If $a_1, a_2, a_3, \ldots, a_r$ are in G.P., then prove that the determinant $\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}$ is independent of $r$.

If a₁, a₂, a₃, …, ar are in G.P., then prove that the determinant ar+1 ar+5 ar+9 ar+7 ar+11 ar+15 ar+11 ar+17

NCERT Exemplar · Q14

a_1, a_2, a_3, \ldots, a_r

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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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Writing $a_n=a\,k^{\,n-1}$ and factoring the row powers leaves rows $1$ and $2$ both equal to $(1,\,k^4,\,k^8)$ ; two equal rows make the determinant $0$ , which carries no $r$ , so it is independent of $r$ .

Intuition

In a G.P. every term is $a\,k^{\,n-1}$ . Inside the determinant that means each row is a power of $k$ times a fixed pattern. Once you pull those powers out, you can read off whether any two rows coincide — and here two of them do, which is what makes the value $0$ regardless of $r$ .

Setting up

Let the first term be $a$ and the common ratio $k$ , so $a_n=a\,k^{\,n-1}$ :

\Delta = \begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}.

Working the steps

1. Replace each entry. Using $a_{r+t}=a\,k^{\,r+t-1}$ :

\Delta = \begin{vmatrix} a k^{r} & a k^{r+4} & a k^{r+8} \\ a k^{r+6} & a k^{r+10} & a k^{r+14} \\ a k^{r+10} & a k^{r+16} & a k^{r+20} \end{vmatrix}.

2. Factor a common power from each row. Row 1 has $a k^{r}$ , row 2 has $a k^{r+6}$ , row 3 has $a k^{r+10}$ :

$\Delta = a^3 k^{3r+16}\begin{vmatrix} 1 & k^4 & k^8 \\ 1 & k^4 & k^8 \\ 1 & k^6 & k^{10} \end{vmatrix}.$ …

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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 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-2025-04294 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-2024-06074 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 2021Set eng-2021-P2-B14 marksMCQ
$\begin{vmatrix}4 & 4^2 & 4^3\\ 3 & 3^2 & 3^3\\ 2 & 2^2 & 2^3\end{vmatrix}$
›Reveal solutionSolution
The determinant equals $-48$ .

Concept and Intuition

Each row $(k, k^2, k^3)$ has a common factor $k$ , which can be pulled out of the determinant. The remaining determinant is a small $3\times 3$ evaluation.

Step-by-Step Solution

Factor $4,3,2$ from rows 1,2,3: value $=4\cdot 3\cdot 2\,\begin{vmatrix}1 & 4 & 16\\ 1 & 3 & 9\\ 1 & 2 & 4\end{vmatrix}=24\,D$ .

Expand $D=1(3\cdot4-9\cdot2)-4(1\cdot4-9\cdot1)+16(1\cdot2-3\cdot1)$ . …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\begin{bmatrix} 1 & 4 & 8 \\ 1 & 9 & 27 \\ 1 & 16 & 64 \end{bmatrix}$
›Reveal solutionSolution
Expanding along the first row gives determinant 52.

Concept and Intuition

A direct cofactor expansion along the first row is quickest for a 3×3 determinant.

Step-by-Step Solution

det = 1(9·64 − 27·16) − 4(1·64 − 27·1) + 8(1·16 − 9·1).

= 1(576 − 432) − 4(64 − 27) + 8(16 − 9). …
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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-2026-04194 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-2026-04184 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 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 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 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 2022Set eng-2022-P2-B14 marksMCQ
$\begin{vmatrix} \sin\alpha & \cos(\alpha+\theta) & \cos\alpha \\ \sin\beta & \cos(\beta+\theta) & \cos\beta \\ \sin\gamma & \cos(\gamma+\theta) & \cos\gamma \end{vmatrix} =$
›Reveal solutionSolution
The determinant equals $0$ .

Concept and Intuition

A determinant is zero when one column is a linear combination of the others.

Step-by-Step Solution

Expand $\cos(\phi+\theta)=\cos\theta\cos\phi-\sin\theta\sin\phi$ for each row's middle entry.

So column 2 $=\cos\theta\,(\text{column of }\cos\phi)-\sin\theta\,(\text{column of }\sin\phi)$ .

Column of $\cos\phi$ is column 3, column of $\sin\phi$ is column 1.

Thus $C_2=\cos\theta\,C_3-\sin\theta\,C_1$ : columns are linearly dependent. …
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Q.If $a_1, a_2, a_3, \ldots, a_r$ are in G.P., then prove that the determinant $\begin{vmatrix} a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21} \end{vmatrix}$ is independent of $r$ .

Concept understanding — Determinant Evaluation Using Identities

Determinant Evaluation Using Identities

The geometric intuition

The three row (or column) operations

The strategy

Worked example

Intuition

Setting up

Working the steps

Method: Factoring Out G.P. Powers to Expose Equal (or Proportional) Rows

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Intuition

Setting up

Working the steps

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