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 :\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)
Answer» Let Δ =\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\) Multiplying C₁, C₂ and C₃ with z, y and x respectively we get, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& xy & yx \\[0.3em]-xz & 0 & zx \\[0.3em]-yz & -zy &0\end{vmatrix}\) Now, Taking y, x and z common from R₁,R₂ and R₃ gives, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & 0 & z \\[0.3em]-y & -y &0\end{vmatrix}\) Applying C₂ → C₂ – C₃ gives, ⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & -z & z \\[0.3em]-y & -y &0\end{vmatrix}\) As, C₁ = C₂, Therefore determinant is zero.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)

Answer»

Let Δ =\(\begin{vmatrix}0& x & y \\[0.3em]-x & 0 & z \\[0.3em]-y & -z &0\end{vmatrix}\)

Multiplying C₁, C₂ and C₃ with z, y and x respectively we get,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& xy & yx \\[0.3em]-xz & 0 & zx \\[0.3em]-yz & -zy &0\end{vmatrix}\)

Now,

Taking y, x and z common from R₁,R₂ and R₃ gives,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & 0 & z \\[0.3em]-y & -y &0\end{vmatrix}\)

Applying C₂ → C₂ – C₃ gives,

⇒ Δ = \((\frac{1}{xyz})\)\(\begin{vmatrix}0& x & x \\[0.3em]-z & -z & z \\[0.3em]-y & -y &0\end{vmatrix}\)

As,

C₁ = C₂,

Therefore determinant is zero.