Question

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

• A. $0$
• B. $1+a^2+b^2$
• C. ${(1+a^2+b^2)}^2$
• D. ${(1+a^2+b^2)}^3$

Answer 1

The correct option is D ${(1+a^2+b^2)}^3$

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

Apply $C_1\to C_1-2C_3$ and $C_2\to a{C}_3+C_2$

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

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

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

$=(1+a^2+b^2)\left\{\right(1-a^2-b^2+2a^2)+2b^2$

$=(1+a^2+b^2)\left(\right(1+a^2+b^2\left)\right)={(1+a^2+b^2)}^3$