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A First Course in Linear Algebra, 2017a

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545<br />

2.2.10 An LU factorization of the coefficient matrix is<br />

[ ] [ 1 2 1 0<br />

=<br />

2 3 2 1<br />

][ 1 2<br />

0 −1<br />

]<br />

<strong>First</strong> solve [ ][ 1 0 u<br />

2 1 v<br />

[ ] [ ]<br />

u 5<br />

which gives = . Then solve<br />

v −4<br />

[ ][<br />

1 2 x<br />

0 −1 y<br />

which says that y = 4andx = −3.<br />

] [ ] 5<br />

=<br />

6<br />

] [<br />

=<br />

5<br />

−4<br />

]<br />

2.2.11 An LU factorization of the coefficient matrix is<br />

⎡ ⎤ ⎡<br />

1 2 1 1 0 0<br />

⎣ 0 1 3 ⎦ = ⎣ 0 1 0<br />

2 3 0 2 −1 1<br />

<strong>First</strong> solve<br />

⎡<br />

1 0 0<br />

⎣ 0 1 0<br />

2 −1 1<br />

which yields u = 1,v = 2,w = 6. Next solve<br />

This yields z = 6,y = −16,x = 27.<br />

⎡<br />

⎣<br />

1 2 1<br />

0 1 3<br />

0 0 1<br />

⎤⎡<br />

⎦⎣<br />

⎤⎡<br />

⎦⎣<br />

2.2.13 An LU factorization of the coefficient matrix is<br />

⎡ ⎤ ⎡<br />

1 2 3 1 0 0<br />

⎣ 2 3 1 ⎦ = ⎣ 2 1 0<br />

3 5 4 3 1 1<br />

<strong>First</strong> solve<br />

⎡<br />

Solution is: ⎣<br />

u<br />

v<br />

w<br />

⎤<br />

⎡<br />

⎦ = ⎣<br />

5<br />

−4<br />

0<br />

⎤<br />

⎡<br />

⎣<br />

⎦. Nextsolve<br />

⎡<br />

⎣<br />

1 0 0<br />

2 1 0<br />

3 1 1<br />

1 2 3<br />

0 −1 −5<br />

0 0 0<br />

⎤⎡<br />

⎦⎣<br />

u<br />

v<br />

w<br />

x<br />

y<br />

z<br />

u<br />

v<br />

w<br />

⎤⎡<br />

⎦⎣<br />

⎤<br />

⎤<br />

⎤⎡<br />

⎦⎣<br />

⎡<br />

⎦ = ⎣<br />

⎡<br />

⎦ = ⎣<br />

⎤<br />

⎤⎡<br />

⎦⎣<br />

⎦ = ⎣<br />

x<br />

y<br />

z<br />

⎤<br />

1<br />

2<br />

6<br />

1 2 1<br />

0 1 3<br />

0 0 1<br />

1<br />

2<br />

6<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

1 2 3<br />

0 −1 −5<br />

0 0 0<br />

⎡<br />

⎡<br />

⎦ = ⎣<br />

5<br />

6<br />

11<br />

⎤<br />

⎦<br />

5<br />

−4<br />

0<br />

⎤<br />

⎦<br />

⎤<br />

⎦<br />

⎤<br />

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