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find the det(A), where A is a square matrix satisfying the property that A(transpose)A= I
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\[A^{t}A=I\]
for square matrices DET(AB)=Det(A)Det(B)
I guess I should add that Det(I)=1 and Det(A)=Det(A^T)
so i have \[\det(A ^{t} A)=\det(I)\] \[\det(A ^{t}) \det(A) =\det(I)\]
\[\det(A ^{t}) \det(A) =\det(I)=1\] \[\det(A) \det(A) =1\] \[\det(A)^2=1\] \[\det(A)=\pm\sqrt{1}=\pm1\]
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