8. ORTHOGONAL MATRICES :
A square matrix is said to be orthogonal matrix, if AAT = I = AT A
Note that :
(a) If A and B are orthogonal matrix then show that AB is also orthogonal. i.e. if AAT = I and BBT = I then to prove that (AB) (AB)T = I = (AB)T (AB)
(b)
AAT=
If AAT = I, then
i=1å3 a2i = å b2i = å c2i
and
i=1å3 ai bi =å bi ci = å ci ai = 0
Example : If li , mi, ni ; i 1, 2, 3 denote the direction cosines of three mutually perpendicular
vectors in space , prove that the matrix A=
is an orthogonal matrix.
Solution :
Since l1 , m1, n1 ; l2 , m2, n2 and l3 , m3, n3 are the direction cosines of three mutually perpendicular vectors.
Therefore, l12 + m12+ n12 = 1, l22 + m22 + n22 = 1, and l32 + m32 + n32 = 1
L1l2 + m1m2 + n1n2 = 0 , l1l3 + m1m3 + n1n3 = 0
L2l3 + m2m3 + n2n3 = 0
Now, AAT =
Similarly, it can be show that AT A = I
Examples :
1. Find a, b, c x and y if the matrix A given by
is orthogonal.
Solution : It is given that the matrix A is orthogonal. Therefore,
AAT = I
substituting the values of a and b in 3ac + 2x + 2y = 0 and 2c + x + 3 by = 0
we get, c + 2x + y = 0 and 2c + x – 2y = 0
Þ c = –x = 2y
Putting c = 2y and x = – 2y in c2 + x2 + y2 = 1,
we get, y= ± 1/3
Thus we have x= ± 2/3, y =±1/3 and c= ± 2/3
Hence, there are two possible solutions :
I a=1/3 , b=-2/3 , c=2/3 ,x= -(2/3) and y=1/3
II a= 1/3 , b=-(2/3) ,c= -(2/3) ,x= 2/3 and y= -(1/3)
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