Question

show that if x^0 is a solution to Ax=b, then rx^0 is a solution to Ax=rb

Answer 1

what do you mean by x^0? this is linear algebra, yes?

Answer 2

yes

Answer 3

r is a constant?

Answer 4

A is a square matrix?

Answer 5

A x b are matrices

Answer 6

if A is a square matrix and Ax=b has a non-trivial solution, i.e., if b = 0, x = 0 is the only solution, then A is invertible.

so \(x=A^{-1}b\) as such, \(rx=rA^{-1}b=A^{-1}rb \Rightarrow Arx=rx \Rightarrow rx \text{ is a solution to } Ax=rb\)

Answer 7

for and non-zero constant, r

Answer 8

"any"

Answer 9

I do not know if A is a square matrix

Answer 10

so A has a pseudo inverse.

Answer 11

what do you know about A?

Answer 12

just matrix

Answer 13

what does x^0 designate?

Answer 14

no there is something wrong with my laptop now
anyway,thanks for ur solution

Answer 15

http://en.wikipedia.org/wiki/Generalized_inverse