Question

Using properties of determinants, prove the following

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

Answer 1

LHS :

$\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|$

Applying $C_1\to C_1-b{C}_3,C_2\to C_2+a{C}_3$, we get,

$=\left|\begin{array}{ccc}1+a^2+b^2& 0& -2b\\ 0& 1+a^2+b^2& 2a\\ b(1+a^2+b^2)& -a(1+a^2+b^2)& 1-a^2-b^2\end{array}\right|$

Taking out $(1+a^2+b^2)$ common from $C_1$ and $C_2$, we get,

$=(1+a^2+b^2{)}^2\left|\begin{array}{ccc}1& 0& -2b\\ 0& 1& 2a\\ b& -a& 1-a^2-b^2\end{array}\right|$

Applying $R_3\to R_3-b{R}_1+a{R}_2$, we get,

$=(1+a^2+b^2{)}^2\left|\begin{array}{ccc}1& 0& -2b\\ 0& 1& 2a\\ 0& 0& 1+a^2+b^2\end{array}\right|$

Expanding along $C_1$, we get,

$=(1+a^2+b^2{)}^2(1+a^2+b^2)=(1+a^2+b^2{)}^3$ = RHS