If $A$, $B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}$ is equal to (A) $0$ (B) $-1$ (C) $1$ (D) None of these

If A, B and C are angles of a triangle, then the determinant -1 cos C cos B ; cos C -1 cos A ; cos B cos A -1

NCERT Exemplar · Q29

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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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Expanding gives $\Delta=-1+\cos^2A+\cos^2B+\cos^2C+2\cos A\cos B\cos C$ ; the triangle identity makes the cosine part equal $1$ , so $\Delta=0$ — option (A).

The idea

Expand the determinant into a symmetric cosine expression, then apply the identity that holds for the angles of any triangle. The expansion lands directly on a known identity — no row-juggling required.

Step 1 — Expand along the first row

\Delta=-1\big[(-1)(-1)-\cos^2A\big]-\cos C\big[\cos C(-1)-\cos A\cos B\big]+\cos B\big[\cos C\cos A-(-1)\cos B\big].

Simplify each bracket:

\Delta=-(1-\cos^2A)+\big(\cos^2C+\cos A\cos B\cos C\big)+\big(\cos A\cos B\cos C+\cos^2B\big).

Step 2 — Collect terms …

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Q.If $A$ , $B$ and $C$ are angles of a triangle, then the determinant $\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}$ is equal to
(A) $0$
(B) $-1$
(C) $1$
(D) None of these

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

The idea

Step 1 — Expand along the first row

Step 2 — Collect terms …

Method: Expanding a Determinant and Matching It to a Known Identity

Steps

Common Mistakes

The idea

Step 1 — Expand along the first row

Step 2 — Collect terms …

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