Question

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

• A. 1
• B. -1
• C. $\pm$1

Answer 1

The correct option is A 1

If $\Delta =\left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|$ other determinant (say ${\Delta }^1$) is the cofactor determinant
$\Delta {\Delta }^1={\Delta }^3$ (for 3rd order det)
$\Delta =\lambda ({\lambda }^2+a^2+b^2+c^2)$ by comparing
$\lambda =1$