Question

[tex] If A=\left[\begin{array}{ccc}Cos α &sin α\\-sin α& cos α\end{array}\right], Prove AAᵀ=I. Deduce A⁻¹ =Aᵀ. [\tex]

Answer 1

Step-by-step explanation: If

$A=\left[\begin{array}{ccc}cos \alpha &sin \alpha\\-sin \alpha& cos \alpha\end{array}\right]$

To find the transpose,we know that columns and rows are interchanged

$A^T=\left[\begin{array}{ccc}cos \alpha &-sin \alpha\\sin \alpha& cos \alpha\end{array}\right]$

Now multiply both

$AA^T=\left[\begin{array}{ccc}cos \alpha &sin \alpha\\-sin \alpha& cos \alpha\end{array}\right] \times\left[\begin{array}{ccc}cos \alpha &-sin \alpha\\sin \alpha& cos \alpha\end{array}\right]$

$AA^T=\left[\begin{array}{ccc}cos^2 \alpha+sin^2 \alpha &cos \alpha\:sin \alpha-cos \alpha\:sin \alpha\\-sin \alpha cos \alpha +cos \alpha\:sin \alpha & cos^2 \alpha+sin^2 \alpha\end{array}\right]$

$AA^T=\left[\begin{array}{ccc}1 &0\\0& 1\end{array}\right]\\\\AA^T=I$

Hence proved.

Now to Deduce A⁻¹ =Aᵀ

Determinant of A

$cos^2\alpha+sin^2 \alpha=1\\\\A^{-1}=\frac{A^T}{|A|}\\\\A^{-1}=\left[\begin{array}{ccc}cos \alpha &-sin \alpha\\sin \alpha& cos \alpha\end{array}\right]$

so,

$A^{-1}=A^T$

Hence proved.

Hope it helps you.