An orthogonal matrix is
A. $\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{2\sin \alpha }\\{ - 2\sin \alpha }&{\cos \alpha }\end{array}} \right]$
B. $\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$
C.$ \left[ {\begin{array}{*{20}{l}}{\cos \alpha }&{\sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{l}}1&1\\1&1\end{array}} \right]$

Hint: In our case, we have been asked to determine the orthogonal matrix. For that we have to first find the A matrix and then the transpose of the Matrix A. Now, after finding the transpose of the “A’ matrix, we have to multiply the transpose of matrix A and matrix A. Now, we have to perform matrix multiplication by multiplying rows and columns of each element of the matrix which gives the ‘I’ matrix. That is the identity matrix, which results in getting the desired answer.

Formula Used: $A \cdot {A^T} = {A^T} \cdot A = I$
Here,
A can be any square matrix of order $n x n$.
The inverse of matrix 'A' is ${A^T}$
The order n x n identity matrix is I

Complete step by step solution: We have been provided in the question that a set of matrixes.
And are asked to determine which of the given matrix is orthogonal.
We have been already known that the condition for orthogonal matrix is
${A^T} \cdot A = I$
Now, we have to determine the transpose of the given matrix, we have
We know that the matrix A is
A = $\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$
And we can write the transpose of the above matrix A as,
${A^ \top } = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$
As, we know that the condition for orthogonal matrix is,
$A \cdot {A^T} = I$
$A \cdot {A^ \top } = \left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{ - \sin \alpha }\\{\sin \alpha }&{\cos \alpha }\end{array}} \right]$
To multiply the matrix, the elements in the first matrix's column and rows are multiplied by one another, and then the results are added.
Therefore, we get
$\left[ {\left[ {\begin{array}{*{20}{c}}{(\cos (\alpha )) \cdot (\cos (\alpha )) + (\sin (\alpha )) \cdot (\sin (\alpha ))}&{(\cos (\alpha )) \cdot ( - \sin (\alpha )) + (\sin (\alpha )) \cdot (\cos (\alpha ))}\\{( - \sin (\alpha )) \cdot (\cos (\alpha )) + (\cos (\alpha )) \cdot (\sin (\alpha ))}&{( - \sin (\alpha )) \cdot ( - \sin (\alpha )) + (\cos (\alpha )) \cdot (\cos (\alpha ))}\end{array}} \right]} \right.$
Now, we have to solve each terms of the row from the above matrix, we get = $\left[ {\begin{array}{*{20}{c}}{{{\cos }^2}\alpha + {{\sin }^2}\alpha }&{ - \sin \alpha \cos \alpha + \sin \alpha \cos \alpha }\\{ - \sin \alpha \cos \alpha + \cos \alpha \sin \alpha }&{{{\cos }^2}\alpha + {{\sin }^2}\alpha }\end{array}} \right]$
Now, by using the trigonometry identity we can simplify the expressions in the above matrix.
Since we know that ${\sin ^2}x + {\cos ^2}x = 1$
The above matrix becomes,
 = $\left[ {\begin{array}{*{20}{c}}1&{ - \sin \alpha \cos \alpha + \sin \alpha \cos \alpha }\\{ - \sin \alpha \cos \alpha + \cos \alpha \sin \alpha }&1\end{array}} \right]$
Now, we have to cancel the similar terms in the each rows of the above matrix.
Thus, we get
 = $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
The above obtained matrix is an Identity matrix.
Since, we have been already known that
$I = \left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$
Therefore, an orthogonal matrix is $\left[ {\begin{array}{*{20}{c}}{\cos \alpha }&{\sin \alpha }\\{ - \sin \alpha }&{\cos \alpha }\end{array}} \right]$

Option ‘B’ is correct

Note: Students are likely to make mistakes in these types of problems because; it includes matrix multiplication and also some trigonometry identities to remember. As we already know that, matrix multiplication is difficult to perform. So, students should keep in mind that applying wrong trigonometry formula will not give the desired answer.