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# If A is a matrix such that $A = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]$, then find ${A^2}$.

Last updated date: 27th Mar 2023
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The given matrix is $A = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right]$. For finding ${A^2}$, we will multiply A with itself. So, we’ll get:
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {\cos \alpha }&{\sin \alpha } \\ { - \sin \alpha }&{\cos \alpha } \end{array}} \right], \\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{\cos \alpha \sin \alpha + \sin \alpha \cos \alpha } \\ { - \sin \alpha \cos \alpha + \cos \alpha \left( { - \sin \alpha } \right)}&{ - {{\sin }^2}\alpha + {{\cos }^2}\alpha } \end{array}} \right], \\ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {{{\cos }^2}\alpha - {{\sin }^2}\alpha }&{2\sin \alpha \cos \alpha } \\ { - 2\sin \alpha \cos \alpha }&{{{\cos }^2}\alpha - {{\sin }^2}\alpha } \end{array}} \right], \\$
We know that $2\sin \alpha \cos \alpha = \sin 2\alpha {\text{ and }}{\cos ^2}\alpha - {\sin ^2}\alpha = \cos 2\alpha$, substituting these value above, we’ll get:
$\Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}} {\cos 2\alpha }&{\sin 2\alpha } \\ { - \sin 2\alpha }&{\cos 2\alpha } \end{array}} \right]$.
Thus, matrix ${A^2}$ is $\left[ {\begin{array}{*{20}{c}} {\cos 2\alpha }&{\sin 2\alpha } \\ { - \sin 2\alpha }&{\cos 2\alpha } \end{array}} \right]$.