Question

Find the value of
$\left|\begin{array}{ccc}{a}^2+{\lambda }^2& ab+c\lambda & ca-b\lambda \\ ab-c\lambda & b^2-{\lambda }^2& bc+a\lambda \\ ca+b\lambda & bc-a\lambda & c^2+{\lambda }^2\end{array}\right|\times \left|\begin{array}{ccc}\lambda & c& -b\\ -c& \lambda & a\\ b& -a& \lambda \end{array}\right|$.

Answer 1

By multiplication, the constituents of the first column of the determinant-product are;-

$\lambda (a^2+{\lambda }^2)+c(ab+c\lambda )-b(ac-b\lambda ),$

$-c(a^2+{\lambda }^2)+\lambda (ab+c\lambda )+a(ca-b\lambda ),b(a^2+{\lambda }^2)-a(ab+c\lambda )+\lambda (ca-b\lambda )$

First expression$={\lambda }^3+\lambda (a^2+b^2+c^2)$

Second expression and third expression becomes $0$

Similarly the product consists of elements $0$ except in the leading diagonal where expression$=\lambda ({\lambda }^2+a^2+b^2+c^2)$

$=\left|\begin{array}{ccc}\lambda ({\lambda }^2+a^2+b^2+c^2)& 0& 0\\ 0& \lambda ({\lambda }^2+a^2+b^2+c^2)& 0\\ 0& 0& \lambda ({\lambda }^2+a^2+b^2+c^2)\end{array}\right|$

$={\lambda }^3{({\lambda }^2+a^2+b^2+c^2)}^3$