Which one of the following does NOT equal 1 & x & x2 1 & y & y2 1 & z & z2 ? A) 1 & x(x+1) & x+1 1 & y(y+1) & y+1 1 & z(z+1) & z+1 B) 1 & x+1 & x2+1 1 & y+1 & y2+1 1 & z+1 & z2+1 C) 0 & x-y & x2-y2 0 & y-z & y2-z2 1 & z & z2 D) 2 & x+y & x2+y2 2 & y+z & y2+z2 1 & z & z2

Official answer: A) 1 & x(x+1) & x+1 1 & y(y+1) & y+1 1 & z(z+1) & z+1. Explanation: Let the given Vandermonde determinant be = 1 & x & x2 1 & y & y2 1 & z & z2 = (y-x)(z-x)(z-y). Let us analyze the determinant in Option A: A = 1 & x(x+1) & x+1 1 & y(y+1) & y+1 1 & z(z+1) & z+1 = 1 & x2+x & x+1 1 & y2+y & y+1 1 & z2+z & z+1 Perform the column operations: 1. C3 → C3 - C1 yields column 3 as x y z . 2. C2 → C2 - C3 (using the new column 3) yields column 2 as x2 y2 z2 . Now the matrix is: 1 & x2 & x 1 & y2 & y 1 & z2 & z Interchanging columns C2 and C3 introduces a negative sign: 1 & x2 & x 1 & y2 & y 1 & z2 & z = - 1 & x & x2 1 & y & y2 1 & z & z2 = - Thus, option A does NOT equal . Options B, C, and D are all equivalent to through standard elementary row/column operations that preserve the determinant.

B) 1 & x+1 & x2+1 1 & y+1 & y2+1 1 & z+1 & z2+1

C) 0 & x-y & x2-y2 0 & y-z & y2-z2 1 & z & z2

D) 2 & x+y & x2+y2 2 & y+z & y2+z2 1 & z & z2

Linear Algebra

GATE CSE 2013 · Q3

MCQ1 markMedium

Time00:00

Which one of the following does NOT equal $\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix}$ ?

Select one answerSingle correct

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Solution

Step-by-step verified solution

Official answer: A) 1 & x(x+1) & x+1 1 & y(y+1) & y+1 1 & z(z+1) & z+1

Linear Algebra

Let the given Vandermonde determinant be $\Delta = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (y-x)(z-x)(z-y)$ .

Let us analyze the determinant in Option A:

\Delta_A = \begin{vmatrix} 1 & x(x+1) & x+1 \\ 1 & y(y+1) & y+1 \\ 1 & z(z+1) & z+1 \end{vmatrix} = \begin{vmatrix} 1 & x^2+x & x+1 \\ 1 & y^2+y & y+1 \\ 1 & z^2+z & z+1 \end{vmatrix}

Perform the column operations:

$C_3 \to C_3 - C_1$ yields column 3 as $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ .
$C_2 \to C_2 - C_3$ (using the new column 3) yields column 2 as $\begin{pmatrix} x^2 \\ y^2 \\ z^2 \end{pmatrix}$ .

Now the matrix is:

\begin{vmatrix} 1 & x^2 & x \\ 1 & y^2 & y \\ 1 & z^2 & z \end{vmatrix}

Interchanging columns $C_2$ and $C_3$ introduces a negative sign:

\begin{vmatrix} 1 & x^2 & x \\ 1 & y^2 & y \\ 1 & z^2 & z \end{vmatrix} = -\begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = -\Delta

Thus, option A does NOT equal $\Delta$ .

Options B, C, and D are all equivalent to $\Delta$ through standard elementary row/column operations that preserve the determinant.

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This Linear Algebra previous year question is Q3 of the GATE CSE 2013 paper. Find more GATE CSE PYQs on Linear Algebra in Engineering Mathematics, or browse all GATE CSE PYQs.

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Independent practice explanation verified by the GateAI team. GATE is conducted by the IITs; question text follows the official paper.