
Simplify:
${\text{cos}}\theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$
Answer
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Hint: Before attempting this question, one should have prior knowledge about the matrix such as the addition two matrix is given as; $\left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
p&q \\
r&s
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + p}&{b + q} \\
{c + r}&{d + s}
\end{array}} \right]$also remember to use trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$to simplify the value of the given matrix.
Complete step-by-step answer:
According to the given information we have matrix
${\text{cos}}\theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$
Further simplifying the matrix, we get
$\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta }&{{{\cos }^2}\theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta }&{ - \sin \theta \cos \theta } \\
{\sin \theta \cos \theta }&{{{\sin }^2}\theta }
\end{array}} \right]$
Now we know that the addition of two matrix is given as; $\left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
p&q \\
r&s
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + p}&{b + q} \\
{c + r}&{d + s}
\end{array}} \right]$
Therefore, $\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta + {{\sin }^2}\theta }&{\sin \theta \cos \theta - \sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }
\end{array}} \right]$
Since we know that by the trigonometric identity, ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Therefore, using this identity in the above matrix we get
$\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]$
As we know that the square matrix where diagonal elements have value 1 and other than diagonal all elements are 0 this type of matrix is called identity matrix
So, we can say that the above matrix is an ${\text{2}} \times {\text{2}}$ identity matrix i.e. ${{\text{I}}_{2 \times 2}}$
therefore, ${\text{cos}}\theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$ = ${{\text{I}}_{2 \times 2}}$
So, this is the required solution.
Note: In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
p&q \\
r&s
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + p}&{b + q} \\
{c + r}&{d + s}
\end{array}} \right]$also remember to use trigonometric identity ${\cos ^2}\theta + {\sin ^2}\theta = 1$to simplify the value of the given matrix.
Complete step-by-step answer:
According to the given information we have matrix
${\text{cos}}\theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$
Further simplifying the matrix, we get
$\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta }&{\sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta }&{{{\cos }^2}\theta }
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
{{{\sin }^2}\theta }&{ - \sin \theta \cos \theta } \\
{\sin \theta \cos \theta }&{{{\sin }^2}\theta }
\end{array}} \right]$
Now we know that the addition of two matrix is given as; $\left[ {\begin{array}{*{20}{c}}
a&b \\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
p&q \\
r&s
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{a + p}&{b + q} \\
{c + r}&{d + s}
\end{array}} \right]$
Therefore, $\left[ {\begin{array}{*{20}{c}}
{{{\cos }^2}\theta + {{\sin }^2}\theta }&{\sin \theta \cos \theta - \sin \theta \cos \theta } \\
{ - \sin \theta \cos \theta + \sin \theta \cos \theta }&{{{\cos }^2}\theta + {{\sin }^2}\theta }
\end{array}} \right]$
Since we know that by the trigonometric identity, ${\cos ^2}\theta + {\sin ^2}\theta = 1$
Therefore, using this identity in the above matrix we get
$\left[ {\begin{array}{*{20}{c}}
1&0 \\
0&1
\end{array}} \right]$
As we know that the square matrix where diagonal elements have value 1 and other than diagonal all elements are 0 this type of matrix is called identity matrix
So, we can say that the above matrix is an ${\text{2}} \times {\text{2}}$ identity matrix i.e. ${{\text{I}}_{2 \times 2}}$
therefore, ${\text{cos}}\theta \left[ {\begin{array}{*{20}{c}}
{\cos \theta }&{\sin \theta } \\
{ - \sin \theta }&{\cos \theta }
\end{array}} \right] + \sin \theta \left[ {\begin{array}{*{20}{c}}
{\sin \theta }&{ - \cos \theta } \\
{\cos \theta }&{\sin \theta }
\end{array}} \right]$ = ${{\text{I}}_{2 \times 2}}$
So, this is the required solution.
Note: In the above solution we used the trigonometric identities which are the expressions which involve trigonometric functions where the term “function” can be explained as relation between the provided inputs and the outputs of the given inputs such that each input is directly related to the one output. The representation of a function is given by supposing if there is a function “f” that belongs from X to Y then the function is represented by $f:X \to Y$ examples of function are one-one functions, onto functions, bijective functions, trigonometric function, binary function, etc.
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