Transcript Triangular Form and Gaussian Elimination

Triangular Form and
Gaussian Elimination
Boldly on to Sec. 7.3a…
Hey Bain, what the heck are they???
Well, I’m glad you asked…
Triangular Form – of a system of equations has the leading
term of each equation with coefficient 1, the final equation has
only one variable, and each higher equation has one additional
variable
Example:
x  2y  z  7
y  2 z  7
z 3
Hey Bain, what the heck are they???
Well, I’m glad you asked…
Gaussian Elimination – the process of transforming a system
to triangular form
Steps that can be used in Gaussian Elimination (all of which
produce equivalent systems of linear equations):
1. Interchange any two equations of the system.
2. Multiply (or divide) one of the equations by any nonzero real
number.
3. Add a multiple of one equation to any other equation in the
system.
Back to our original example  Solve by substitution!
x  2y  z  7
y  2 z  7
z 3
x  2y  z  7
y  2  3  7  y  1
z 3
x  2  1  3  7  x  2
Solution:
y  1
x, y, z    2, 1,3

z 3
Another Example – Solve using Gaussian Elimination:
x  2y  z  7
3x  5 y  z  14
2x  2 y  z  3
3E1  E 2
Multiply the first equation by –3 and
add the result to the second equation,
replacing the second equation
x  2y  z  7
y  2 z  7
2x  2 y  z  3
Another Example – Solve using Gaussian Elimination:
x  2y  z  7
y  2 z  7
2x  2 y  z  3
2E1  E3
Multiply the first equation by –2 and
add the result to the third equation,
replacing the third equation.
x  2y  z  7
y  2 z  7
2 y  3 z  11
Another Example – Solve using Gaussian Elimination:
x  2 y  z  7 Multiply the second equation by –2
and add the result to the third equation,
y  2 z  7 replacing the third equation.
2 y  3 z  11
2E 2  E3
x  2y  z  7
y  2 z  7
z 3
 This is our first example!!!
Solve using Gaussian Elimination:
x  3y  z  4
 x  2 y  5z  3
5 x  13 y  13z  8
x  3y  z  4
 y  4z  7
02
E1  E 2
5E1  E3
2E 2  E3
Steps:
This last equation is never true…
 No Solution!!!
Solve using Gaussian Elimination:
2x  y  0
x  3 y  z  3
3y  z  8
E12
x  3 y  z  3
2x  y  0
3y  z  8
x  3 y  z  3 3 E  E x  3 y  z  3
3
2E1  E 2 7 y  2 z  6 7 2
7 y  2 z  6
3y  z  8
13 7  z  74 7
Solution: (x, y, z) = (5/13, 10/13, 74/13)
Solve using Gaussian Elimination:
0.5 x  y  z  w  1
0.5 x  y  z  w  1
 x  y  z  2w  3 E 3  E 2
y  2 w  1
xz 2
xz 2
yw0
yw0
 y  1.5 z  w  0
0.5 x  y  z  w  1
1
w  1
w  1  2 E 3  E1
E 4  E 2
xz 2
xz 2
yw0
yw0
Solution: (x, y, z, w) = (2, 1, 0, –1)