Question

If $\left|\begin{array}{ccc}{a}^2+{\lambda }^2& ab+c\lambda & ca-b\lambda \\ ab-c\lambda & b^2+{\lambda }^2& bc+a\lambda \\ ca+b\lambda & bc-a\lambda & c^2+{\lambda }^2\end{array}\right|\left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|={(1+a^2+b^2+c^2)}^3$, then the value of $\lambda$ is

• A. 8
• B. 27
• C. 1
• D. -1

Answer 1

The correct option is C 1

Let $D=\left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|$

determinant of cofactors is

$D^c=\left|\begin{array}{ccc}{a}^2+{\lambda }^2& ab+c\lambda & ca-b\lambda \\ ab-c\lambda & b^2+{\lambda }^2& bc+a\lambda \\ ca+b\lambda & bc-a\lambda & c^2+{\lambda }^2\end{array}\right|=D^2$

$\left|\begin{array}{ccc}{a}^2+{\lambda }^2& ab+c\lambda & ca-b\lambda \\ ab-c\lambda & b^2+{\lambda }^2& bc+a\lambda \\ ca+b\lambda & bc-a\lambda & c^2+{\lambda }^2\end{array}\right|\left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|={(1+a^2+b^2+c^2)}^3$

$\Rightarrow D^3={(1+a^2+b^2+c^2)}^3$ -------(1)
Now,
$D=\left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|$

$=\lambda ({\lambda }^2+a^2)-c(-\lambda c-ab)-b(ac-b\lambda )$
$=\lambda ({\lambda }^2+a^2+b^2+c^2)$
from (1)
${\left(\lambda \right({\lambda }^2+a^2+b^2+c^2\left)\right)}^3={(1+a^2+b^2+c^2)}^3$
comparing on both sides gives
${\lambda }^3=1$ and ${\lambda }^2=1$
$\therefore \lambda =1$
Hence, option C.