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Evaluate the following :\(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\) |
Answer» Let Δ = \(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\) = 0(0 – y3z3) – xy2(0 – x2yz3) + xz2(x2y3z – 0) = 0 + x3y3z3 + x3y3z3 = 2x3y3z3 So, Δ = 2x3y3z3 |
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