If ${\rm{A}} = \left[ {\begin{array}{*{20}{c}}
{\cos {\rm{\alpha }}}&{ - \sin {\rm{\alpha }}}\\
{\sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}}
\end{array}} \right]$ is identity matrix, then write the value of ${\rm{\alpha }}$.

Hint:
We know the Identity matrix is a matrix which on multiplying with another matrix, return that matrix as a result. Identity matrix is $\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$. Compare the elements of given matrix with $\left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$ to get the value of ${\rm{\alpha }}$.

Complete step by step solution:
Given: ${\rm{A}} = \left[ {\begin{array}{*{20}{c}}
{\cos {\rm{\alpha }}}&{ - \sin {\rm{\alpha }}}\\
{\sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}}
\end{array}} \right]$ is identity matrix.
In linear algebra, the identity matrix (sometimes called a unit matrix) of size n is the n x n square matrix with ones on the main diagonal and zeroes elsewhere, it is denoted by In.
Here A is $2 \times 2$ ,matrix, so identity matrix is ${{\rm{I}}_2}$
       ${{\rm{I}}_2} = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$
Now it is given that ${\rm{A}} = {{\rm{I}}_2}$
So, $\left[ {\begin{array}{*{20}{c}}
{\cos {\rm{\alpha }}}&{ - \sin {\rm{\alpha }}}\\
{\sin {\rm{\alpha }}}&{\cos {\rm{\alpha }}}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0\\
0&1
\end{array}} \right]$
Comparing the corresponding elements
We get,
     $\cos {\rm{\alpha }} = 1,{\rm{\;}} - \sin {\rm{\alpha }} = 0,\sin {\rm{\alpha }} = 0,\cos {\rm{\alpha }} = 1$
Here, we get
 $\cos {\rm{\alpha }} = 1{\rm{\;or}}\sin {\rm{\alpha }} = 0$
Solving both of them
 $ \Rightarrow \cos {\rm{\alpha }} = \cos 0^\circ {\rm{\;and}}\sin {\rm{\alpha }} = \sin 0^\circ $
Since $\cos 0^\circ = 1{\rm{and}}\sin 0^\circ = 0$
So are get
${\rm{\alpha }} = 0{\rm{\;}}$

Therefore, answer is ${\rm{\alpha }} = 0$

Note:
Equality of two matrices says, two matrices are equal if and only if they have the same element at the same place. This is the reason for comparing matrix elementwise in our solution. Don’t ever confuse the identity matrix with a scaler or diagonal matrices. An identity matrix is always a diagonal matrix but the converse is not true.