Question

Using properties of determinant, 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

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

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

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

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

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

$(1+a^2+b^2)\left|\begin{array}{ccc}1& 0& -b\\ 0& 1+a^2+b^2& a+a^3+a{b}^2\\ 2b& -2a& 1-a^2-b^2\end{array}\right|$

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

Solving the above matrix, we get,

$(1+a^2+b^2{)}^2\cdot (1+a^2+b^2)$

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