Question

If $A = \left( {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right)$ , find $\alpha $ satisfying $0 < \alpha < \dfrac{\pi }{2}$ when $A + {A^T} = \sqrt 2 {I_2}$ ; where ${A^T}$ is transpose of A.

Answer 1

Hint:This question is a combination of matrix and trigonometry terms. First we have to compute the transpose matrix of A. Further we compute $A + {A^T}$ which will be compared to the identity matrix with some computation. This comparison will give the result.

Complete step-by-step answer:
Given matrix A is,
$A = \left( {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right)$
Now we will compute its transpose matrix, with the required rearrangement of rows and column values as follows:
${A^T} = \left( {\begin{array}{*{20}{c}}
  {\cos \alpha }&{ - \sin \alpha } \\
  {\sin \alpha }&{\cos \alpha }
\end{array}} \right)$
Now, identity matrix of order 2 means ${I_2}$ is as given here,
\[{I_2} = \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right)\]
We multiply the above matrix by$\sqrt 2 $, we get
$
  \sqrt 2 {I_2} = \sqrt 2 \times \left( {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right) \\
   = \left( {\begin{array}{*{20}{c}}
  {\sqrt 2 }&0 \\
  0&{\sqrt 2 }
\end{array}} \right) \\
 $
Now we will evaluate the expression:
$A + {A^T} = \sqrt 2 {I_2}$
Substitute the terms in above equation, we have
$\left( {\begin{array}{*{20}{c}}
  {\cos \alpha }&{\sin \alpha } \\
  { - \sin \alpha }&{\cos \alpha }
\end{array}} \right) + \left( {\begin{array}{*{20}{c}}
  {\cos \alpha }&{ - \sin \alpha } \\
  {\sin \alpha }&{\cos \alpha }
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\sqrt 2 }&0 \\
  0&{\sqrt 2 }
\end{array}} \right)$
We will add two matrices in LHS,
$\left( {\begin{array}{*{20}{c}}
  {2\cos \alpha }&0 \\
  0&{2\cos \alpha }
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
  {\sqrt 2 }&0 \\
  0&{\sqrt 2 }
\end{array}} \right)$
Compare the corresponding terms on both the sides will give,
$2\cos \alpha = \sqrt 2 $
Now divide the RHS by 2 we get
$\cos \alpha = \dfrac{1}{{\sqrt 2 }}$
From the table of trigonometry ratios cosine will be $\dfrac{1}{{\sqrt 2 }}$ for the angle $\dfrac{\pi }{4}$ which will lie in the interval $0 < \alpha < \dfrac{\pi }{2}$ .
$\therefore $Value of $\alpha $ will be $\dfrac{\pi }{4}$ .

Additional Information:In matrix, transpose matrix can be obtained by interchanging the rows and columns values with each other. Transpose matrix is possible for matrices of any order. With the help of the transpose matrix we can find the inverse of the matrix.

Note:Here we have used the concept of transpose matrix and the one to one comparison of the terms of two matrices. Also a suitable identity matrix is very important to know and use in the question. Since values in the matrix are trigonometry ratios. So, knowledge of those ratios and their values for specific angles are important to know. Such problems are always requiring the careful observation of the concept and utilization of the theories properly.