Using the properties of determinants, prove that: $\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \end{vmatrix} = 0$

Using the properties of determinants, prove that: y² z² yz y + z z² x² zx z + x x² y² xy x + y = 0

NCERT Exemplar · Q7

\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \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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Multiply the rows by $x,y,z$ and factor $xyz$ from the first two columns; then $C_3 \to C_3+C_1$ makes the third column a multiple of the second, so the determinant is $0$ .

Intuition

A determinant is zero exactly when its columns are linearly dependent. Here the first two columns look unrelated to the third — until you scale the rows by $x,y,z$ . That scaling turns the entries into a symmetric pattern in which one column operation exposes two proportional columns.

Setting up

\Delta = \begin{vmatrix} y^2z^2 & yz & y+z \\ z^2x^2 & zx & z+x \\ x^2y^2 & xy & x+y \end{vmatrix}.

Working the steps

1. Scale the rows. Multiply $R_1,R_2,R_3$ by $x,y,z$ respectively. This multiplies the whole determinant by $xyz$ :

xyz\,\Delta = \begin{vmatrix} xy^2z^2 & xyz & x(y+z) \\ x^2yz^2 & xyz & y(z+x) \\ x^2y^2z & xyz & z(x+y) \end{vmatrix}.

2. Factor the columns. Column 1 is $xyz\,(yz,\,zx,\,xy)$ and column 2 is $xyz\,(1,1,1)$ :

xyz\,\Delta = x^2y^2z^2\begin{vmatrix} yz & 1 & x(y+z) \\ zx & 1 & y(z+x) \\ xy & 1 & z(x+y) \end{vmatrix}.

3. One column operation. Apply $C_3 \to C_3+C_1$ :

x(y+z)+yz = y(z+x)+zx = z(x+y)+xy = xy+yz+zx.

So every entry of the new third column is $xy+yz+zx$ , i.e. $C_3=(xy+yz+zx)\,C_2$ : …

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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 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 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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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 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 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 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 2021Set eng-2021-P2-B14 marksMCQ
$\begin{vmatrix} x-1 & 2 & 1 \\ 2 & x-1 & 2 \\ 1 & x+2 & x-1 \end{vmatrix} = ax^3 + bx^2 + cx + d$
›Reveal solutionSolution
Setting $x=0$ gives the constant term $d=16$ .

Concept and Intuition

Writing the determinant as $ax^3+bx^2+cx+d$ , the constant $d$ equals the value at $x=0$ , since all $x$ -bearing terms vanish there.

Step-by-Step Solution

At $x=0$ the matrix is $\begin{bmatrix}-1 & 2 & 1\\ 2 & -1 & 2\\ 1 & 2 & -1\end{bmatrix}$ .

Expand: $-1((-1)(-1)-2\cdot2)-2(2\cdot(-1)-2\cdot1)+1(2\cdot2-(-1)\cdot1)$ . …
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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-2026-04184 marksMCQ
$f(x)=\begin{vmatrix}-1 & x & 3\\ 0 & 1 & 2x\\ 1 & -1 & 1\end{vmatrix}$
›Reveal solutionSolution

At $x=-1$ the determinant evaluates to $0$ .

At $x=-1$ the matrix is $\begin{pmatrix}-1&-1&3\\0&1&-2\\1&-1&1\end{pmatrix}$ . Expanding along the first column: …
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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 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 2024Set eng-2024-06094 marksMCQ
$A=(a_{ij})$
›Reveal solutionSolution

Convert minors to cofactors with alternating signs, then expand: $|A|=-80$ .

Cofactors: $C_{11}=+M_{11}=-40$ , $C_{12}=-M_{12}=10$ , $C_{13}=+M_{13}=35$ .

Expanding along the first row: …
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←Previous questionQ6 — Using the properties of determinants, evaluate: a - b - c 2a 2a \2b b …→Next questionQ8 — Using the properties of determinants, prove that: y + z z y z z + x x …

Q.Using the properties of determinants, prove that: $\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \end{vmatrix} = 0$

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: Proving a Determinant is Zero via Linear Dependence

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Q.Using the properties of determinants, prove that: ∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣=0\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \end{vmatrix} = 0​y2z2z2x2x2y2​yzzxxy​y+zz+xx+y​​=0

Intuition

Setting up

Working the steps

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Q.Using the properties of determinants, prove that: $\begin{vmatrix} y^2 z^2 & yz & y + z \\ z^2 x^2 & zx & z + x \\ x^2 y^2 & xy & x + y \end{vmatrix} = 0$

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