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Engineering Maths · Q140

Linear Algebra

Engineering and GATE · Engineering Maths · question 140

Q140

For the given orthogonal matrix Q\[ Q = [ array*20c 3/7&2/7&6/7 \\ - 6/7&3/7&2/7 \\ 2/7&6/7& - 3/7 array ]\] The inverse is

For the given orthogonal matrix Q
{\text{Q}} = \left[ {\begin{array}{*{20}{c}} {\frac{3}{7}}&{\frac{2}{7}}&{\frac{6}{7}} \\ { - \frac{6}{7}}&{\frac{3}{7}}&{\frac{2}{7}} \\ {\frac{2}{7}}&{\frac{6}{7}}&{ - \frac{3}{7}} \end{array}} \right]
The inverse is
A.
\left[ {\begin{array}{*{20}{c}} {\frac{3}{7}}&{\frac{2}{7}}&{\frac{6}{7}} \\ { - \frac{6}{7}}&{\frac{3}{7}}&{\frac{2}{7}} \\ {\frac{2}{7}}&{\frac{6}{7}}&{ - \frac{3}{7}} \end{array}} \right]
B.
\left[ {\begin{array}{*{20}{c}} { - \frac{3}{7}}&{ - \frac{2}{7}}&{ - \frac{6}{7}} \\ {\frac{6}{7}}&{ - \frac{3}{7}}&{ - \frac{2}{7}} \\ { - \frac{2}{7}}&{ - \frac{6}{7}}&{\frac{3}{7}} \end{array}} \right]
C.
\left[ {\begin{array}{*{20}{c}} {\frac{3}{7}}&{ - \frac{6}{7}}&{\frac{2}{7}} \\ {\frac{2}{7}}&{\frac{3}{7}}&{\frac{6}{7}} \\ {\frac{6}{7}}&{\frac{2}{7}}&{ - \frac{3}{7}} \end{array}} \right]
Answer
D.
\left[ {\begin{array}{*{20}{c}} { - \frac{3}{7}}&{\frac{6}{7}}&{ - \frac{2}{7}} \\ { - \frac{2}{7}}&{ - \frac{3}{7}}&{ - \frac{6}{7}} \\ { - \frac{6}{7}}&{ - \frac{2}{7}}&{\frac{3}{7}} \end{array}} \right]

Answer: Option C

Solution

Answer: Option C
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