If $$A = \left[ {\matrix{ 0 & { - \tan \left( {{\theta \over 2}} \right)} \cr {\tan \left( {{\theta \over 2}} \right)} & 0 \cr } } \right]$ and $({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$, then $13({a^2} + {b^2})$$ is equal to

Question

If $$A = \left[ {\matrix{ 0 & { - \tan \left( {{\theta \over 2}} \right)} \cr {\tan \left( {{\theta \over 2}} \right)} & 0 \cr } } \right]$ and
$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$, then $13({a^2} + {b^2})$$ is equal to

Correct Answer

Solution

Please login to view the detailed solution steps...

Go to DASH