Homework SourseConsider the m x n homogeneous sustem of linear equations Ax
Consider the m x n homogeneous sustem of linear equations
Ax = 0
(a) if x = [x1, x2 ... xn]T and y = [y1, y2 ... yn]T are solutions to Ax = 0,
show that z = x + y and w = cx are solutions, where c is an arbitrary scalar
(b) is the result to (a) true when x and y are solutions to the nonhomogeneous system Ax = b? Explain.
Note: the answer to (b) is no... I just don\'t know how to get that or the proof for (a)
Solution
a)
z = x + y = [(x1+y1),(x2+y2),...,(xn+yn)]^T
Ax = 0 => a11x1 + a12x2 + a13x3 + .... + a1nxn = 0
Ay = 0 => a11y1 + a12y2 + a13y3 + .... + a1nyn = 0
Az = a11(x1+y1) + a12(x2+y2) + ... a1n(xn+yn) = (a11x1 + a12x2 + .. + a1nxn) + (a11y1 + a12y2 + ... + a1nyn) = 0
Hence z = (x+y) is the solution to Ax=0
w = cx
Aw = 0 => ca11x1 + ca12x2 + ca13x3 + .... + ca1nxn = c(a11x1 + a12x2 + a13x3 + .... + a1nxn) = c(0) = 0
Hence w=cx is the solution to Ax=0
b)
z = x + y = [(x1+y1),(x2+y2),...,(xn+yn)]^T
Ax = b => a11x1 + a12x2 + a13x3 + .... + a1nxn = b
Ay = b => a11y1 + a12y2 + a13y3 + .... + a1nyn = b
Az = a11(x1+y1) + a12(x2+y2) + ... a1n(xn+yn) = (a11x1 + a12x2 + .. + a1nxn) + (a11y1 + a12y2 + ... + a1nyn) = b + b = 2b
Hence z = (x+y) is not the solution of Ax=b
w = cx
Aw => ca11x1 + ca12x2 + ca13x3 + .... + ca1nxn = c(a11x1 + a12x2 + a13x3 + .... + a1nxn) = c(b) = cb
Hence w=cx is not the solution to Ax=b
![Consider the m x n homogeneous sustem of linear equations Ax = 0 (a) if x = [x1, x2 ... xn]T and y = [y1, y2 ... yn]T are solutions to Ax = 0, show that z = x +](/WebImages/35/consider-the-m-x-n-homogeneous-sustem-of-linear-equations-ax-1104801-1761584504-0.webp "Consider the m x n homogeneous sustem of linear equations Ax = 0 (a) if x = [x1, x2 ... xn]T and y = [y1, y2 ... yn]T are solutions to Ax = 0, show that z = x +")
