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 :\(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\) Applying, C₂→C₂ + C₁ and C₃→C₃ + C₁ ⇒ Δ = \(\begin{vmatrix}a & b & c \\[0.3em]2a+2x& 2b+2y & 2c+2z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\) Taking 2 common from R₂ we get, ⇒ Δ = 2\(\begin{vmatrix}a & b & c \\[0.3em]a+x& b+y & c+z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\) As, R₂ = R₃, hence value of determinant is zero.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)

Applying, C₂→C₂ + C₁ and C₃→C₃ + C₁

⇒ Δ = \(\begin{vmatrix}a & b & c \\[0.3em]2a+2x& 2b+2y & 2c+2z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\)

Taking 2 common from R₂ we get,

⇒ Δ = 2\(\begin{vmatrix}a & b & c \\[0.3em]a+x& b+y & c+z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\)

As,

R₂ = R₃, hence value of determinant is zero.