Solutions 221

(b) Let us take the first three columns of A 2 as the basis:

B

"2 13"

1 32

32 1

, N =

"10"

01

_10_

, XB

Xi

X2

Xi

, XN =

X4

X 5

Let XN — 0. Then, BXB — b 2 is solved by LU decomposition as above:

"2

Lc = b 2 & k 10 c 2 = 19 & < => c 2 = 18

ci = 3

r

2

• 1
  ii
  3 7 L
  3 10
  1 00_

Cl"

C2

.C3.

_

" 8"

19

3

=*> C3 — y-

£/x = c<£>

"321"

0 1 s U

3 3

(^00) T
Xi
«2
Xz

3"
18
60

. 7.

=> Xl = --

<=>

^3^10 3

X 2 = f

XS =

XJV = [1,1]

11 16 10]T

3 ' 3 ' 3 J •

"T Then,

If XJV ^ 6, then x# = [• -j±,f,f]T -B^NXN. Let

xB

Xl

X2

.X3.

=

• 11 •
  3
  16
  3
  10

3. 
   

• 
  

_5_

18

_7_

18

_i_

18

1

L J^1

1

_ 6

"-23"

31

19_

Final check:

A 2 x =

A 2 x =

2 13 10

13201

32 110

2 1310

1320 1

32 110

• 11 •
  3
  16
  3
  10
  3
  0
  0_

=

8"

19

3

/

-23

31

19

6

6

19

3

/

(c)

A* =

" 1

4

7

2"

5

8

.10 11.

J A 3

147 10

258 11

AJA 3 =

166 188

188 214

A^A^ is clearly invertible, and (A3A3) l =

107

90

_47

45

47

"45

83

90