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 14 on this concept.

GUJCET 2022Set 081 markMCQ
$x,y,z$
›Reveal solutionSolution

Split the last column into 1 and x³; the determinant factors as (Vandermonde)(1+xyz).

Concept. $\begin{vmatrix}x&x^2&1+x^3\\y&y^2&1+y^3\\z&z^2&1+z^3\end{vmatrix}=\begin{vmatrix}x&x^2&1\\y&y^2&1\\z&z^2&1\end{vmatrix}+\begin{vmatrix}x&x^2&x^3\\y&y^2&y^3\\z&z^2&z^3\end{vmatrix}$ . …
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GUJCET 2020Set 071 markMCQ
$x, y \in R$
›Reveal solutionSolution

Column 1 − Column 2 $= 4\times$ (Column 3), so the determinant is $0$ ; $2y+6=0$ gives $y=-3$ .

Concept — spot the linear dependence. For any base $a$ , let $p=a^x,\ q=a^{-x}$ , so $pq = a^x a^{-x} = 1$ . Then
$(a^x+a^{-x})^2 - (a^x-a^{-x})^2 = 4\,a^x a^{-x} = 4$
This holds for every row (with bases $a,b,c$ ). So in the matrix,

$C_1 - C_2 = 4 = 4\,C_3$ …
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GUJCET 2020Set 071 markMCQ
$\triangle ABC$
›Reveal solutionSolution

Using $A+B+C=\pi$ the matrix becomes skew-symmetric; a $3\times3$ (odd-order) skew-symmetric determinant is always $0$ .

Concept — trig identities in a triangle. Since $A+B+C=\pi$ : $\sin(B+C)=\sin(\pi-A)=\sin A$ and $\tan(A+C)=\tan(\pi-B)=-\tan B$ .

Substituting, the matrix is …
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GUJCET 2025Set 031 markMCQ
$\begin{vmatrix} \cos^2\theta & -\sin^2\theta \\ \sin^2\theta & \cos^2\theta \end{vmatrix} =$
›Reveal solutionSolution

Expand the 2×2 determinant, then use double/quadruple-angle identities.

$\begin{vmatrix}\cos^2\theta&-\sin^2\theta\\\sin^2\theta&\cos^2\theta\end{vmatrix}=\cos^4\theta+\sin^4\theta=1-2\sin^2\theta\cos^2\theta=1-\tfrac12\sin^2 2\theta.$
Using $\sin^2 2\theta=\tfrac{1-\cos4\theta}{2}$ : …
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GUJCET 2019Set 171 markMCQ
$\begin{vmatrix} \sin^2\theta & \cos^2\theta \\ -\cos^2\theta & \sin^2\theta \end{vmatrix} = \underline{\hspace{2cm}}$
›Reveal solutionSolution

$\begin{vmatrix}\sin^2\theta&\cos^2\theta\\-\cos^2\theta&\sin^2\theta\end{vmatrix}=\sin^4\theta+\cos^4\theta$ , which equals $\tfrac12(1+\cos^2 2\theta)$ .

Concept: Expand: $\sin^2\theta\cdot\sin^2\theta-\cos^2\theta\cdot(-\cos^2\theta)=\sin^4\theta+\cos^4\theta$ .

Now $\sin^4\theta+\cos^4\theta=1-2\sin^2\theta\cos^2\theta=1-\tfrac12\sin^2 2\theta$ . Using $\sin^2 2\theta=1-\cos^2 2\theta$ : …
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GUJCET 2024Set 131 markMCQ
$\begin{vmatrix} 2017 & 2018 \\ 2019 & 2020 \end{vmatrix} + \begin{vmatrix} 2021 & 2022 \\ 2023 & 2024 \end{vmatrix} = 2k$
›Reveal solutionSolution

Both determinants evaluate to $-2$ ; their sum $-4=2k$ gives $k=-2$ and $k^3=-8$ .

Concept. Evaluate each $2\times2$ determinant $ad-bc$ .

Steps.

$\begin{vmatrix}2017&2018\\2019&2020\end{vmatrix}=2017\cdot2020-2018\cdot2019=-2,$ …
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GUJCET 2020Set 071 markMCQ
$f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\tan t & t & 2t \\ \tan t & t & t \end{vmatrix}$
›Reveal solutionSolution

The determinant equals $t(-t\cos t+\tan t)$ , so $f(t)/t^2 = -\cos t + \tfrac{\tan t}{t}\to -1+1 = 0$ .

Concept — simplify the determinant first. Column 2 is $t[1,1,1]^T$ , so pull out $t$ :
$f(t) = t\,\begin{vmatrix}\cos t & 1 & 1\\ 2\tan t & 1 & 2t\\ \tan t & 1 & t\end{vmatrix} = t\, g(t)$
Expand $g(t)$ along the first column's cofactors (about row 1):
$g(t) = \cos t\,(t-2t) - 1\,(2t\tan t - 2t\tan t) + 1\,(2\tan t-\tan t)$ $= -t\cos t + 0 + \tan t$
Therefore …
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GUJCET 2021Set 151 markMCQ
$\begin{vmatrix} 2 & 3 & 5 \\ 1 & 0 & 7 \\ -1 & -2 & 4 \end{vmatrix}$
›Reveal solutionSolution

Element $7$ sits at position $(2,3)$ ; cofactor $=(-1)^{2+3}\times$ minor.

Concept: Deleting row 2 and column 3:

$M_{23}=\begin{vmatrix}2&3\\-1&-2\end{vmatrix}=2(-2)-3(-1)=-1.$ …
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GUJCET 2019Set 171 markMCQ
$\begin{vmatrix} 1! & 2! & 3! \\ 2! & 3! & 4! \\ 3! & 4! & 5! \end{vmatrix} = 2016K$
›Reveal solutionSolution

Evaluating the determinant of factorials gives $24$ ; with $24=2016K$ , $K=\dfrac{1}{84}$ .

Concept: Write out the values: $1!=1,2!=2,3!=6,4!=24,5!=120$ . …
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GUJCET 2023Set 091 markMCQ
$\begin{vmatrix} \sin\frac{11\pi}{36} & \cos\frac{11\pi}{36} \\ \sin\frac{2\pi}{9} & \cos\frac{2\pi}{9} \end{vmatrix} =$
›Reveal solutionSolution

A determinant of this sin/cos form collapses to a single sine of the angle difference.

Concept. $\begin{vmatrix}\sin P&\cos P\\\sin Q&\cos Q\end{vmatrix}=\sin P\cos Q-\cos P\sin Q=\sin(P-Q)$ .

Solution. $P=\tfrac{11\pi}{36}$ , $Q=\tfrac{2\pi}{9}=\tfrac{8\pi}{36}$ .

$\sin\!\left(\tfrac{11\pi}{36}-\tfrac{8\pi}{36}\right)=\sin\tfrac{3\pi}{36}=\sin\tfrac{\pi}{12}.$ …
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GUJCET 2019Set 171 markMCQ
$A_r = \begin{bmatrix} r & r-1 \\ r-1 & r \end{bmatrix}$
›Reveal solutionSolution

$|A_r|=r^2-(r-1)^2=2r-1$ ; the sum of the first 100 odd numbers is $100^2=10000=(\sqrt{10})^8$ , so $K=8$ .

Concept: $|A_r|=\begin{vmatrix}r&r-1\r-1&r\end{vmatrix}=r^2-(r-1)^2=2r-1$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$f(\theta) = \begin{vmatrix} \cos\theta & -\sin\theta \\ \sin\theta & -\cos\theta \end{vmatrix}$
(a) $-\frac{1}{2}$
(b) $\frac{1}{2}$
(c) $\frac{\sqrt{3}}{2}$
(d) $-\frac{\sqrt{3}}{2}$
›Reveal solutionSolution

Evaluate the $2\times2$ determinant, simplify with a double-angle identity, then substitute.

$f(\theta)=\cos\theta(-\cos\theta)-(-\sin\theta)(\sin\theta)=-\cos^2\theta+\sin^2\theta=-\cos2\theta$ .

…

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

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

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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