If $\cos 2\theta = 0$, then $\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 = $ ________ .

If cos 2θ = 0, then 0 cosθ sinθ ; cosθ sinθ 0 ; sinθ 0 cosθ ² = ____ .

NCERT Exemplar · Q41

\cos 2\theta = 0

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Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

Expanding a $4\times4$ or $5\times5$ determinant term by term is painful and error-prone. The smarter route is to transform the determinant into an easy form using properties (the "identities") that change its value in a known, controlled way — then read the answer off a triangular matrix.

The geometric intuition

A determinant measures the signed "volume" of the box spanned by the rows in $n$ -dimensional space. Sliding one row parallel to another doesn't change that volume; swapping two rows flips its sign; scaling a row scales the volume. The algebraic identities are just these facts translated into rules.

The three row (or column) operations

\det \to -\det

The identical rules hold for columns. There is also row-wise linearity: if a row is a sum $R_i = R_i' + R_i''$ , the determinant splits into the sum of two determinants with all other rows fixed.

Watch out

Row-wise linearity is not $\det(A+B)=\det A+\det B$ — that is false. The splitting works one row at a time.

The strategy

Use operation 3 to create zeros in a row or column (value unchanged).
Factor out common factors with operation 2.
Swap rows if needed to reach upper-triangular form (track the sign change).
The determinant is then the product of the diagonal entries.

Worked example

\det\begin{pmatrix}1&2&3\\4&5&6\\7&8&10\end{pmatrix}.

Apply $R_2\to R_2-4R_1$ and $R_3\to R_3-7R_1$ (no change), then $R_3\to R_3-2R_2$ : …

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The determinant equals $-(\cos^3\theta+\sin^3\theta)$ ; with $\cos 2\theta = 0$ its square is $\tfrac12$ .

Set up

Write $a=\cos\theta$ and $b=\sin\theta$ , so the determinant is

\Delta = \begin{vmatrix} 0 & a & b \\ a & b & 0 \\ b & 0 & a \end{vmatrix}.

Expand along the first row

\Delta = 0\cdot(\dots) - a\begin{vmatrix} a & 0 \\ b & a \end{vmatrix} + b\begin{vmatrix} a & b \\ b & 0 \end{vmatrix} = -a(a^2) + b(-b^2) = -(a^3+b^3).

So $\Delta = -(\cos^3\theta+\sin^3\theta)$ and $\Delta^2 = (\cos^3\theta+\sin^3\theta)^2$ .

Use the condition

$\cos 2\theta = 0 \Rightarrow \cos^2\theta = \sin^2\theta = \tfrac12$ , hence $\cos\theta\sin\theta = \pm\tfrac12$ . Expanding the square,

$\Delta^2 = \cos^6\theta+\sin^6\theta + 2\cos^3\theta\sin^3\theta = \big(1 - 3\cos^2\theta\sin^2\theta\big) + 2\cos^3\theta\sin^3\theta.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 18 on this concept.

KEAM 2026Set eng-2026-04184 marksMCQ
$A$
›Reveal solutionSolution

$|A|=-1/7$ , and $|B|=|AB|/|A|=21/(-1/7)=-147$ .

Since $|A^{-1}|=\dfrac{1}{|A|}=-7$ , we have $|A|=-\dfrac17$ .

Using $|AB|=|A|\,|B|$ : …
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KEAM 2026Set eng-2026-04184 marksMCQ
$f(x)=\begin{vmatrix}-1 & x & 3\\ 0 & 1 & 2x\\ 1 & -1 & 1\end{vmatrix}$
›Reveal solutionSolution

At $x=-1$ the determinant evaluates to $0$ .

At $x=-1$ the matrix is $\begin{pmatrix}-1&-1&3\\0&1&-2\\1&-1&1\end{pmatrix}$ . Expanding along the first column: …
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KEAM 2026Set eng-2026-04194 marksMCQ
$\begin{vmatrix} 11 & 1 & 1 \\ 1 & 21 & 1 \\ 1 & 1 & 31 \end{vmatrix}$
›Reveal solutionSolution

Cofactor expansion along the first row gives $7100$ .

Expansion. For $\begin{vmatrix}11&1&1\\1&21&1\\1&1&31\end{vmatrix}$ :

$=11(21\cdot31-1\cdot1)-1(1\cdot31-1\cdot1)+1(1\cdot1-21\cdot1)$ …
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KEAM 2026Set eng-2026-04204 marksMCQ
$\begin{vmatrix} (10^5 + 10^{-5})^2 & (10^5 - 10^{-5})^2 & 1 \\ (100^6 + 100^{-6})^2 & (100^6 - 100^{-6})^2 & 1 \\ (6^{100} + 6^{-100})^2 & (6^{100} - 6^{-100})^2 & 1 \end{vmatrix}$
›Reveal solutionSolution
In every row the two entries are $(P+Q)^2$ and $(P-Q)^2$ with $PQ=1$ , so $C_1-C_2=4PQ=4$ for all rows. That makes column $C_1-C_2$ a multiple of the all-ones column $C_3$ ; two proportional columns force the determinant to be $0$ .

In each row the entries have the form $(P+Q)^2,\ (P-Q)^2,\ 1$ , where:

Row 1: $P=10^5,\ Q=10^{-5}$ , $PQ=1$ .

Row 2: $P=100^6,\ Q=100^{-6}$ , $PQ=1$ .

Row 3: $P=6^{100},\ Q=6^{-100}$ , $PQ=1$ .

Apply the column operation $C_1\to C_1-C_2$ . For every row, …
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KEAM 2026Set eng-2026-04224 marksMCQ
$\begin{vmatrix} \sin 30^\circ & \cos 30^\circ & \sin(30^\circ + 75^\circ) \\ \sin 45^\circ & \cos 45^\circ & \sin(45^\circ + 75^\circ) \\ \sin 60^\circ & \cos 60^\circ & \sin(60^\circ + 75^\circ) \end{vmatrix}$
›Reveal solutionSolution

The third column is a fixed linear combination of the first two, forcing a zero determinant.

Each entry of column 3 is $\sin(\theta+75^\circ)=\sin\theta\cos75^\circ+\cos\theta\sin75^\circ$ . …
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KEAM 2025Set eng-2025-04254 marksMCQ
$a_1, a_2, a_3, a_4, a_5$
›Reveal solutionSolution

Two columns are proportional, forcing the determinant to zero.

With $a_1 = 2, r = \tfrac12$ : the terms are $2, 1, \tfrac12, \tfrac14, \tfrac18, \tfrac{1}{16}$ .

In the determinant

$\begin{vmatrix} a_1 & a_2 & 1 \\ a_3 & a_4 & 1 \\ a_5 & a_6 & 1 \end{vmatrix},$ …
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KEAM 2025Set eng-2025-04284 marksMCQ
$\Delta = \begin{vmatrix} x & y & 1 \\ x+y & y+1 & x+1 \\ 1 & x & y \end{vmatrix}$
›Reveal solutionSolution
Expanding $\Delta$ and substituting $x+y=-1$ , every term reduces to $0$ because it carries a factor $x+y+1=0$ .

Expanding along the first row:
$\Delta=x\big[(y+1)y-(x+1)x\big]-y\big[(x+y)y-(x+1)\big]+\big[(x+y)x-(y+1)\big].$
With $x+y=-1$ :

First term: $x\big[(y^2-x^2)+(y-x)\big]=x(y-x)(x+y+1)=x(y-x)(0)=0$ . …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\alpha + \beta + \gamma = 0$
›Reveal solutionSolution

Let $a=e^{\alpha},b=e^{\beta},c=e^{\gamma}$ , so $abc=e^{\alpha+\beta+\gamma}=1$ . Split the third column $a^3-1=a^3+(-1)$ ; the two resulting Vandermonde determinants are equal and cancel, giving $0$ .

Write the rows as $(a,a^2,a^3-1)$ etc. By column-linearity the determinant splits as $D_1-D_2$ where

$D_1=\begin{vmatrix}a&a^2&a^3\\\\ b&b^2&b^3\\\\ c&c^2&c^3\end{vmatrix}=abc\begin{vmatrix}1&a&a^2\\\\1&b&b^2\\\\1&c&c^2\end{vmatrix}=abc\\,V,$ …
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KEAM 2024Set eng-2024-06064 marksMCQ
$A=\begin{pmatrix}2 & -1 & 1\\ -1 & 0 & 2\\ 1 & -2 & -1\end{pmatrix}$
›Reveal solutionSolution

For a $3\times3$ matrix, scaling by $1/|A|$ scales the determinant by $(1/|A|)^3$ . …

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KEAM 2024Set eng-2024-06074 marksMCQ
$A=\begin{pmatrix}a_1&b_1&c_1\\a_2&b_2&c_2\\a_3&b_3&c_3\end{pmatrix}$
›Reveal solutionSolution

Pull common factors out of each row and each column of $B$ .

Rows of $B$ are $(a_1,2b_1,4c_1)$ , $2(a_2,2b_2,4c_2)$ , $4(a_3,2b_3,4c_3)$ , giving a row factor $1\cdot2\cdot4=8$ . The remaining matrix has columns with factors $1,2,4$ , giving a column factor $1\cdot2\cdot4=8$ , and what remains i …
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KEAM 2024Set eng-2024-06074 marksMCQ
$A=\begin{pmatrix}-7&3\\3&-1\end{pmatrix}$
›Reveal solutionSolution

$\det(A^5)=(\det A)^5$ .

$\det A=(-7)(-1)-(3)(3)=7-9=-2$ . Then $\det(A^5)=(\det A)^5=(-2)^5=-32$ …
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KEAM 2024Set eng-2024-06094 marksMCQ
$A=(a_{ij})$
›Reveal solutionSolution

Convert minors to cofactors with alternating signs, then expand: $|A|=-80$ .

Cofactors: $C_{11}=+M_{11}=-40$ , $C_{12}=-M_{12}=10$ , $C_{13}=+M_{13}=35$ .

Expanding along the first row: …
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←Previous questionQ40 — If x, y, z ∈ mathbbR, then the value of determinant (2x + 2-x)² (2x - …→Next questionQ42 — If A is a matrix of order 3 × 3, then (A²)⁻¹ = ____ .

Q.If $\cos 2\theta = 0$ , then $\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 =$ ________ .

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Set up

Expand along the first row

Use the condition

Method: Evaluating a Trigonometric Determinant Under a Given Angle Condition

Steps

Common Mistakes

Q.If cos⁡2θ=0\cos 2\theta = 0cos2θ=0, then ∣0cos⁡θsin⁡θcos⁡θsin⁡θ0sin⁡θ0cos⁡θ∣2=\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 = ​0cosθsinθ​cosθsinθ0​sinθ0cosθ​​2= ________ .

Set up

Expand along the first row

Use the condition

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Q.If $\cos 2\theta = 0$ , then $\begin{vmatrix} 0 & \cos\theta & \sin\theta \\ \cos\theta & \sin\theta & 0 \\ \sin\theta & 0 & \cos\theta \end{vmatrix}^2 =$ ________ .

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