Question

Using properties of determinants, prove that:

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

Answer 1

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

Operating $R_1\to R_1+b{R}_3$,

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

Taking $(1+a^2+b^2)$ common from $R_1$,

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

Operating $R_2\to R_2-a{R}_3$,

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

Taking $(1+a^2+b^2)$ common from $R_2$,

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

Now, on expanding along $R_1$, we get

$=(1+a^2+b^2{)}^2[(1-a^2-b^2)+(2a^2+2b^2)]$

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

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