How to determine #a# and #b# so that the real matrix #((a,1,b),(b,a,1),(1,b,a))# is orthogonal matrix ?
2 Answers
Apr 9, 2017
Explanation:
A matrix
#MM^T = M^TM = I#
Evaluating this directly with the given form of matrix:
#((a,1,b),(b,a,1),(1,b,a))((a,1,b),(b,a,1),(1,b,a))^T#
#= ((a,1,b),(b,a,1),(1,b,a))((a,b,1),(1,a,b),(b,1,a))#
#=((a^2+b^2+1,ab+a+b,ab+a+b),(ab+a+b,a^2+b^2+1,ab+a+b),(ab+a+b,ab+a+b,a^2+b^2+1))#
In order for this to be the identity matrix, we require:
#a^2+b^2+1 = 1#
So if
Then we find:
#ab+a+b=0#
satisfying the requirement that the off diagonal elements be
Apr 9, 2017
See below
Explanation:
Also, an orthogonal matrix has columns and rows that are orthogonal unit vectors.
For the first column vector: