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 :\(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\) Applying C₃ → C₃ – C₂, we get, Δ = \(\begin{vmatrix}a+b & 2a+b & a \\[0.3em]2a+b & 3a+b & a \\[0.3em]4a+b & 5a+b& a\end{vmatrix}\) Applying C₂→C₂ – C₁ gives, Δ = \(\begin{vmatrix}a+b & a & a \\[0.3em]2a+b & a & a \\[0.3em]4a+b & a& a\end{vmatrix}\) As, C₂ = C_3, So the value of the determinant is zero.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\)

Applying C₃ → C₃ – C₂, we get,

Δ = \(\begin{vmatrix}a+b & 2a+b & a \\[0.3em]2a+b & 3a+b & a \\[0.3em]4a+b & 5a+b& a\end{vmatrix}\)

Applying C₂→C₂ – C₁ gives,

Δ = \(\begin{vmatrix}a+b & a & a \\[0.3em]2a+b & a & a \\[0.3em]4a+b & a& a\end{vmatrix}\)

As,

C₂ = C_3,

So the value of the determinant is zero.