Question

If $\left|\begin{array}{c}{-a}^2\\ ab\\ ac\end{array}\begin{array}{c}ab\\ {-b}^2\\ bc\end{array}\begin{array}{c}ac\\ bc\\ {-c}^2\end{array}\right|=\lambda a^2{b}^2{c}^2$, then the value of $\lambda$ is

• A. $1$
• B. $2$
• C. $4$
• D. $3$

Answer 1

Answer: (C)

$4$

Consider the L.H.S as shown below:

Let $\Delta =\left|\begin{array}{ccc}-a^2& ab& ac\\ ab& -b^2& bc\\ ac& bc& -c^2\end{array}\right|$

Expand the above determinant along $R_1$ as shown below:

$\Delta =-a^2[(-b^2)(-c^2)-\left(bc\right)\left(bc\right)]-ab[\left(ab\right)(-c^2)-\left(ac\right)\left(bc\right)]+ac[\left(ab\right)\left(bc\right)-\left(ac\right)(-b^2)]$

$\Delta =-a^2[b^2{c}^2-b^2{c}^2]-ab[-c^{2}ab-c^{2}ab]+ac[b^{2}ac+b^{2}ac]$

$\Delta =0-ab[-2c^{2}ab]+ac\left[2b^{2}ac\right]$

$\Delta =2a^2{b}^2{c}^2+2a^2{b}^2{c}^2$

$\Delta =4a^2{b}^2{c}^2$

Now, compare L.H.S and R.H.S to get the value of $\lambda$ as shown below:

$4a^2{b}^2{c}^2=\lambda a^2{b}^2{c}^2$

$\Rightarrow$$\lambda =4$

Hence option $C$ is correct.