If $\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc$, then the value of $k$ is: (A) $0$ (B) $1$ (C) $2$ (D) $4$

> [!TLDR] > Expand the determinant using cofactor expansion or row operations to find that it equals $4abc$, so $k = 4$. The problem asks us to evaluate a $3 \times 3$ determinant with a specific symmetric structure and express it as a multiple of $abc$. The key insight is recognizing that this determinant has a pattern: each diagonal entry is the negative of a variable, while off-diagonal entries in each row are positive. This structure often yields clean results. We can compute this determinant by expanding along the first row using cofactors. 1. **Set up the cofactor expansion along row 1:** $$\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = -a \begin{vmatrix} -b & c \\ b & -c \end{vmatrix} - b \begin{vmatrix} a & c \\ a & -c \end{vmatrix} + c \begin{vmatrix} a & -b \\ a & b \end{vmatrix}$$ 2. **Evaluate the first $2 \times 2$ determinant:** $$\begin{vmatrix} -b & c \\ b & -c \end{vmatrix} = (-b)(-c) - (c)(b) = bc - bc = 0$$ So the first term contributes $-a \cdot 0 = 0$. 3. **Evaluate the second $2 \times 2$ determinant:** $$\begin{vmatrix} a & c \\ a & -c \end{vmatrix} = a(-c) - c(a) = -ac - ac = -2ac$$ The second term contributes $-b \cdot (-2ac) = 2abc$. 4. **Evaluate the third $2 \times 2$ determinant:** $$\begin{vmatrix} a & -b \\ a & b \end{vmatrix} = a(b) - (-b)(a) = ab + ab = 2ab$$ The third term contributes $c \cdot 2ab = 2abc$. 5. **Sum all contributions:** $$0 + 2abc + 2abc = 4abc$$ Therefore, the determinant equals $4abc$, which means $k = 4$. > [!TIP] > When a determinant has this "negative diagonal, positive off-diagonal" pattern, it often simplifies nicely. Notice how the first cofactor vanished completely because its $2 \times 2$ minor had identical rows up to sign. > [!ANSWER] > The value of $k$ is $\boxed{4}$, so the correct option is **(D)**.

If -a b c a -b c a b -c = kabc, then the value of k is: (A) 0 (B) 1 (C) 2 (D) 4

Question

\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

✓ Free question

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 $R_3\to R_3-2R_2$ :

\det\begin{pmatrix}1&2&3\\0&-3&-6\\0&0&1\end{pmatrix}=1\times(-3)\times 1=-3.

No cofactor was ever expanded — we just slid rows around.

Tip

Aim your zeros at a row or column that already contains a $1$ to keep the arithmetic clean. And remember operation 3 needs a different row: adding a multiple of a row to itself rescales it and changes the value.

Evaluating determinants using row and column operations rather than direct expansion is a core skill in the CBSE Class 12 Determinants chapter, and "properties of determinants class 12 with examples" is one of the most searched topics for board exam revision. This technique of reducing a determinant to triangular form is also a favourite approach in JEE Main and JEE Advanced problems involving higher-order determinants.

Expand the determinant using cofactor expansion or row operations to find that it equals $4abc$ , so $k = 4$ .

The problem asks us to evaluate a $3 \times 3$ determinant with a specific symmetric structure and express it as a multiple of $abc$ . The key insight is recognizing that this determinant has a pattern: each diagonal entry is the negative of a variable, while off-diagonal entries in each row are positive. This structure often yields clean results.

We can compute this determinant by expanding along the first row using cofactors.

Set up the cofactor expansion along row 1:

\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = -a \begin{vmatrix} -b & c \\ b & -c \end{vmatrix} - b \begin{vmatrix} a & c \\ a & -c \end{vmatrix} + c \begin{vmatrix} a & -b \\ a & b \end{vmatrix}

Evaluate the first $2 \times 2$ determinant:

\begin{vmatrix} -b & c \\ b & -c \end{vmatrix} = (-b)(-c) - (c)(b) = bc - bc = 0

So the first term contributes $-a \cdot 0 = 0$ .

Evaluate the second $2 \times 2$ determinant:

\begin{vmatrix} a & c \\ a & -c \end{vmatrix} = a(-c) - c(a) = -ac - ac = -2ac

The second term contributes $-b \cdot (-2ac) = 2abc$ .

Evaluate the third $2 \times 2$ determinant:

\begin{vmatrix} a & -b \\ a & b \end{vmatrix} = a(b) - (-b)(a) = ab + ab = 2ab

The third term contributes $c \cdot 2ab = 2abc$ .

Sum all contributions:

0 + 2abc + 2abc = 4abc

Therefore, the determinant equals $4abc$ , which means $k = 4$ .

Tip

When a determinant has this "negative diagonal, positive off-diagonal" pattern, it often simplifies nicely. Notice how the first cofactor vanished completely because its $2 \times 2$ minor had identical rows up to sign.

✓Final answer

k

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 59 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix}$
›Reveal solutionSolution

The first determinant is diagonal; the second requires one row interchange to reach diagonal form. Each interchange flips the sign; evaluating both determinants gives $\Delta_2 = -2\Delta_1$ . The answer is (B).

Why determinants change under row operations

A determinant measures the signed volume of the parallelepiped spanned by the row vectors. When you swap two rows, you reflect the figure across a hyperplane—the volume stays the same in magnitude but the orientation reverses, flipping the sign.

The diagonal determinant is the easiest to compute: the product of the diagonal entries. The second determinant looks scrambled, but a single row swap will bring it into a form we recognize.

Step-by-step evaluation

1. Compute $\Delta_1$ directly.

The matrix is diagonal:
$\Delta_1 = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{vmatrix} = 1 \cdot 2 \cdot 3 = 6.$
2. Recognize the structure of $\Delta_2$ .
$\Delta_2 = \begin{vmatrix} 0 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The first two rows are out of order compared to a diagonal form. Swap rows 1 and 2 to bring the $1$ into the top-left position.

3. Apply the row-interchange property.

Swapping rows 1 and 2:
$\Delta_2 = -\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix}.$
The negative sign comes from the single interchange.

4. Evaluate the new diagonal determinant.
$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 6 \end{vmatrix} = 1 \cdot 2 \cdot 6 = 12.$
So $\Delta_2 = -12$ .

5. Relate $\Delta_2$ to $\Delta_1$ .

We have $\Delta_1 = 6$ and $\Delta_2 = -12$ . Notice that
$\Delta_2 = -12 = -2 \cdot 6 = -2\Delta_1.$
This matches option (B).

Watch out

A common mistake is to forget the sign change from the row swap. Without it, you'd incorrectly conclude $\Delta_2 = 12$ and miss the negative relationship.

Tip

For small determinants, you can also expand along the first row or column. For $\Delta_2$ , expanding along row 1 gives $0 \cdot C_{11} + 2 \cdot C_{12} + 0 \cdot C_{13}$ , where $C_{12} = -\begin{vmatrix}1 & 0 \\ 0 & 6\end{vmatrix} = -6$ , so $\Delta_2 = 2 \cdot (-6) = -12$ .

✓Final answer
$\Delta_2 = -2\Delta_1$
CBSE 2026Set A1 markMCQ
$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix} =$
(a) $1$
(b) $-1$
(c) $0$
(d) $2$
›Reveal solutionSolution

The determinant equals $0$ because one column is the sum of the other two.

Inspect the columns of

$\begin{vmatrix} 23 & 12 & 11 \\ 36 & 10 & 26 \\ 63 & 26 & 37 \end{vmatrix}.$

Check: $12+11 = 23$ , $10+26 = 36$ , $26+37 = 63$ .

So $C_1 = C_2 + C_3$ . When one column is a linear combination of the others, the columns are linearly dependent and the determinant is $0$ .

✓Final answer
$0$
CBSE 2026Set A1 markMCQ
$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} =$
(a) $1$
(b) $0$
(c) $-1$
(d) $\frac{1}{2}$
›Reveal solutionSolution

The determinant equals $\cos 90^\circ = 0$ .

Expand:

$\begin{vmatrix} \cos 15^\circ & \sin 15^\circ \\ \sin 75^\circ & \cos 75^\circ \end{vmatrix} = \cos 15^\circ\cos 75^\circ - \sin 15^\circ\sin 75^\circ.$

By the cosine addition formula $\cos A\cos B - \sin A\sin B = \cos(A+B)$ :

$= \cos(15^\circ + 75^\circ) = \cos 90^\circ = 0.$

✓Final answer
$0$
CBSE 2026Set A1 markMCQ
$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} =$
(a) $a^2 + b^2 + c^2 + d^2$
(b) $a^2 - b^2 - c^2 - d^2$
(c) $a^2 - b^2 + c^2 + d^2$
(d) $a^2 + b^2 + c^2 - d^2$
›Reveal solutionSolution

Expand the $2\times2$ determinant and simplify the complex products.

$\begin{vmatrix} a+ib & c+id \\ -c+id & a-ib \end{vmatrix} = (a+ib)(a-ib) - (c+id)(-c+id).$

First term: $(a+ib)(a-ib) = a^2 - (ib)^2 = a^2 + b^2$ .

Second term: $(c+id)(-c+id) = -c^2 + icd - icd + (id)^2 = -c^2 - d^2$ .

So the determinant $= (a^2 + b^2) - (-c^2 - d^2) = a^2 + b^2 + c^2 + d^2$ .

✓Final answer
$a^2 + b^2 + c^2 + d^2$
CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix}x^2-x+1 & x-1\\ x+1 & x+1\end{vmatrix}$
(a) $x^2-x+2$
(b) $x^3+x^2-2$
(c) $x^3-x^2+2$
(d) $x^3+x^2+4$
›Reveal solutionSolution

Expand the $2\times2$ determinant using $\begin{vmatrix}a&b\\c&d\end{vmatrix}=ad-bc$ .

$\Delta = (x^2-x+1)(x+1) - (x-1)(x+1)$

$(x^2-x+1)(x+1) = x^3+1$ (the middle terms cancel).

$(x-1)(x+1) = x^2-1$ .

$\Delta = (x^3+1)-(x^2-1) = x^3-x^2+2$ .

✓Final answer
$x^3-x^2+2$
CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row.

$\Delta=\begin{vmatrix}1&2&4\\-1&3&0\\4&1&0\end{vmatrix}$

Expanding along row 1:

$\Delta = 1(3\cdot0-0\cdot1) - 2(-1\cdot0-0\cdot4) + 4(-1\cdot1-3\cdot4)$

$= 1(0) - 2(0) + 4(-1-12) = 4(-13) = -52$ .

✓Final answer
$\Delta = -52$
CBSE 2026Set ANNUAL1 mark
$\Delta = \begin{vmatrix}0 & \sin\alpha & -\cos\alpha\\ -\sin\alpha & 0 & \sin\beta\\ \cos\alpha & -\sin\beta & 0\end{vmatrix}$
›Reveal solutionSolution

The matrix is skew-symmetric (each $a_{ij}=-a_{ji}$ ) and every odd-order skew-symmetric matrix has determinant $0$ .

Check: $a_{12}=\sin\alpha=-a_{21}$ , $a_{13}=-\cos\alpha=-a_{31}$ , $a_{23}=\sin\beta=-a_{32}$ , and all diagonal entries are $0$ — so the matrix is skew-symmetric.

For a skew-symmetric matrix $A$ of odd order $n$ , $\det A = \det A^T = \det(-A) = (-1)^n\det A = -\det A$ , so $\det A=0$ .

✓Final answer
$\Delta = 0$
CBSE 2026Set ANNUAL1 markMCQ
$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} =$
(a) $\dfrac{1}{2}$
(b) $\dfrac{\sqrt{3}}{2}$
(c) 0
(d) None of these
›Reveal solutionSolution

This determinant has the form $\cos^2\theta - \sin^2\theta = \cos 2\theta$ .

$\begin{vmatrix} \cos 30^\circ & \sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{vmatrix} = \cos 30^\circ \cdot \cos 30^\circ - \sin 30^\circ \cdot \sin 30^\circ = \cos^2 30^\circ - \sin^2 30^\circ$

Using $\cos^2\theta - \sin^2\theta = \cos 2\theta$ : this equals $\cos 60^\circ = \dfrac{1}{2}$ .

✓Final answer
$\dfrac{1}{2}$
CBSE 2026Set ANNUAL1 mark
$Q.Evaluate the determinant \Delta = \begin{vmatrix}1 & 2 & 4\ -1 & 3 & 0\ 4 & 1 & 0\end{vmatrix}.$
›Reveal solutionSolution

Expand the $3\times3$ determinant along the first row (or any row/column) using cofactors.

Working: Expanding along Row 1:

$\Delta = \begin{vmatrix}1 & 2 & 4\\ -1 & 3 & 0\\ 4 & 1 & 0\end{vmatrix}$

$= 1\begin{vmatrix}3 & 0\\ 1 & 0\end{vmatrix} - 2\begin{vmatrix}-1 & 0\\ 4 & 0\end{vmatrix} + 4\begin{vmatrix}-1 & 3\\ 4 & 1\end{vmatrix}$

$= 1(3\cdot0 - 0\cdot1) - 2(-1\cdot0 - 0\cdot4) + 4(-1\cdot1 - 3\cdot4)$

$= 1(0) - 2(0) + 4(-1-12) = 4(-13) = -52$

✓Final answer
$\Delta = -52$
CBSE 2025Set 65/4/11 markMCQ
$\det(M) = m$
›Reveal solutionSolution
The key idea is that $MN = mI$ implies $N = m M^{-1}$ , so $\det(N) = m^3 \det(M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$ . The correct option is (D).

The problem gives us two square matrices $M$ and $N$ of order 3, with $\det(M) = m$ and $MN = mI$ , where $I$ is the $3 \times 3$ identity matrix. We need $\det(N)$ .

The central concept here is the relationship between matrix multiplication and determinants. When two matrices multiply to give a scalar times the identity, that scalar is intimately connected to the determinant of the first matrix. The equation $MN = mI$ is not just a product — it tells us that $N$ is essentially a scaled inverse of $M$ .

Why? Because if $MN = mI$ , then multiplying both sides on the left by $M^{-1}$ (assuming $M$ is invertible) gives $N = m M^{-1}$ . But we must first check: is $M$ invertible? Yes — since $\det(M) = m \neq 0$ (the problem doesn't state $m \neq 0$ explicitly, but if $m = 0$ , then $MN = 0$ , which would make $N$ singular and the answer ambiguous; in standard exam contexts, $m$ is taken as a non-zero scalar, often a real number, and the options suggest $m \neq 0$ ). So $M^{-1}$ exists.

Now, the determinant of a scalar multiple of a matrix: for an $n \times n$ matrix $A$ , $\det(kA) = k^n \det(A)$ . Here $n = 3$ , so $\det(m M^{-1}) = m^3 \det(M^{-1})$ .

And we know $\det(M^{-1}) = \frac{1}{\det(M)} = \frac{1}{m}$ .

Putting it together:

From $MN = mI$ , take determinant on both sides: $\det(MN) = \det(mI)$ .

$\det(MN) = \det(M) \cdot \det(N) = m \cdot \det(N)$ .

$\det(mI)$ : $mI$ is a diagonal matrix with all diagonal entries $m$ , so its determinant is $m^3$ (since it's $3 \times 3$ ).

So $m \cdot \det(N) = m^3$ .

Divide both sides by $m$ (non-zero): $\det(N) = m^2$ .

Tip

A faster route: from $MN = mI$ , multiply both sides on left by $M^{-1}$ to get $N = m M^{-1}$ . Then $\det(N) = \det(m M^{-1}) = m^3 \cdot \frac{1}{m} = m^2$ . This avoids the determinant-of-product step, but both are equivalent.

Watch out

A common mistake is to forget the exponent on $m$ when taking $\det(mI)$ . Since $I$ is $3 \times 3$ , $\det(mI) = m^3$ , not $m$ . Also, do not confuse $MN = mI$ with $MN = I$ — the scalar $m$ changes the scaling factor.

Thus, the determinant of $N$ is $m^2$ .

✓Final answer
$\boxed{m^2}$
CBSE 2025Set E1 markMCQ
$\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix} =$
(a) $1190$
(b) $841$
(c) $0$
(d) $1$
›Reveal solutionSolution

A column that is the sum of the other two makes the determinant zero.

Examine the columns of

$\begin{vmatrix} 21 & 11 & 10 \\ 25 & 15 & 10 \\ 64 & 27 & 37 \end{vmatrix}.$

Check $C_2+C_3$ against $C_1$ :

$11+10=21,\quad 15+10=25,\quad 27+37=64.$

So $C_1=C_2+C_3$ , i.e. the columns are linearly dependent. A determinant with linearly dependent columns is $0$ (apply $C_1\to C_1-C_2-C_3$ to get a zero column).

✓Final answer
$0$
CBSE 2025Set E1 markMCQ
$\begin{vmatrix} 10 & 4 \\ 13 & 5 \end{vmatrix} =$
(a) $102$
(b) $2$
(c) $-2$
(d) $-102$
›Reveal solutionSolution

Cross-multiply: $ad-bc=-2$ .

For a $2\times2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}=ad-bc$ :

$\begin{vmatrix} 10 & 4 \\ 13 & 5 \end{vmatrix}=(10)(5)-(4)(13)=50-52=-2.$

✓Final answer
$-2$

Q.If ∣−abca−bcab−c∣=kabc\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc​−aaa​b−bb​cc−c​​=kabc, then the value of kkk is: (A) 000 (B) 111 (C) 222 (D) 444

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Q.If $\begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} = kabc$ , then the value of $k$ is: (A) $0$ (B) $1$ (C) $2$ (D) $4$

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