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?

Assuming (A'A) is invertible, then (A'A)-1 exists.

A(A'A)-1A'
By the property of inverse of product of matrices,
(A'A)-1
=A-1 A'-1

Therefore
C=A(A'A)-1A'
=A(A-1 A'-1)A'
=(A A-1) (A'-1A')
= (I) (I)
=I
after application of associativity and the properties of inverse of matrices.
Since I is idempotent and invertible, so is C.

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)x-2/3(C) Express A,B and C with k.Please solve this problem.please

To understand why the matrix C = A(A'A)^(-1)A' is idempotent and symmetric, let's break it down step by step:

1. Idempotency:
A matrix is idempotent if it remains unchanged when multiplied by itself. In this case, let's multiply C by itself:

CC = (A(A'A)^(-1)A')(A(A'A)^(-1)A')
= A(A'A)^(-1)A'A(A'A)^(-1)A'
= A(A'A)^(-1)(A'A)(A'A)^(-1)A'
= A(A'A)^(-1)A'

We can see that CC is equal to C itself. Therefore, the matrix C is idempotent.

2. Symmetry:
A matrix is symmetric if it is equal to its own transpose. In this case, let's check if C = C':

C' = (A(A'A)^(-1)A')'
= (A')'(A(A'A)^(-1))'A'
= AA'(A')'(A'A)^(-1)A'
= A(A'A)^(-1)A'

Since C = C', the matrix C is symmetric.

3. Invertibility:
To determine whether the matrix C is invertible, we need to check if its inverse exists. The inverse of C, denoted as C^(-1), can be found by:

C^(-1) = (A(A'A)^(-1)A')^(-1)
= (A')^(-1)((A'A)^(-1))^(-1)A^(-1)
= A(A'A)A'^(-1)

Since the inverse of C can be found, the matrix C is invertible.

In summary, the matrix C = A(A'A)^(-1)A' is idempotent because CC = C, symmetric because C = C', and invertible because its inverse C^(-1) = A(A'A)A'^(-1) exists.