Question

Solve the following matrix
$\cos \theta \left[ {\begin{array}{*{20}{c}}
  {\cos \theta }&{\sin \theta } \\
  { - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
  {\sin \theta }&{ - \cos \theta } \\
  {\cos \theta }&{\sin \theta }
\end{array}} \right]$

Answer 1

Hint – In this particular type of question use the concept that when we multiply something into the square matrix it will multiply in all the rows of the square matrix and use the basic trigonometric identity (${\sin ^2}\theta + {\cos ^2}\theta = 1$) so use these concepts to reach the solution of the question.

Complete step by step solution:
Given matrix
$\cos \theta \left[ {\begin{array}{*{20}{c}}
  {\cos \theta }&{\sin \theta } \\
  { - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
  {\sin \theta }&{ - \cos \theta } \\
  {\cos \theta }&{\sin \theta }
\end{array}} \right]$
Now multiply cos$\theta $ and sin$\theta $ inside the matrix (when we multiply cos$\theta $ and sin$\theta $ is multiplied in both of the rows respectively).
So multiply this we have,
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {\cos \theta \cos \theta }&{\cos \theta \sin \theta } \\
  { - \cos \theta \sin \theta }&{\cos \theta \cos \theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {\sin \theta \sin \theta }&{ - \sin \theta \cos \theta } \\
  {\sin \theta \cos \theta }&{\sin \theta \sin \theta }
\end{array}} \right]$
Now simplify we have,
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {{{\cos }^2}\theta }&{\cos \theta \sin \theta } \\
  { - \cos \theta \sin \theta }&{{{\cos }^2}\theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
  {{{\sin }^2}\theta }&{ - \sin \theta \cos \theta } \\
  {\sin \theta \cos \theta }&{{{\sin }^2}\theta }
\end{array}} \right]$
Now it is simple addition of matrix, so add the above matrix using the rule that the first element of the first matrix is added only with the first element of the second matrix and same procedure is for the rest of the elements, and the resultant is a single matrix
So add the above matrix we have,
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  {{{\cos }^2}\theta + {{\sin }^2}\theta }&{\cos \theta \sin \theta - \cos \theta \sin \theta } \\
  { - \cos \theta \sin \theta + \cos \theta \sin \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }
\end{array}} \right]$
Now as we all know the basic trigonometric identity (i.e. ${\sin ^2}\theta + {\cos ^2}\theta = 1$) so use this property in the above equation we have,
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&{\cos \theta \sin \theta - \cos \theta \sin \theta } \\
  { - \cos \theta \sin \theta + \cos \theta \sin \theta }&1
\end{array}} \right]$
And the rest of the terms is cancel out as there are two terms which are equal but in opposite sign so these terms are cancel out with itself so we have,
$ \Rightarrow \left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
So as we see that the resultant is an identity matrix.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always recall how to add the two or more than two matrix which is stated above so first simplify the matrix as above then apply matrix addition as above and then apply basic trigonometric identity and simplify we will get the required answer.