Question

If $\Delta = \left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|$ and $A_{1},B_{1},C_{1}$ denote the cofactors of $a_{1},b_{1},c_{1}$ respectively, then the value of the determinant $\left| \begin{matrix} A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \\ A_{3} & B_{3} & C_{3} \end{matrix} \right|$ is

• A. $\Delta$
• B. $\Delta^{2}$
• C. $\Delta^{3}$
• D. $0$

Answer 1

Answer: (B)

$\Delta^{2}$

Answer 2

We know that $\Delta.\Delta' = \left| \begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix} \right|$.$\left| \begin{matrix} A_{1} & B_{1} & C_{1} \\ A_{2} & B_{2} & C_{2} \\ A_{3} & B_{3} & C_{3} \end{matrix} \right|$

\Sigma a_{1}A_{1} & 0 & 0 \\ 0 & \Sigma a_{2}A_{2} & 0 \\ 0 & 0 & \Sigma a_{3}A_{3} \end{matrix} \right| = \left| \begin{matrix} \Delta & 0 & 0 \\ 0 & \Delta & 0 \\ 0 & 0 & \Delta \end{matrix} \right| = \Delta^{3} \Rightarrow \Delta' = \Delta^{2}$$ Trick : According to property of cofactors $\Delta' = \Delta^{n - 1} = \Delta^{2}$ $$(\because\text{Hence}n = 3)$$