Question

COS X -sin x 0
if F(x) =
sin x
COS
0
then show that
0
0
1
F(x). F (y) = F (x + y)​

Answer 1

Appropriate Question

$\sf \: If \: F(x) \: \begin{gathered}\sf =\left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right]\end{gathered}$

Show that, F(x) × F(y) = F(x + y)

$\large\underline{\sf{Solution-}}$

Given that

$\sf \: \: F(x) \: \begin{gathered}\sf =\left[\begin{array}{ccc}cos&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right]\end{gathered}$

So,

$\sf \: \: F(y) \: \begin{gathered}\sf =\left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right]\end{gathered}$

Consider,

$\rm :\longmapsto\:F(x) \: F(y)$

$\sf \:=\begin{gathered}\sf \left[\begin{array}{ccc}cosx&-sinx&0\\sinx&cosx&0\\0&0&1\end{array}\right]\end{gathered}\begin{gathered}\sf \left[\begin{array}{ccc}cosy&-siny&0\\siny&cosy&0\\0&0&1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf =\left[\begin{array}{ccc}cosxcosy - sinxsiny&-sinxcosy - sinycosx&0\\sinxcosy + sinycosx&cosxcosy - sinxsiny&0\\0&0&1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf =\left[\begin{array}{ccc}cosxcosy-sinxsiny&-(sinxcosy+ sinycosx)&0\\sinxcosy + sinycosx&cosxcosy-sinxsiny&0\\0&0&1\end{array}\right]\end{gathered}$

$\begin{gathered}\sf =\left[\begin{array}{ccc}cos(x + y)&-sin(x + y)&0\\sin(x + y)&cos(x + y)&0\\0&0&1\end{array}\right]\end{gathered}$

$\: \: \: \: \: \: \boxed{ \because \bf \: cosxcosy - sinxsiny = cos(x + y)} \\ \: \: \: \: \: \: \boxed{ \because \bf \: sinxcosy + cosxsiny = sin(x + y)}$

$\sf \: = \: F(x + y)$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. Commutative Property :-

• Matrix multiplication may or may not be Commutative.

2. Associative Property :-

• Matrix multiplication is Associative.

• i.e. (AB)C = A(BC)

3. Distributive Property :-

• Matrix multiplication is Distributive over Addition.

• i.e. A(B + C) = AB + AC

4. There exist an identity matrix I such that

• AI = IA = A