Correction: For #3a) x+2y+3z = 6 ------------------> x+2y+z = 6 I would apreciat
ID: 2971160 • Letter: C
Question
Correction: For #3a) x+2y+3z = 6 ------------------> x+2y+z = 6
I would apreciate a detailed answer for part a.
The system of equations x + 2y + 3z = 6.2x + 5y + 3z = 15 has general solution = (z, 3 - z, z)T where z is a free variable. Show the 3 ways to find the minimum-length solution. The system of equations x+y = 2,x + 2y - l,x + 3y = 2 has no solutions, but you can try to minimize the sum of the "squared errors" E(x, y) = (x + y - 2)2 + (x + 2y - l)2 +(x + 3y -2)2. Use the calculus method of setting E/dx and E/dy both = 0 and solving for x,y. Also use the Normal Equation method of solving ATAx = ATb and check that they give the same equations.Explanation / Answer
(a)
1st method:
Let length^2 be denoted by L
L = z^2 + (3-z)^2 + z^2
= 3 z^2 - 6z + 9
dL/dz = 6z - 6 = 0
hence z = 1
thus the minimum length solution of the original system of equations is
(1, 2, 1)^T
2nd method:
A =
| 1 2 1 |
| 2 5 3 |
A *A tranpose =
| 6 15 |
| 15 38 |
hence (u v)^T = ( 1 0)^T
hence minimum length solution =
| 1 2 | * | 1 |
| 2 5 | | 0 |
| 1 3 |
= ( 1 2 1) transpose
3rd method:
pseudoinverse A =
| 8/3 -1 |
| 1/3 0 |
| -7/3 1 |
hence minimum length solution =
| 8/3 -1 | * | 6 |
| 1/3 0 | | 15 |
| -7/3 1 |
= ( 1 2 1) transpose