Q150
Given an orthogonal matrix \[ A = [ array*20c 1&1&1&1 \\ 1&1& - 1& - 1 \\ 1& - 1&0&0 \\ 0&0&1& - 1 array ],\, [ A A^ T ]^ - 1\, is\]
Given an orthogonal matrix {\text{A}} = \left[ {\begin{array}{*{20}{c}}
1&1&1&1 \\
1&1&{ - 1}&{ - 1} \\
1&{ - 1}&0&0 \\
0&0&1&{ - 1}
\end{array}} \right],\,{\left[ {{\text{A}}{{\text{A}}^{\text{T}}}} \right]^{ - 1}}\,{\text{is}}
A.
\left[ {\begin{array}{*{20}{c}}
{\frac{1}{4}}&0&0&0 \\
0&{\frac{1}{4}}&0&0 \\
0&0&{\frac{1}{2}}&0 \\
0&0&0&{\frac{1}{2}}
\end{array}} \right]
B.
\left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}}&0&0&0 \\
0&{\frac{1}{2}}&0&0 \\
0&0&{\frac{1}{2}}&0 \\
0&0&0&{\frac{1}{2}}
\end{array}} \right]
C.
\left[ {\begin{array}{*{20}{c}}
1&0&0&0 \\
0&1&0&0 \\
0&0&1&0 \\
0&0&0&1
\end{array}} \right]
AnswerD.
\left[ {\begin{array}{*{20}{c}}
{\frac{1}{4}}&0&0&0 \\
0&{\frac{1}{4}}&0&0 \\
0&0&{\frac{1}{4}}&0 \\
0&0&0&{\frac{1}{4}}
\end{array}} \right]
Answer: Option C
Solution
Answer: Option C
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