Without expanding, show that the value of each of the following determ

1.	Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\) Multiplying R₁, R₂ and R₃ with a, b and c respectively we get, ⇒ Δ = \(\begin{vmatrix} 1& a^3 & abc \\[0.3em] 1& b^3 &abc \\[0.3em] 1 &c^3& abc\end{vmatrix}\) Taking, a b c common from C₃ gives, ⇒ Δ = \(\begin{vmatrix} 1& a^3 & 1 \\[0.3em] 1& b^3 &1 \\[0.3em] 1 &c^3& 1\end{vmatrix}\) As, C₁ = C₃ hence value of determinant is zero.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\)

Multiplying R₁, R₂ and R₃ with a, b and c respectively we get,

⇒ Δ = \(\begin{vmatrix} 1& a^3 & abc \\[0.3em] 1& b^3 &abc \\[0.3em] 1 &c^3& abc\end{vmatrix}\)

Taking, a b c common from C₃ gives,

⇒ Δ = \(\begin{vmatrix} 1& a^3 & 1 \\[0.3em] 1& b^3 &1 \\[0.3em] 1 &c^3& 1\end{vmatrix}\)

As,

C₁ = C₃ hence value of determinant is zero.