Question

If $A=\left[\begin{array}{cc}1& tanx\\ -tanx& 1\end{array}\right],A^T{A}^{-1}=$ ___

• A. $\left[\begin{array}{cc}cos2x& -sin2x\\ sin2x& cos2x\end{array}\right]$
• B. $\left[\begin{array}{cc}cos2x& sin2x\\ -sin2x& cos2x\end{array}\right]$
• C. $\left[\begin{array}{cc}1-ta{n}^{2}x& 2tanx\\ -2tanx& 1-ta{n}^{2}x\end{array}\right]$
• D. $\left[\begin{array}{cc}1-ta{n}^{2}x& -2tanx\\ 2tanx& 1-ta{n}^{2}x\end{array}\right]$

Answer 1

The correct option is A

$\left[\begin{array}{cc}cos2x& -sin2x\\ sin2x& cos2x\end{array}\right]$

$A=\left[\begin{array}{cc}1& tanx\\ -tanx& 1\end{array}\right]|A|=1+ta{n}^{2}xadj\left(A\right)=\left[\begin{array}{cc}1& -tanx\\ tanx& 1\end{array}\right]A^{-1}=\frac{1}{1+ta{n}^{2}x}\left[\begin{array}{cc}1& -tanx\\ tanx& 1\end{array}\right]A^T=\left[\begin{array}{cc}1& -tanx\\ tanx& 1\end{array}\right]\therefore A^T{A}^{-1}=\frac{1}{1+ta{n}^{2}x}\left[\begin{array}{cc}1& -tanx\\ tanx& 1\end{array}\right]\left[\begin{array}{cc}1& -tanx\\ tanx& 1\end{array}\right]=\frac{1}{1+ta{n}^{2}x}\left[\begin{array}{cc}1-ta{n}^{2}x& -2tanx\\ 2tanx& 1-ta{n}^{2}x\end{array}\right]=\left[\begin{array}{cc}\frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}& \frac{-2tanx}{1+ta{n}^{2}x}\\ \frac{2tanx}{1+ta{n}^{2}x}& \frac{1-ta{n}^{2}x}{1+ta{n}^{2}x}\end{array}\right]=\left[\begin{array}{cc}cos2x& -sin2x\\ sin2x& cos2x\end{array}\right]$