - (\sqrt{2} r - s), X_{12} & = \sqrt{2} (p + q / \sqrt{2}) - (r + s / \sqrt{2}), X_{21} & = (\sqrt{2} p - q) + \frac{1}{\sqrt{2}} (\sqrt{2} r - s), X_{22} & = (p + q / \sqrt{2}) + \frac{1}{\sqrt{2}} (r + s / \sqrt{2}) .

Substitute the named entries, then collect rational and radical terms separately. The final matrix is:

X = [\begin{matrix} 129 / 4 + 23 \sqrt{2} / 2 & 9 / 2 + 67 \sqrt{2} / 8 \\ 45 / 2 + 83 \sqrt{2} / 8 & 57 / 8 + 19 \sqrt{2} / 4 \end{matrix}] .

This grouping is valid because matrix multiplication is associative. It changes brackets, not factor order:

(A^{T} C^{2}) A = A^{T} (C^{2} A) .

Moving the transpose beside the final factor would require a separate justification. Matrices do not generally commute.

Why does the determinant equal 64 when the source reports 729?

The determinant properties give 64 from the displayed matrices, so the source’s 729 conflicts with the calculation. Complete the supplied solution’s fifth step by retaining one determinant for every matrix factor.

det (A) = \sqrt{2} \times \frac{1}{\sqrt{2}} - 1 \times (- 1) = 2, det (B) = 1 \times 1 - 0 \times 1 = 1 .

A transpose has the same determinant as the original matrix. Apply that fact to the middle product: $det (A^{T}) = det (A) = 2 .$

det (C) = det (A) det (B) det (A^{T}) = 2 \times 1 \times 2 = 4 .

For the final product, square the middle determinant and retain both outer determinants. This avoids expanding the radical-heavy final matrix:

det (X) = det (A^{T}) [det (C)]^{2} det (A) = 2 \times 4^{2} \times 2 = 64 .

Result for the displayed matrices: 64. Answer reported by the supplied source: option B, 729. No valid simplification of these inputs establishes 729.

Do not change a matrix entry or invent algebra to force agreement. Neither a corrected question nor a corrected option label can be established without source evidence.

What happens if I drop an outer determinant factor?

Dropping the final factor produces 32, an illustrative incorrect result, not a verified wrong option. The error occurs when the rightmost matrix disappears during conversion from the matrix product to scalar determinants:

det (X) = det (A^{T}) [det (C)]^{2} = 2 \times 16 = 32 (incorrect) .

Repair the setup by writing all three determinant factors before inserting values. The rightmost factor must remain:

det (X) = \underset{left factor}{\underset{⏟}{det (A^{T})}} \underset{middle factor}{\underset{⏟}{det (C^{2})}} \underset{right factor}{\underset{⏟}{det (A)}} .

Explaining an actual distractor requires the missing option list. There is no evidence here that 32 is commonly chosen or especially tempting.

Which three practice questions test the same determinant skills?

These questions test the middle product, a changed power and the signs of the outer determinants. They are original related practice questions, not additional verified PYQs. Attempt each before reading its short solution.

Question 1. Using the same original matrices, find: $det (A B A^{T}) .$

Solution. Multiply the three determinant factors:

det (A B A^{T}) = 2 \times 1 \times 2 = 4 .

Question 2. Using the same original matrix and middle matrix, find: $det (A^{T} C^{3} A) .$

Solution. Cube the middle determinant, but retain both outer factors:

det (A^{T} C^{3} A) = 2 \times 4^{3} \times 2 = 256 .

Question 3. Two square matrices of the same order satisfy:

det (P) = - 2, det (Q) = 3 .

Find: $det (P^{T} Q^{2} P) .$

Solution. The transpose preserves the negative determinant, and the two negative outer factors give a positive product:

det (P^{T} Q^{2} P) = (- 2) \times 3^{2} \times (- 2) = 36 .

For your next attempt, write one determinant factor for every matrix factor before evaluating anything. Check the power separately, then substitute the numbers.

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

How do I find det(A^T C^2 A) without multiplying the matrices?

Use det(A^T) = det(A) and multiply one determinant for every matrix factor. For the displayed matrices, det(A) = 2 and det(C) = det(ABA^T) = 4. Therefore, det(A^T C^2 A) = 2 × 4² × 2 = 64.

Is the answer 64 or 729 for the nested-product question?

The displayed matrices give 64, while the supplied source reports option B, 729. No valid simplification of these inputs gives 729. The missing option list prevents assigning 64 an option label.

Why am I getting 32 instead of 64?

Dropping the rightmost determinant factor gives 2 × 4² = 32 instead of 64. Write det(A^T C^2 A) = det(A^T) × det(C^2) × det(A), retaining both outer factors. Here, the omitted det(A) equals 2.

Is this a verified JEE Main 2024 question?

The supplied record labels the question JEE Main 2024 but gives the date 2023-01-24. That conflict prevents verified year-and-shift attribution. Source correction is required before treating it as a verified PYQ.