Transcript Echelon Method - EGG Math Help

Echelon Method
Problem 2.1 # 23
Prepared by E. Gretchen Gascon
Step 1: The problem
r1  x  y  z  2
r 2  2x  y  z  5
r 3  x  y  z  2
Step 2: Eliminate the x term
Use equation 1 and 2
r1  x  y  z  2
r 2  2x  y  z  5
Use the
multiplication rule
to get the
coefficients of the
x term opposites
Next Add the two
equations to get a
new equation.
2(r1)  2 x  2 y  2 z  4
r 2  2 x  y  z  5
r 4  y  3z  1
Remember this equation it will be
used later
Step 2: Eliminate the x term (cont.)
Use equation 2 and 3
r2  2x  y  z  5
r 3  x  y  z  2
Use the
multiplication rule  r 2  2 x  y  z  5
to get the
coefficients of the 2( r 3)  2 x  2 y  2 z  4
x term opposites
r 5  3 y  3z  9
Next Add the two
equations to get a
new equation.
Remember this equation it will be
used later
Step 3: Use the two equations to
eliminate the y term
r 4  y  3z  1
r 5  3 y  3z  9
Use the
multiplication rule
to get the
coefficients of the
y term opposites
Next Add the two
equations to get a
new equation.
3(r 4)  3 y  9 z  3
r 5  3 y  3z  9
r 6  12 z  12
You are now ready to create this new set of equations
The new system of equations
r1  x  y  z  2
r 4  y  3z  1
r6 
12 z  12
One equations has 3
variables, one has 2
variables, and one
equation has only 1
variable.
Step 6
Solve the last equation (the equation with only one variable)
r 6 /12 
12 z  12 
z  1
Step 7: Back Substitution
r1  x  y  z  2
r 4  y  3z  1
r6 
z  1
Start with the last equation
(the simplest one)
r4 
y  3 z  1
y  3( 1)  1
y  3  1
y2
r1  x  y  z  2
x  2  ( 1)  2
x  2 1  2
x 1  2
x 1
Ans. (1, 2, 1)
There is more than one way to solve
this problem
The next slides will assume that you
worked with r1 and r2, then when to use
r1 and r3. The work looks a bit different,
however the end results are the same.
(alternative solution)
Step 2: Eliminate the x term (cont.)
As demonstrated previous in step 2(slide #4) , you can use equation 2 and
3, however, you could have used equations 1 and 3 as well.
This example uses 1 and 3
r1  x  y  z  2
r 3  x  y  z  2
Multiply the r3 by
-1 to get the
coefficients as
opposites
Next Add the two
equations to get a
new equation.
r1  x  y  z  2
r 3   x  y  z  2
2y  4
Notice that this equation can be solved for y already without further
adding of equations .
Same Solutions
2y  4
y2
r 4  y  3 z  1
2  3 z  1
3 z  3
z  1
r1  x  y  z  2
x  2  (1)  2
x  2 1  2
x 1  2
x 1
Ans. (1, 2, 1)
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