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
Answer (integer)
13
Solution
$$A = \left[ {\matrix{
0 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 0 \cr
} } \right]$$<br><br>$$ \Rightarrow I + A = \left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$$ \Rightarrow I - A = \left[ {\matrix{
1 & {\tan {\theta \over 2}} \cr
{ - \tan {\theta \over 2}} & 1 \cr
} } \right]$$ { $\therefore$ $\left| {I - A} \right| = {\sec ^2}\theta /2$}<br><br>$$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$\Rightarrow (1 + A){(I - A)^{ - 1}}$
<br><br>$$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]\left[ {\matrix{
1 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 1 \cr
} } \right]$$<br><br>$$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{
{1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr
{2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr
} } \right]$$<br><br>$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$<br><br>$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$<br><br>$\therefore$ ${a^2} + {b^2} = 1$
<br><br>$\Rightarrow$ $13({a^2} + {b^2})$ = 13
About this question
Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations
This question is part of PrepWiser's free JEE Main question bank. 274 more solved questions on Matrices and Determinants are available — start with the harder ones if your accuracy is >70%.