Lesson 16 Linear Systems - Bruce E. Shapiro

106 LESSON 16. LINEAR SYSTEMS
⎛
⎝
1 2 3
0 −3 −10
0 4 −1
⎞ ⎛
⎠ ⎝
x
y
z
⎞
⎛
⎠ = ⎝
5
−10
5
Now the first column is all zeroes (except for the first row). The next step is to
subtract a multiple of the second row from the third row to get a zero in the second
entry of the third row. Since the coefficient of y is -3 in the second row and 4 in the
third row, we can add 4/3 times the second row to the third row.
⎛
⎝
1 2 3
0 −3 −10
0 4 + (4/3)(−3) −1 + (4/3)(−10)
⎛
⎝
1 2 3
0 −3 −10
0 0 −43/3
⎞ ⎛
⎠ ⎝
⎞ ⎛
⎠ ⎝
x
y
z
⎞
x
y
z
⎞
⎠ = ⎝
⎛
⎠ = ⎝
⎛
⎞
⎠
5
−10
−25/3
5
−10
5 + (4/3)(−10)
This completes the Gaussian elimination. We can then read off the solution by backsubstitution.
From the third row of the matrix,
From the second row of the matrix,
z = (−25/3)/(−43/3) = 25/43
−3y − 10z = −10
hence
y = − 1 60
(−10 + 10(25/43)) =
3 43
Finally, from the first row, we have
x + 2y + 3z = 5
( ) ( )
60 25
x = 5 − 2 − 3 = 20
43 43 43
We can write a simple recursive algorithm for Gaussian elimination as
Algorithm LinearSolve
Input: A, b
If n > 1,
{A ′ , b ′ } = Reduce(A, b)
LinearSolve (A ′ , b ′ )
End if
x 1 = (b 1 − a 12 x 2 − a 13 x 3 − · · · − a 1n x n )/a 11
⎞
⎠
⎞
⎠
Math 481A
California State University Northridge
2008, B.E.Shapiro
Last revised: November 16, 2011