Multiplication of matrices E and F is G. Matrices E and G are\(E = \le

1.	Multiplication of matrices E and F is G. Matrices E and G are\(E = \left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) and \(G= \left[ {\begin{array}{{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\). What is the matrix F?1. \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)2. \(\left[ {\begin{array}{{20}{c}} {sin\theta }&{cos\theta }&0\\ { - cos\theta }&{sin\theta }&0\\ 0&0&1 \end{array}} \right]\)3. \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)4. \(\left[ {\begin{array}{{20}{c}} {sin\theta }&{ - cos\theta }&0\\ {cos\theta }&{sin\theta }&0\\ 0&0&1 \end{array}} \right]\)
Answer» Correct Answer - Option 3 : \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) \(E = \left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) & \(G = \left[ {\begin{array}{{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\) According to problem, E × F = G OR, \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) × \(F = \left[ {\begin{array}{{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\) Hence it can be seen that product of (E × F) is unit matrix. So F will be the inverse of E. \(\Rightarrow F = {E^{ - 1}} = \left[ {\begin{array}{{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)

Multiplication of matrices E and F is G. Matrices E and G are\(E = \left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) and \(G= \left[ {\begin{array}{{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\). What is the matrix F?1. \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)2. \(\left[ {\begin{array}{{20}{c}} {sin\theta }&{cos\theta }&0\\ { - cos\theta }&{sin\theta }&0\\ 0&0&1 \end{array}} \right]\)3. \(\left[ {\begin{array}{{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)4. \(\left[ {\begin{array}{{20}{c}} {sin\theta }&{ - cos\theta }&0\\ {cos\theta }&{sin\theta }&0\\ 0&0&1 \end{array}} \right]\)

Answer» Correct Answer - Option 3 : \(\left[ {\begin{array}{*{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)

\(E = \left[ {\begin{array}{*{20}{c}} {cos\theta }&{ - sin\theta }&0\\ {sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) & \(G = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

According to problem, E × F = G

OR, \(\left[ {\begin{array}{*{20}{c}} {cos\theta }&{ - sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\) × \(F = \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\)

Hence it can be seen that product of (E × F) is unit matrix. So F will be the inverse of E.

\(\Rightarrow F = {E^{ - 1}} = \left[ {\begin{array}{*{20}{c}} {cos\theta }&{sin\theta }&0\\ { - sin\theta }&{cos\theta }&0\\ 0&0&1 \end{array}} \right]\)

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