Question

If $A=\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]$ is orthogonal, then $|abc|$ is equal to

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{6}$
• D. $1$

Answer 1

The correct option is C $\frac{1}{6}$

$A=\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]$

$A$ is orthogonal

$\therefore A{A}^T=I$


$\left[\begin{array}{ccc}0& 2b& c\\ a& b& -c\\ a& -b& c\end{array}\right]\left[\begin{array}{ccc}0& a& a\\ 2b& b& -b\\ c& -c& c\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$\left[\begin{array}{ccc}4b^2+c^2& 2b^2-c^2& -2b^2+c^2\\ 2b^2-c^2& a^2+b^2+c^2& a^2-b^2-c^2\\ -2b^2+c^2& a^2-b^2-c^2& a^2+b^2+c^2\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]$

$4b^2+c^2=1$ and $2b^2-c^2=0\Rightarrow c=\pm \frac{1}{\sqrt{3}},b=\pm \frac{1}{\sqrt{6}}$

$a^2-b^2-c^2=0$
$a^2=b^2+c^2=3b^2$
$a=\pm \frac{1}{\sqrt{2}}$

$\therefore$ $|abc|=\frac{1}{6}$