If two sets of data x1, X2 , xn and Y1, Y2 The average number of YN is x, y respectively. Find a group of new data: ax1 + by1, AX2 + BY2 The average of axn + byn

If two sets of data x1, X2 , xn and Y1, Y2 The average number of YN is x, y respectively. Find a group of new data: ax1 + by1, AX2 + BY2 The average of axn + byn

X1 + x2 +. + xn = nx
Y1 + Y2 +. + yn = NY
A (x1 + x2 +. + xn) = anx
B (Y1 + Y2 +. + yn) = bny
So (ax1 + by1) + (axn + byn) = anx + bny
So the average is ax + by

If two sets of data: x1, X2 , xn and Y1, Y2 The average of, yn is a, B respectively, then a new set of data: 4x1 + Y1, 4x2 + Y2 The average of 4xN + yn is ()
A. 4A B.B c.4a + B D
Teach me to write clearly the reason for each step, please

The sum of the second group of data is 4x1 + Y1 + 4x2 + Y2 + +4xn+yn=4(x1+x2+x3+…… xn)+(y1+y2+y3+…… +Yn) = 4An + BN, then the average is (4An + BN) / N = 4A + B
Choose C

In linear algebra, if B is an invertible matrix, then R (AB) = R (a), why?

Because the rank of matrix is not changed by elementary column transformation
Right multiplication of an invertible matrix is equivalent to a series of elementary column transformations

Linear algebra problem: let matrices A and B and a + B be invertible, prove that a '+ B' is also invertible, and find its inverse matrix (temporarily use a 'to represent the inverse matrix of a, and the others are similar)
The answer to this question should be a (a + b)'b or B (a + b)'a or both. Why

A '+ B' = a '(a + b) B' = B '(a + b) a', so a '+ B' is invertible, and its inverse matrix is the inverse matrix B (a + b) 'a of a' (a + b) B ', or the inverse matrix A (a + b)' B of B '(a + b) a'
So the inverse matrix of a '+ B' is B (a + b) 'a, which can also be written as a (a + b)' B