Posted by Lisa on Wednesday, September 8, 2010 at 4:18am.
For a given square matrix A the predicted values of matrix B are:
predicted B=A(A'A)^(1)A'B
why is the matrix C=A(A'A)^(1)A' an idempotent and symmetric matrix? and is this matrix invertible?

MATHSMatrix  MathMate, Wednesday, September 8, 2010 at 11:26am
Assuming (A'A) is invertible, then (A'A)^{1} exists.
A(A'A)^{1}A'
By the property of inverse of product of matrices,
(A'A)^{1}
=A^{1} A'^{1}
Therefore
C=A(A'A)^{1}A'
=A(A^{1} A'^{1})A'
=(A A^{1}) (A'^{1}A')
= (I) (I)
=I
after application of associativity and the properties of inverse of matrices.
Since I is idempotent and invertible, so is C.

Equation of parabola  Bayarbold, Friday, June 10, 2011 at 11:10am
The equation of the parabola which contains 2 points (1,1) and (2,2) and whose tangent at the point (1,1) has the slope k is y=1/3(A)x^2+1/3(B)x2/3(C) Express A,B and C with k.Please solve this problem.please
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