JEE MAIN - Mathematics Hindi (2021 - 25th February Morning Shift - No. 17)
यदि $$A = \left[ {\matrix{
0 & { - \tan \left( {{\theta \over 2}} \right)} \cr
{\tan \left( {{\theta \over 2}} \right)} & 0 \cr
} } \right]$$ और
$$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$$, तो $$13({a^2} + {b^2})$$ के बराबर है
$$({I_2} + A){({I_2} - A)^{ - 1}} = \left[ {\matrix{ a & { - b} \cr b & a \cr } } \right]$$, तो $$13({a^2} + {b^2})$$ के बराबर है
Answer
13
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