Which homogeneous 2D matrix transforms the figure (a) on the left side to the figure (b) on the right?

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  1. \(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}&6\\ 1&0&2\\ 0&0&1 \end{array}} \right]\)
  2. \(\left[ {\begin{array}{*{20}{c}} 0&{ - 2}&6\\ 1&0&1\\ 0&0&1 \end{array}} \right]\)
  3. \(\left[ {\begin{array}{*{20}{c}} 0&2&6\\ 1&0&1\\ 0&0&1 \end{array}} \right]\)
  4. \(\left[ {\begin{array}{*{20}{c}} 0&2&{ - 6}\\ 2&0&1\\ 0&0&1 \end{array}} \right]\)

Answer (Detailed Solution Below)

Option 2 : \(\left[ {\begin{array}{*{20}{c}} 0&{ - 2}&6\\ 1&0&1\\ 0&0&1 \end{array}} \right]\)
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Detailed Solution

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To transform figure on left side into figure on right side, two things must be done in this.

1) It requires anticlockwise rotation by 90 degree.

2) It requires translation along x-axis and y-axis.

In case of 2-D, rotation is done by:

\(P = \;\left[ {\begin{array}{*{20}{c}} {\cos \theta }&{ - \;sin\theta }\\ {sin\theta }&{cos\theta } \end{array}} \right]\)

In case of 3-D rotation, also rotation is with respect to 2 on y-axis

\(P = \;\left[ {\begin{array}{*{20}{c}} {\cos 90}&{ - sin90}&0\\ {sin90}&{cos90}&0\\ 0&0&1 \end{array}} \right] = \;\left[ {\begin{array}{*{20}{c}} 0&{ - 1}&0\\ 1&0&0\\ 0&0&1 \end{array}} \right]\)

So, final matrix after rotation = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}&0\\ 1&0&0\\ 0&0&1 \end{array}} \right]\)

Next step is translation, along x-axis translation factor is 6 units and along y- axis translation factor is 1 unit.

Translation matrix = \(\left[ {\begin{array}{*{20}{c}} 1&0&{tx}\\ 0&1&{ty}\\ 0&0&1 \end{array}} \right]\)

By putting tx = 6 and ty = 1

Final matrix will be = \(\left[ {\begin{array}{*{20}{c}} 1&{ - 2}&{tx}\\ 1&0&{ty}\\ 0&0&1 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0&{ - 2}&6\\ 1&0&1\\ 0&0&1 \end{array}} \right]\)
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