The value of the determinant $Δ=\begin{vmatrix}1+a_1b_1 & 1+a_1b_2 & 1+a_1b_3\\1+a_2b_1 & 1+a_2b_2 & 1+a_2b_3\\1+a_3b_1 & 1+a_3b_2 & 1+a_3b_3\end{vmatrix},$ is |
$a_1a_2a_3+b_1b_2b_3$ $(a_1a_2a_3)(b_1b_2b_3)$ $a_1a_2b_1b_2+a_2a_3b_2b_3+a_3a_1b_3b_1$ none of these |
none of these |
The correct answer is option (4) : none of these We have, $Δ=\begin{vmatrix}1 & 1+a_1b_2 & 1+a_1b_3\\1 & 1+a_2b_2 & 1+a_2b_3\\1 & 1+a_3b_2 & 1+a_3b_3\end{vmatrix}+\begin{vmatrix}a_1b_1 & 1+a_1b_2 & 1+a_1b_3\\a_2b_1 & 1+a_2b_2 & 1+a_2b_3\\a_3b_1 & 1+a_3b_2 & 1+a_3b_3\end{vmatrix}$ Applying $C_2→C_2-C_1, C_3 →C_3 -C_1$ in first determinant and $C_2 → C_2 - \left(\frac{b_2}{b_1}\right) C_1, C_3 → C_3-\frac{b_3}{b_1}C_1 $ in second determinant, we get $Δ=\begin{vmatrix}1 & a_1b_2 & a_1b_3\\1 & a_2b_2 & a_2b_3\\1 & a_3b_2 & a_3b_3\end{vmatrix}+\begin{vmatrix}a_1b_1 & 1 & 1\\a_2b_1 & 1 & 1\\a_3b_1 &1 & 1\end{vmatrix}$ $⇒Δ=b_1b_2\begin{vmatrix}1 & a_1 & a_1\\1 & a_2 & a_2\\1 & a_3 & a_3\end{vmatrix}+0$ $⇒Δ= 0+0=0$ |