GATE CS 2013 Q.3 (CS Q.3): Linear Algebra

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

$\begin{vmatrix} 1 & x(x+1) & x+1 \\ 1 & y(y+1) & y+1 \\ 1 & z(z+1) & z+1 \end{vmatrix}$

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

$\begin{vmatrix} 0 & x-y & x^2-y^2 \\ 0 & y-z & y^2-z^2 \\ 1 & z & z^2 \end{vmatrix}$

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

Official answer and solution

Solution

Apply elementary row and column operations to determine which determinant is not identical to the given Vandermonde determinant.

The reference determinant is the standard $3 \times 3$ Vandermonde determinant:

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

Let us evaluate each option:

In Option B: Subtracting 1 from columns 2 and 3 corresponds to column operations $C_2 \leftarrow C_2 - C_1$ and $C_3 \leftarrow C_3 - C_1$ , which preserve the determinant. Thus it equals $D$ .
In Option C: Applying $R_1 \leftarrow R_1 - R_2$ and $R_2 \leftarrow R_2 - R_3$ produces the exact matrix in C. Elementary row operations do not change the determinant, so it equals $D$ .
In Option D: Row 1 is $R_1 + R_2$ , Row 2 is $R_2 + R_3$ , Row 3 is $R_3$ . Row reduction shows its determinant is also equal to $D$ .
In Option A: Columns are $[1, x(x+1), x+1]^T$ . Swapping columns or evaluating with $x = 0, y = 1, z = 2$ gives $D = 2$ , whereas the determinant in A evaluates to $-2 \neq 2$ .

Hence, matrix A does NOT equal $D$ .

Final answer

Option (A)

Common trap

Overlooking the column swap and sign change when comparing Option A.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.