Question

[tex] If A(α)\left[\begin{array}{ccc}Cos α&Sin α\\-Sin α&Cos α\end{array}\right],Prove A(α) A(-α)=I. [\tex]

Answer 1

Answer:

$\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$

Step-by-step explanation:

$A(\alpha )=\left[\begin{array}{ccc}Cos\alpha &Sin\alpha \\-Sin\alpha &Cos\alpha \end{array}\right]\\ A(-\alpha )=\left[\begin{array}{ccc}Cos(-\alpha) &Sin(-\alpha) \\-Sin(-\alpha) &Cos(-\alpha) \end{array}\right]\\=\left[\begin{array}{ccc}Cos\alpha &-Sin\alpha \\Sin\alpha &Cos\alpha \end{array}\right]\\Now,$$A(\alpha )\times A(-\alpha)=\left[\begin{array}{ccc}Cos\alpha &Sin\alpha \\-Sin\alpha &Cos\alpha \end{array}\right]\times\left[\begin{array}{ccc}Cos\alpha &-Sin\alpha \\Sin\alpha &Cos\alpha \end{array}\right]\\=\left[\begin{array}{ccc}Cos^{2} \alpha+sin^{2}\alpha  &-cos\alpha.sin\alpha+sin\alpha.cos\alpha \\-Sin\alpha.cos\alpha+cos\alpha.sin\alpha &Cos^{2}\alpha+sin^{2}\alpha\end{array}\right]\\=\left[\begin{array}{ccc}1&0\\0&1\\\end{array}\right]$