If A = [ * 20 c 1& /2 - /2 &1 ] and AB = I, then B = - Vidyadip

MCQ

If $A = \left[ {\begin{array}{*{20}{c}}1&{\tan \theta /2}\\{ - \tan \theta /2}&1\end{array}} \right]$ and $AB = I$, then $B = $

A
${\cos ^2}\frac{\theta }{2}.A$
✓
${\cos ^2}\frac{\theta }{2}.{A^T}$
C
${\cos ^2}\frac{\theta }{2}.I$
D
None of these

✓

Answer

Correct option: B.

${\cos ^2}\frac{\theta }{2}.{A^T}$

b
(b) $|A| = 1 + {\tan ^2}\frac{\theta }{2} = {\sec ^2}\frac{\theta }{2}$

$AB = I$ $ \Rightarrow $ $B = I{A^{ - 1}}$

$\frac{{\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]\,\,\left[ {\begin{array}{*{20}{c}}1&{ - \tan \frac{\theta }{2}}\\{\tan \frac{\theta }{2}}&1\end{array}} \right]}}{{{{\sec }^2}\frac{\theta }{2}}}$= ${\cos ^2}\frac{\theta }{2}\,.\,{A^T}$.

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