If $A =\begin{bmatrix}1&tanx\\-tanx&1\end{bmatrix}$, then the value of $|A^TA^{-1}|$ is
Answer & explanation
Correct answer: option 4
$A^T =\begin{bmatrix}1&-tanx\\tanx&1\end{bmatrix}$
and $A^{-1} =\frac{1}{1+tan^2x}\begin{bmatrix}1&-tanx\\tanx&1\end{bmatrix}$
$⇒A^TA^{-1}\begin{bmatrix}cos2x&-sin2x\\sin2x&cos2x\end{bmatrix}$
Hence (D) is the correct answer.
$⇒|A^TA^{-1}|=1$
Alternatively,
$|A^TA^{-1}| = |A^T| |A^{-1}| = |A| |A^{-1}| = |AA^{-1}| = | I | = 1$