(Move to 3 by 3) Forward elimination changes Ax = b to a triangular Ux = c: x+y+z = 5
Chapter 1, Problem 1.5.21(choose chapter or problem)
(Move to 3 by 3) Forward elimination changes Ax = b to a triangular Ux = c: x+y+z = 5 x+2y+3z = 7 x+3y+6z = 11 x+y+z = 5 y+2z = 2 2y+5z = 6 x+y+z = 5 y+2z = 2 z = 2. The equation z = 2 in Ux = c comes from the original x+3y+6z = 11 in Ax = b by subtracting `31 = times equation 1 and `32 = times the final equation 2. Reverse that to recover [1 3 6 11] in [A b] from the final [1 1 1 5] and [0 1 2 2] and [0 0 1 2] in [U c]: Row 3 of h A bi = (`31 Row 1 +`32 Row 2+1 Row 3) of h U ci . In matrix notation this is multiplication by L. So A = LU and b = Lc.
Unfortunately, we don't have that question answered yet. But you can get it answered in just 5 hours by Logging in or Becoming a subscriber.
Becoming a subscriber
Or look for another answer