The value of \( \begin{vmatrix}0 & xy^2 & xz^2\\x^2y & 0 & yz^2\\x^2z & zy^2 & 0\end{vmatrix} \) is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 2x^3y^3z^3 \)
Taking the common factors \( x^2 \) out of C1, \( y^2 \) out of C2 and \( z^2 \) out of C3:
\( \displaystyle \Delta=x^2y^2z^2\begin{vmatrix}0 & x & x\\y & 0 & y\\z & z & 0\end{vmatrix} \)
Now taking x out of R1, y out of R2 and z out of R3:
\( \displaystyle \Delta=x^3y^3z^3\begin{vmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 & 0\end{vmatrix} \)
Expanding along R1:
\( \displaystyle \begin{vmatrix}0 & 1 & 1\\1 & 0 & 1\\1 & 1 & 0\end{vmatrix}=0-1(0-1)+1(1-0) \)
\( \displaystyle =2 \)
\( \displaystyle \Delta=2x^3y^3z^3 \)
Why the other options are incorrect
Option (1) 0 — Incorrect. The diagonal is zero, but the matrix is not skew-symmetric (for example \( a_{12}=xy^2 \) while \( a_{21}=x^2y \)), so the determinant need not be 0.
Option (2) \( x^3y^3z^3 \) — Incorrect. Only one of the two non-zero terms of the expansion is counted.
Option (4) \( -2x^3y^3z^3 \) — Incorrect. The sign is wrong: both terms of the expansion are \( +x^3y^3z^3 \).