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 ${A^2}$.

Hint: Multiply the matrix A with itself using the multiplication rule of matrices.

Complete step-by-step answer:
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 $\left[ {\begin{array}{*{20}{c}}
  {\cos 2\alpha }&{\sin 2\alpha } \\
  { - \sin 2\alpha }&{\cos 2\alpha }
\end{array}} \right]$.

Note: For matrix multiplication to exist, it is necessary that the column of the first matrix must be the same as the row of the second matrix otherwise multiplication will not be defined.