JEE Main 2021 (Online) 25th February Morning ShiftQuestion 1
If $A=\begin{bmatrix}0&-\tan\left(\frac{\theta}{2}\right)\\ \tan\left(\frac{\theta}{2}\right)&0\end{bmatrix}$ and $(I_2+A)(I_2-A)^{-1}=\begin{bmatrix}a&-b\\ b&a\end{bmatrix}$, then $13(a^{2}+b^{2})$ is equal to $\_\_\_\_\_\_$ .
Matrices
Answer$13$
Solution
$(I+A)(I-A)^{-1}$ has determinant $\dfrac{\det(I+A)}{\det(I-A)}=\dfrac{1+t^2}{1+t^2}=1$ with $t=\tan\tfrac\theta2$. And the matrix $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ has determinant $a^2+b^2$. So $a^2+b^2=1$ and $13(a^2+b^2)=13$.