Question

$a^2{b}^2=1$ . Let $S$ be the least value of $\Delta$.
$\Delta =\left|\begin{array}{ccc}1+a^2-b^2& 2ab& -2b\\ 2ab& 1-a^2+b^2& 2a\\ 2b& -2a& 1-a^2-b^2\end{array}\right|$
Find $\frac{4S}{3}$

Answer 1

Given: $a^2{b}^2=1$

To find: $\frac{S}{3}$

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

Applying $R_3\to b{R}_3,R_1\to R_1+b{R}_3$

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

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

Applying $R_3\to a{R}_3,R_2\to R_2+a{R}_3$

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

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

Now apply $R_1\to 2b{R}_1,R_3\to R_3-2b{R}_1$

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

$R_2\to 2a{R}_2,R_3\to R_3+2a{R}_2$

$\Delta =\frac{{(1+a^2+b^2)}^2}{4a^2{b}^2}\left|\begin{array}{ccc}1& 0& -b\\ 0& 1& a\\ 0& 0& 1-a^2+b^2\end{array}\right|$

$\Delta =\frac{{(1+a^2+b^2)}^3}{4a^2{b}^2},a^2{b}^2=1$

$\Delta =\frac{{(1+a^2+b^2)}^3}{4}$

Applying AP, GP comparison

$AP\ge GP$

$\frac{a^2+b^2}{2}\ge \sqrt{a^2{b}^2}$

$\Rightarrow \frac{a^2+b^2}{2}\ge 1$

$\Rightarrow a^2+b^2\ge 2$

Least value of $a^2+b^2=2$

$S=\frac{{(1+2)}^3}{4}=\frac{3^3}{4}$

$\frac{4S}{3}=4\times \frac{9}{4}=9$