Page 252 - Advanced engineering mathematics

232 CHAPTER 7 Matrices and Linear Systems

⎛ ⎞
1 1 −3 In each of Problems 11 through 15, use a matrix inverse to
6. ⎝ 2 16 1 ⎠ find the unique solution of the system.
0 0 4
11. x 1 − x 2 + 3x 3 − x 4 = 1
x 2 − 3x 3 + 5x 4 = 2
⎛ ⎞
−3 4 1
x 1 − x 3 + x 4 = 0
7. ⎝ 1 2 0 ⎠
x 1 + 2x 3 − x 4 =−5
1 1 3
12. 8x 1 − x 2 − x 3 = 4
−2 1 −5
⎛ ⎞ x 1 + 2x 2 − 3x 3 = 0
8. ⎝ 1 1 4 ⎠ 2x 1 − x 2 + 4x 3 = 5
0 3 3 13. 2x 1 − 6x 2 + 3x 3 =−4
−x 1 + x 2 + x 3 = 5
−2 1 1 2x 1 + 6x 2 − 5x 3 = 8
⎛ ⎞
9. ⎝ 0 1 1 ⎠ 14. 12x 1 + x 2 − 3x 3 = 4
−3 0 6
x 1 − x 2 + 3x 3 =−5
−2x 1 + x 2 + x 3 = 0
⎛ ⎞
12 1 14
15. 4x 1 + 6x 2 − 3x 3 = 0
10. ⎝−3 2 ⎟
⎜
2x 1 + 3x 2 − 4x 3 = 0
0 ⎠
0 9 14 x 1 − x 2 + 3x 3 =−7
7.8 Least Squares Vectors and Data Fitting

In this section, we will develop an approach to the method of least squares as it applies to a data
fitting problem.

n
Let A be an n × m matrix of numbers and B a vector in R . The system AX = B may or
∗
may not have a solution. Define an m-vector X to be a least squares vector for the system
AX = B if
∗
AX − B ≤ AX − B (7.2)
for every X in R .
m

Thus X is a least squares vector for AX = B if AX is at least as close to B as AX is to B,
∗
∗
for every m-vector X. This means that, for every X,
AX − B ≤ AX − B .
∗
We will develop a method for finding all least squares vectors for a given system AX = B.
n
The key lies in the column space S of A. S is a subspace of R , spanned by the columns C 1 , ···,
n
C m of A. S consists of exactly those vectors B in R for which the system AX=B has a solution.
This is because, if
⎛ ⎞
x 1
x 2
⎜ ⎟
⎜ ⎟
X = ⎜ . ⎟
.
⎝ . ⎠
x m

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October 14, 2010 14:23 THM/NEIL Page-232 27410_07_ch07_p187-246