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Home  >>  CBSE XII  >>  Math  >>  Matrices
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if $ F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}, $ show that $ F( x )\; F( y ) = F( x + y ). $

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Toolbox:
  • Given $ F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, to arrive at $F(y)$, we replace $x$ with $y$, i.e., $ F( y ) = \begin{bmatrix} cos\;y & -sin\;y & 0 \\ sin\;y & cos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  • Given $ F( x ) = \begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}$, to arrive at $F(x+y)$, we replace $x$ with $x+y$, i.e., $F(x+y)= \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
  • Multiplication of two matrices is defined only if the number of columns of the left matrix is the same as the number of rows of the right matrix.
  • If A is an m-by-n matrix and B is an n-by-p matrix, then their matrix product AB is the m-by-p matrix whose entries are given by dot product of the corresponding row of A and the corresponding column of B:
  • $\begin{bmatrix}AB\end{bmatrix}_{i,j} = A_{i,1}B_{1,j} + A_{i,2}B_{2,j} + A_{i,3}B_{3,j} ... A_{i,n}B_{n,j}$
  • $sin (x+y) = sinx.cosy + cosx.siny$
  • $cos(x+y) = cosx.cosy - sinx.siny$
We have to show that:$F(x)F(y)=F(x+y). LHS: F(x)F(y)$ and $RHS: F(x+y).$
Evaluating $LHS:F(x)F(y)=\begin{bmatrix} cos\;x & -sin\;x & 0 \\ sin\;x & cos\;x & 0 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} cos\;y & -sin\;y & 0 \\ sin\;y & cos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$:
$\Rightarrow \begin{bmatrix} cos\;xcos\;y-sin\;xsin\;y+0 & cos\;x(-sin\;y)+(-sin\;xcos\;y)+0 &0+0+0 \\ sin\;xcos\;y+cos\;xsin\;y+0 & -sin\;xsin\;y+cos\;xcos\;y+0 & 0+0+0 \\ 0+0+0 & 0+0+0 & 0+0+1 \end{bmatrix}$
$\Rightarrow \begin{bmatrix} cos\;xcos\;y-sin\;xsin\;y & cos\;x(-sin\;y)+(-sin\;xcos\;y) &0\\ sin\;xcos\;y+cos\;xsin\;y & -sin\;xsin\;y+cos\;xcos\;y & 0 \\ 0 & 0 & 1 \end{bmatrix}$
We know that $sin (x+y) = sinx.cosy + cosx.siny$ and $cos(x+y) = cosx.cosy - sinx.siny$.
$\Rightarrow LHS: \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$RHS: F(x+y)= \begin{bmatrix} cos (x+y) & -sin(x+y) &0\\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{bmatrix}$, which is the equal to $LHS$.
answered Feb 13, 2013 by sreemathi.v
edited Feb 27, 2013 by balaji.thirumalai
 

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