Question

Without expanding, show that the value of the following determinant is zero:

\[\begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]

Answer 1

\[ ∆ = \begin{vmatrix}a + b & 2a + b & 3a + b \\ 2a + b & 3a + b & 4a + b \\ 4a + b & 5a + b & 6a + b\end{vmatrix}\]
\[ = \begin{vmatrix}a & a & a \\ 2a & 2a & 2a \\ 4a + b & 5a + b & 6a + b\end{vmatrix} \left[ \text{ Applying } R_1 \to R_2 - R_1  \text{ and }  R_2 \to R_3 - R_2 \right]\]
\[ = 2\begin{vmatrix}a & a & a \\ a & a & a \\ 4a + b & 5a + b & 6a + b\end{vmatrix} = 0\]