JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 17)
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
$$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$$, then $$13({a^2} + {b^2})$$ is equal to
Answer
13
Explanation
$$A = \left[ {\matrix{
0 & { - \tan {\theta \over 2}} \cr
{\tan {\theta \over 2}} & 0 \cr
} } \right]$$
$$ \Rightarrow I + A = \left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ \Rightarrow I - A = \left[ {\matrix{ 1 & {\tan {\theta \over 2}} \cr { - \tan {\theta \over 2}} & 1 \cr } } \right]$$ { $$\therefore$$ $$\left| {I - A} \right| = {\sec ^2}\theta /2$$}
$$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ \Rightarrow (1 + A){(I - A)^{ - 1}} $$
$$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ {1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr {2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr } } \right]$$
$$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$
$$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$
$$\therefore$$ $${a^2} + {b^2} = 1$$
$$ \Rightarrow $$ $$13({a^2} + {b^2})$$ = 13
$$ \Rightarrow I + A = \left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ \Rightarrow I - A = \left[ {\matrix{ 1 & {\tan {\theta \over 2}} \cr { - \tan {\theta \over 2}} & 1 \cr } } \right]$$ { $$\therefore$$ $$\left| {I - A} \right| = {\sec ^2}\theta /2$$}
$$ \Rightarrow {(I - A)^{ - 1}} = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ \Rightarrow (1 + A){(I - A)^{ - 1}} $$
$$= {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]\left[ {\matrix{ 1 & { - \tan {\theta \over 2}} \cr {\tan {\theta \over 2}} & 1 \cr } } \right]$$
$$ = {1 \over {{{\sec }^2}{\theta \over 2}}}\left[ {\matrix{ {1 - {{\tan }^2}{\theta \over 2}} & { - 2\tan {\theta \over 2}} \cr {2\tan {\theta \over 2}} & {1 - {{\tan }^2}{\theta \over 2}} \cr } } \right]$$
$$a = {{1 - {{\tan }^2}{\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$
$$b = {{2\tan {\theta \over 2}} \over {{{\sec }^2}{\theta \over 2}}}$$
$$\therefore$$ $${a^2} + {b^2} = 1$$
$$ \Rightarrow $$ $$13({a^2} + {b^2})$$ = 13
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