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.

COMEDK 2025Set 2025-A1 markMCQ
$a_{21}$
›Reveal solutionSolution
The cofactor of $a_{21}$ is found by taking $(-1)^{2+1}$ times the determinant of the submatrix obtained by deleting row 2 and column 1. The result is $-4$ , so the correct option is (C).

The cofactor of an element in a matrix is not just the minor (the determinant of the submatrix left after removing that element’s row and column). It also includes a sign factor $(-1)^{i+j}$ , where $i$ and $j$ are the row and column indices. This sign alternates like a chessboard pattern. For $a_{21}$ (row 2, column 1), the sign is negative because $2+1=3$ is odd. So we compute the minor and then flip its sign.

Identify the element and its position. The element $a_{21}$ is in row 2, column 1. In the given matrix

$\Delta = \begin{vmatrix} 1 & 4 & 4 \\ -3 & 5 & 9 \\ 2 & 1 & 2 \end{vmatrix},$
$a_{21} = -3$ . But the cofactor depends only on position, not on the value of the element itself.

Delete row 2 and column 1. Removing row 2 and column 1 leaves the submatrix:

$\begin{pmatrix} 4 & 4 \\ 1 & 2 \end{pmatrix}.$

Compute the minor $M_{21}$ . The minor is the determinant of that $2\times2$ submatrix:

$M_{21} = \begin{vmatrix} 4 & 4 \\ 1 & 2 \end{vmatrix} = (4)(2) - (4)(1) = 8 - 4 = 4.$

Apply the sign factor. …
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COMEDK 2024Set 2024-A1 markMCQ
$\left|\begin{array}{ccc} \cos (\alpha+\beta) & -\sin (\alpha+\beta) & \cos 2 \beta \\ \sin \alpha & \cos \alpha & \sin \beta \\ -\cos \alpha & \sin \alpha & \cos \beta \end{array}\right|$
›Reveal solutionSolution

Expanding along the first row collapses the determinant to $1+\cos 2\beta$ , which contains no $\alpha$ — so it is independent of $\alpha$ : option (D).

Cofactor expansion along row 1

The three minors are
$M_{11}=\begin{vmatrix}\cos\alpha&\sin\beta\\\sin\alpha&\cos\beta\end{vmatrix}=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\cos(\alpha+\beta),$ $M_{12}=\begin{vmatrix}\sin\alpha&\sin\beta\\-\cos\alpha&\cos\beta\end{vmatrix}=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\sin(\alpha+\beta),$ $M_{13}=\begin{vmatrix}\sin\alpha&\cos\alpha\\-\cos\alpha&\sin\alpha\end{vmatrix}=\sin^2\alpha+\cos^2\alpha=1.$
With the cofactor sign pattern $(+,-,+)$ and the row-1 entries $\cos(\alpha+\beta),\ -\sin(\alpha+\beta),\ \cos2\beta$ : …
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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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