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If $F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right],$ show that $F(x) F(y)=F(x+y)$
Solution
$F(x)=\left[\begin{array}{ccc}\cos x & -\sin x & 0 \\ \sin x & \cos x & 0 \\ 0 & 0 & 1\end{array}\right]$, $F(y)=\left[\begin{array}{ccc}\cos y & -\sin y & 0 \\ \sin y & \cos y & 0 \\ 0 & 0 & 1\end{array}\right]$
$R.H.S:$ $F(x+y)=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$
$L.H.S:$ $F(x)$ $F(y)$
$ = \left[ {\begin{array}{*{20}{c}}
{\cos x}&{ – \sin x}&0 \\
{\sin x}&{\cos x}&0 \\
0&0&1
\end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}}
{\cos y}&{ – \sin y}&0 \\
{\sin y}&{\cos y}&0 \\
0&0&1
\end{array}} \right]$
$=\left[\begin{array}{ccc}
\cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \\
\sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y+0 & 0 \\
0 & 0 & 0
\end{array}\right]$
$=\left[\begin{array}{ccc}\cos (x+y) & -\sin (x+y) & 0 \\ \sin (x+y) & \cos (x+y) & 0 \\ 0 & 0 & 1\end{array}\right]$
$=F(x+y)$
$\therefore $ $F(x) F(y)=F(x+y)$
Similar Questions
Consider the following information regarding the number of men and women workers in three factories $I,\,II$ and $III$
Men workers |
Women workers |
|
$I$ | $30$ | $25$ |
$II$ | $25$ | $31$ |
$III$ | $27$ | $26$ |
Represent the above information in the form of a $3 \times 2$ matrix. What does the entry in the third row and second column represent?