$\begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0$, where $a$, $b$, $c$ are in A.P.

> [!TLDR] > The determinant simplifies to a constant expression independent of $x$, and setting it to zero forces $a$, $b$, $c$ to be in arithmetic progression — which is already given. The determinant is identically zero for all $x$ when $a$, $b$, $c$ are in A.P., so the equation holds for all $x$. The key insight here is that the determinant looks messy with $x$ everywhere, but the structure is special. Each row is almost the same — the first two columns are just consecutive integers shifted by $x$, and the third column has $x$ plus a constant. When $a$, $b$, $c$ are in arithmetic progression, the rows become linearly dependent, and the determinant vanishes identically. Let’s see why. 1. **Write the determinant explicitly.** $$ \Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} $$ 2. **Use row operations to eliminate $x$.** Subtract Row 1 from Row 2, and Row 2 from Row 3. These operations don’t change the determinant’s value. $$ R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_2 $$ We get: $$ \Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ 1 & 1 & b-a \\ 1 & 1 & c-b \end{vmatrix} $$ 3. **Notice the pattern in the last two rows.** The first two columns of rows 2 and 3 are identical: both are $(1, 1)$. If $b-a = c-b$, then rows 2 and 3 become identical, making the determinant zero. But $b-a = c-b$ is exactly the condition that $a$, $b$, $c$ are in arithmetic progression. 4. **But we are told $a$, $b$, $c$ are in A.P. — so what happens?** Since $b-a = c-b$, let this common difference be $d$. Then $b-a = d$ and $c-b = d$. The determinant becomes: $$ \Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ 1 & 1 & d \\ 1 & 1 & d \end{vmatrix} $$ Rows 2 and 3 are now identical, so $\Delta = 0$ for **any** $x$. > [!WARNING] > A common mistake is to expand the determinant and try to solve for $x$. That leads to a messy expression that simplifies to

x + 1 x + 2 x + a x + 2 x + 3 x + b x + 3 x + 4 x + c = 0, where a, b, c are in A.P.

NCERT Exemplar · Q53

\begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 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 simplifies to a constant expression independent of $x$ , and setting it to zero forces $a$ , $b$ , $c$ to be in arithmetic progression — which is already given. The determinant is identically zero for all $x$ when $a$ , $b$ , $c$ are in A.P., so the equation holds for all $x$ .

The key insight here is that the determinant looks messy with $x$ everywhere, but the structure is special. Each row is almost the same — the first two columns are just consecutive integers shifted by $x$ , and the third column has $x$ plus a constant. When $a$ , $b$ , $c$ are in arithmetic progression, the rows become linearly dependent, and the determinant vanishes identically.

Let’s see why.

Write the determinant explicitly.

\Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix}

Use row operations to eliminate $x$ . Subtract Row 1 from Row 2, and Row 2 from Row 3. These operations don’t change the determinant’s value.

R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_2

We get:

\Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ 1 & 1 & b-a \\ 1 & 1 & c-b \end{vmatrix}

Notice the pattern in the last two rows.

The first two columns of rows 2 and 3 are identical: both are $(1, 1)$ . If $b-a = c-b$ , then rows 2 and 3 become identical, making the determinant zero. But $b-a = c-b$ is exactly the condition that $a$ , $b$ , $c$ are in arithmetic progression.
But we are told $a$ , $b$ , $c$ are in A.P. — so what happens?

Since $b-a = c-b$ , let this common difference be $d$ . Then $b-a = d$ and $c-b = d$ . The determinant becomes:

\Delta = \begin{vmatrix} x+1 & x+2 & x+a \\ 1 & 1 & d \\ 1 & 1 & d \end{vmatrix}

Rows 2 and 3 are now identical, so $\Delta = 0$ for any $x$ .

Watch out

A common mistake is to expand the determinant and try to solve for $x$ . That leads to a messy expression that simplifies to $0$ only when $a$ , $b$ , $c$ are in A.P. — but the problem already gives that condition, so the equation is an identity in $x$ , not something to solve for $x$ .

What if we didn’t notice the row identity? Let’s verify by expanding. Subtract Row 1 from Row 2 and Row 3 (or use column operations). Another clean approach: subtract Column 1 from Column 2, and Column 2 from Column 3.

C_2 \to C_2 - C_1, \quad C_3 \to C_3 - C_2

This gives: …

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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 59 on this concept.

CBSE 2024Set 65/1/11 markMCQ
$\begin{vmatrix} x+1 & x-1 \\ x^2+x+1 & x^2-x+1 \end{vmatrix}$
›Reveal solutionSolution

Expand the $2\times2$ determinant as $ad-bc$ ; the cube-sum and cube-difference collapse to a constant. The value is $2$ , option (B).

For a $2\times2$ determinant, $\begin{vmatrix} a & b \\c & d \end{vmatrix}=ad-bc$ .
$\Delta=\begin{vmatrix} x+1 & x-1 \\x^2+x+1 & x^2-x+1 \end{vmatrix}=(x+1)(x^2-x+1)-(x-1)(x^2+x+1)$
Use the standard factorisations $a^3+b^3=(a+b)(a^2-ab+b^2)$ and $a^3-b^3=(a-b)(a^2+ab+b^2)$ with $a=x,\ b=1$ : …
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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 $\Delta_2 = -12$ . …
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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: $12+11 = 23$ , $10+26 = 36$ , $26+37 = 63$ .

…

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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:

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

…

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

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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).

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

…

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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: …

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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.

…

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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 $\cos^2\theta - \sin^2\theta = \cos 2\theta$ .

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

…

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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:

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

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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: $\det(MN) = \det(mI)$ .

$\det(MN) = \det(M) \cdot \det(N) = m \cdot \det(N)$ . …
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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 $C_1$ :

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

…

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Q. $\begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0$ , where $a$ , $b$ , $c$ are in A.P.

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Method: Showing a Determinant Vanishes When Entries Follow an Arithmetic Progression

Steps

Common Mistakes

Why determinants change under row operations

Step-by-step evaluation

Q.∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣=0\begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0​x+1x+2x+3​x+2x+3x+4​x+ax+bx+c​​=0, where aaa, bbb, ccc are in A.P.

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Q. $\begin{vmatrix} x + 1 & x + 2 & x + a \\ x + 2 & x + 3 & x + b \\ x + 3 & x + 4 & x + c \end{vmatrix} = 0$ , where $a$ , $b$ , $c$ are in A.P.

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