Question

If \[F\left( x \right) = \left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\] then show that \[F\left( x \right) \cdot F\left( y \right) = F\left( {x + y} \right)\]. Hence prove that \[{\left[ {F\left( x \right)} \right]^{ - 1}} = F\left( { - x} \right)\].

Answer 1

Hint:
Here, we will find the value of \[F\left( y \right)\] by replacing the \[x\] by \[y\] in the matrix. Similarly, we will find the matrix \[F\left( {x + y} \right)\]. Then using the values we will compare the LHS and RHS respectively to show the required relation. Then, we will find the inverse of \[F\left( x \right)\] using the cofactors, adjoint and its determinant to prove that it is equal to \[F\left( { - x} \right)\] and hence, proving the required situation.

Formula Used: We will use the following formulas:
A) \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\]
B) \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
C) \[{\left[ {F\left( x \right)} \right]^{ - 1}} = \dfrac{{adj\left[ {F\left( x \right)} \right]}}{{\left| {F\left( x \right)} \right|}}\]

Complete step by step solution:
According to the question, we are given that,
\[F\left( x \right) = \left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]…………………………..\[\left( 1 \right)\]
Similarly, if we try to find \[F\left( y \right)\] , then,
\[F\left( y \right) = \left[ {\begin{array}{*{20}{l}}{\cos y}&{ - \sin y}&0\\{\sin y}&{\cos y}&0\\0&0&1\end{array}} \right]\]………………………………\[\left( 2 \right)\]
Also, \[F\left( {x + y} \right)\] can also be written as:
\[F\left( {x + y} \right) = \left[ {\begin{array}{*{20}{l}}{\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0\\{\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0\\0&0&1\end{array}} \right]\]
Now, we have to show that \[F\left( x \right) \cdot F\left( y \right) = F\left( {x + y} \right)\]
First we will consider LHS \[ = F\left( x \right) \cdot F\left( y \right)\]
Therefore, Substituting the values from equation \[\left( 1 \right)\] and \[\left( 2 \right)\], we get
 \[F\left( x \right) \cdot F\left( y \right) = \left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{l}}{\cos y}&{ - \sin y}&0\\{\sin y}&{\cos y}&0\\0&0&1\end{array}} \right]\]
Now, multiplying these matrices, we get,
\[ \Rightarrow F\left( x \right) \cdot F\left( y \right) = \left[ {\begin{array}{*{20}{l}}{\cos x\cos y - \sin x\sin y}&{ - \left( {\cos x\sin y + \sin x\cos y} \right)}&0\\{\sin x\cos y + \cos x\sin y}&{\cos x\cos y - \sin x\sin y}&0\\0&0&1\end{array}} \right]\]
We know that, \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\] and \[\sin \left( {A + B} \right) = \sin A\cos B + \cos A\sin B\]
Hence, using these formulas in the above equation, we get, we get,
\[ \Rightarrow F\left( x \right) \cdot F\left( y \right) = \left[ {\begin{array}{*{20}{l}}{\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0\\{\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0\\0&0&1\end{array}} \right]\]
We know that RHS \[ = F\left( {x + y} \right) = \left[ {\begin{array}{*{20}{l}}{\cos \left( {x + y} \right)}&{ - \sin \left( {x + y} \right)}&0\\{\sin \left( {x + y} \right)}&{\cos \left( {x + y} \right)}&0\\0&0&1\end{array}} \right]\]
We can see that \[LHS = RHS\]. Hence,
\[F\left( x \right) \cdot F\left( y \right) = F\left( {x + y} \right)\]
Now we are required to prove that \[{\left[ {F\left( x \right)} \right]^{ - 1}} = F\left( { - x} \right)\].
First, we will consider the LHS of the above equation.
\[F\left( x \right) = \left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]
Determinant of \[F\left( x \right)\] is:
\[\left| {F\left( x \right)} \right| = \left| {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right|\]
\[ \Rightarrow \left| {F\left( x \right)} \right| = \cos x\left( {\cos x - 0} \right) + \sin x\left( {\sin x - 0} \right) + 0\]
Multiplying the terms, we get
\[ \Rightarrow \left| {F\left( x \right)} \right| = {\cos ^2}x + {\sin ^2}x\]
We know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\], so using this identity in the above equation, we get
\[ \Rightarrow \left| {F\left( x \right)} \right| = 1\]
Here, cofactors in the first row are:
\[{A_{11}} = {\left( { - 1} \right)^{1 + 1}}\left[ {\begin{array}{*{20}{l}}{\cos x}&0\\0&1\end{array}} \right] = \cos x\]
\[{A_{12}} = {\left( { - 1} \right)^{1 + 2}}\left[ {\begin{array}{*{20}{l}}{\sin x}&0\\0&1\end{array}} \right] = - \sin x\]
\[{A_{13}} = {\left( { - 1} \right)^{1 + 3}}\left[ {\begin{array}{*{20}{l}}{\sin x}&{\cos x}\\0&0\end{array}} \right] = 0\]
Cofactors in the second row are:

\[{A_{21}} = {\left( { - 1} \right)^{2 + 1}}\left[ {\begin{array}{*{20}{l}}{ - \sin x}&0\\0&1\end{array}} \right] = \sin x\]
\[{A_{22}} = {\left( { - 1} \right)^{2 + 2}}\left[ {\begin{array}{*{20}{l}}{\cos x}&0\\0&1\end{array}} \right] = \cos x\]
\[{A_{23}} = {\left( { - 1} \right)^{2 + 3}}\left[ {\begin{array}{*{20}{l}}{\cos x}&0\\0&0\end{array}} \right] = 0\]
Cofactors in the third row are:
\[{A_{31}} = {\left( { - 1} \right)^{3 + 1}}\left[ {\begin{array}{*{20}{l}}{ - \sin x}&0\\{\cos x}&0\end{array}} \right] = 0\]
 \[{A_{32}} = {\left( { - 1} \right)^{3 + 2}}\left[ {\begin{array}{*{20}{l}}{\cos x}&0\\{\sin x}&0\end{array}} \right] = 0\]
\[{A_{33}} = {\left( { - 1} \right)^{3 + 3}}\left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}\\{\sin x}&{\cos x}\end{array}} \right] = {\cos ^2}x + {\sin ^2}x = 1\]
Hence, \[adj\left[ {F\left( x \right)} \right] = {\left[ {{\rm{Cofactor}}} \right]^T}\]
\[ \Rightarrow adj\left[ {F\left( x \right)} \right] = {\left[ {\begin{array}{*{20}{l}}{\cos x}&{ - \sin x}&0\\{\sin x}&{\cos x}&0\\0&0&1\end{array}} \right]^T}\]
\[ \Rightarrow adj\left[ {F\left( x \right)} \right] = \left[ {\begin{array}{*{20}{l}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]
Hence, inverse of \[F\left( x \right)\] will be:
LHS \[ = {\left[ {F\left( x \right)} \right]^{ - 1}} = \dfrac{{adj\left[ {F\left( x \right)} \right]}}{{\left| {F\left( x \right)} \right|}} = \Rightarrow adj\left[ {F\left( x \right)} \right] = \dfrac{1}{1}\left[ {\begin{array}{*{20}{l}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]
Hence,
\[ \Rightarrow \] LHS \[ = {\left[ {F\left( x \right)} \right]^{ - 1}} = \left[ {\begin{array}{*{20}{l}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]
Also, RHS \[ = F\left( { - x} \right) = \left[ {\begin{array}{*{20}{l}}{\cos \left( { - x} \right)}&{ - \sin \left( { - x} \right)}&0\\{\sin \left( { - x} \right)}&{\cos \left( { - x} \right)}&0\\0&0&1\end{array}} \right]\]
Now, knowing the fact that \[\cos \left( { - x} \right) = \cos \left( {0^\circ - x} \right) = \cos x\] because this lies in the fourth quadrant where cosine is positive.
Also, \[\sin \left( { - x} \right) = \sin \left( {0^\circ - x} \right) = - \sin x\] because this lies in fourth quadrant where sine is negative
Hence,
RHS \[ = F\left( { - x} \right) = \left[ {\begin{array}{*{20}{l}}{\cos x}&{\sin x}&0\\{ - \sin x}&{\cos x}&0\\0&0&1\end{array}} \right]\]
Clearly, LHS \[ = \] RHS
Hence, \[{\left[ {F\left( x \right)} \right]^{ - 1}} = F\left( { - x} \right)\]
Hence proved

Note:
A matrix is a collection of numbers which are arranged into a fixed number of rows and columns where, usually, those numbers are real numbers. In real life, matrices are used for making seismic surveys in geology. They are also used for plotting graphs, statistics and to do scientific research and studies. Most of the data are stored using matrices such as mortality rate, population in the country, etc. Hence, matrices are used in day-to-day life.