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$\Delta=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$ and $A_1, B_1, C_1$ denote the co-factors of $a_1, b_1, c_1$ respectively, then the value of the determinant $\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|$ is
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Verified Answer
The correct answer is:
$\Delta^2$
We know that $\Delta \Delta^{\prime}=\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right| \cdot\left|\begin{array}{lll}A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3\end{array}\right|$
$\begin{aligned} & =\left|\begin{array}{ccc}\Sigma a_1 A_1 & 0 & 0 \\ 0 & \Sigma a_2 A_2 & 0 \\ 0 & 0 & \Sigma a_3 A_3\end{array}\right|=\left|\begin{array}{ccc}\Delta & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & \Delta\end{array}\right|=\Delta^3 \\ & \Rightarrow \Delta^{\prime}=\Delta^2 .\end{aligned}$
$\begin{aligned} & =\left|\begin{array}{ccc}\Sigma a_1 A_1 & 0 & 0 \\ 0 & \Sigma a_2 A_2 & 0 \\ 0 & 0 & \Sigma a_3 A_3\end{array}\right|=\left|\begin{array}{ccc}\Delta & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & \Delta\end{array}\right|=\Delta^3 \\ & \Rightarrow \Delta^{\prime}=\Delta^2 .\end{aligned}$
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