(1) We have given:
Which can be written as: where, and .
Multiplying both side of Ax = b by we get:
.
(2)
We have given which can be written as where and .
Multiplying both side of Ax = b by we get:
solving this system of two equation in two unknows we get:
.
(1 point) Find the least-squares solution x* of the system [10] 0 1 x= 4 Loo]...
1 2-4 6-2 7 (1 point) Find the least-squares solution î of the system 6-6 2 ( -3 2 5 3
Find the least squares solution to the inconsistent system Ax = b 1 1 0 110 3 A = b 101 101 2 8 5 O a. 1 8=t-1 tec 0 Ob. 5 8=t tec Ос. X 5 -3 0 O d. 2. 7 X = t -3 11 + 00 tec 1 O e. 5 X = t -3 tec Of. X =
Find the least squares solution to the inconsistent system a. b. c. d. e. f. Find the least squares solution to the inconsistent system Ax = b 110 110 A= b = 101 [101 1 38 2 Oa. -1 2= 1 1 Ob. 5 X= -3 0 C. 5 X=t + 3 tec 1 0 O d. 7 2 + 8 tec X=t| -3 11 -6 1 5 X=t -1 -3 tec + -1 0 O f. 5 - X=t|...
Find the least squares solution to the inconsistent system Ax = b 1 1 0 110 3 A = b = 101 8 1101 2
12 === 9 1 4 4. Find the least-squares solution of A7 = 5 where A = 0 1 and 6 = 1 0 Hint: Solve the normal equations AT AT = ATV.
Compute the least-squares error associated with the least-squares solution x of Ax = b 1 -3 2 194 139 -1 3 1 A= b= X = 0 2 -4 6 139 3 7 5 The least squares error is (Type an exact answer, using radicals as needed.)
Find the least squares solution to the inconsistent Ax = b system A= 110 110 101 101. a. x = t| -3 0 b. 1 5 tec 2-[i][3cc d. 7 8 tec X=t-3 + 11 -6 e. x = f. 5 X 0
Find the least squares solution to the inconsistent system Ax = b 3 A = 1 1 0 1 1 0 1 0 1 101] b = 8 2
Find the least squares solution to the inconsistent system Ax = b 110 110 3 A = b = 101 8 101 O a. 5 x = t -3 tec 5 Ob. 1 X = t -1 + -3 tec 0 2 Ос. 7 X = t -3 11 + police tec o d. -1 X = O e. 5 -3 X = Of. X = t 1 5 +-3 0 1
Find the least squares solution to the inconsistent system Ax = b 110 110 3 A = b = 101 8 101 O a. 5 x = t -3 tec 5 Ob. 1 X = t -1 + -3 tec 0 2 Ос. 7 X = t -3 11 + police tec o d. -1 X = O e. 5 -3 X = Of. X = t 1 5 +-3 0 1