Target Exam

CUET

Subject

-- Applied Mathematics - Section B2

Chapter

Algebra

Question:

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

Options:

$a_1a_2a_3+b_1b_2b_3$

$(a_1a_2a_3)(b_1b_2b_3)$

$a_1a_2b_1b_2+a_2a_3b_2b_3+a_3a_1b_3b_1$

none of these

Correct Answer:

none of these

Explanation:

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

$Δ=\begin{vmatrix}1 & a_1b_2 & a_1b_3\\1 & a_2b_2 & a_2b_3\\1 & a_3b_2 & a_3b_3\end{vmatrix}+\begin{vmatrix}a_1b_1 & 1 & 1\\a_2b_1 & 1 & 1\\a_3b_1 &1 & 1\end{vmatrix}$

$⇒Δ=b_1b_2\begin{vmatrix}1 & a_1 & a_1\\1 & a_2 & a_2\\1 & a_3 & a_3\end{vmatrix}+0$

$⇒Δ= 0+0=0$