Download LT 10.8 Gaussian Elimination

Mathematical Investigations II
Name
Mathematical Investigations II
Linear Thinking
Gaussian Elimination or Row Reduction
Part I.
Read the example CAREFULLY
Consider how we might solve the following system of equations:

 x  y  z  5
 x  3y  2z  2

 2x  y  z  1
Add equation 1 (or row 1, r1) to equation 2 (or row 2, r2) and copy over the other equations:
[We will denote this by r1+ r2  r2.]
r1  r1
r1  r2  r2
r3  r3

 x  y  z  5
 4y  3z  7

2x  y  z  1
Next, multiply equation 1 (r1) by -2 and add it to equation 3 (r3), copying over the other
equations:
r1  r1
r2  r2
2r1  r3  r3

x  y  z  5
 4y  3z  7

 y  z  9
Next, swap equation 2 and equation 3:
r1  r1
r3  r2
r2  r3

x  y  z  5
 y  z  9

 4y  3z  7
Finally, multiply (equation 2) by (–1):
r1  r1
r2  r2
r3  r3

x  y  z  5
 y  z  9

 4y  3z  7
LinearThinking 10.1
Rev. S06
Mathematical Investigations II
Name
We have used equation 1 (r1), in conjuction with equations 2 and 3 to obtain two new equations
that contain only 2 variables (y and z). Let us now use equation 2 in a similar fashion; that is, to
remove the y variable from equations 1 and 3.
Replace equation 1 with –(equation 2) + (equation 1) and replace (equation 3) with
-4 • (equation 2) + (equation 3) to obtain:
r2  r1  r1
r2  r2
4r2  r3  r3
 x   4

y  z  9

 z  29
Finally replace equation 2 with – (equation 3) + (equation 2) to obtain:
 4
x
y
 20
z  29
Which is the solution to the system?
Describe the general process!
LinearThinking 10.2
Rev. S06
Mathematical Investigations II
Name
Part II.
If you stop to think for a minute, the essential information used to solve the system in Activity I
is contained in the following augmented matrix.
 1 1 1 5


  1 3 2 2
 2 1 1 1


This matrix is obtained from augmenting the matrix of the coefficients with the matrix of the
constant terms. Repeat the operations from the system of equations above to transform that
matrix to the one containing the solution. Remember that in this situation, you must use a ‘0’
wherever the coefficient of a variable is 0.
 1 1 1 5


  1 3 2 2 
 2 1 1 1


1 0 0 4 
 0 1 0 20
0 0 1 29 


This process is called "row reduction" or "Gaussian Elimination".
How can you tell whether or not row reduction is finished?
LinearThinking 10.3
Rev. S06
Mathematical Investigations II
Name
Part III.
Solve the following systems using Gaussian Elimination on the appropriate augmented matrices.
Use the notation described on the previous pages to indicate how you solved each system
(e.g., -r1 + r2  r2).
a)
 x  y  z  1

  x  2z  3

 x  y  3z  1
b)

 x  3y  2z  8
4x  11y  6z  34

 3x  2y  5z  13
c)
 x  y  2z  w  6
 x  y
 w  12
 x  3y  z
 6

 24
 2x  y  z
LinearThinking 10.4
Rev. S06
Mathematical Investigations II
Name
d)

 2x  y  z  3
4x  3y  2z  1
 6x  4y  z  2

f)
What happens when you try to use Gaussian Elimination on the last two sets of systems?
e)

 2x  y  z  3
4x  3y  2z  1
6x  5y  5z  1

Use matrix algebra to solve these systems. What happens?
2 0 5 8 
In problem d), you should have ended up with something like 0 1 4 5;
0 0 0 5


2 0 5 8 
for e) your matrix should look something like 0 1 4 5.
0 0 0 0 


The last equation for d) states that 0x + 0y + 0z = -5, which is impossible. There is no solution to
the system of equations. The system is called inconsistent.
The last equation for e) states that 0x + 0y + 0z = 0, which is true for all (x, y, z). The system has
been reduced from 3 equations in 3 variables to 2 equations in 3 variables. For this system, there
are infinitely many solutions. The system is called dependent.
LinearThinking 10.5
Rev. S06
Mathematical Investigations II
Name
Inconsistent System of Equations
A system of equations that has no solutions.
A system of equations that has at least one solution is called
consistent.
Dependent System of Equations
A system of equations that has infinitely many solutions.
A system of equations that has at most one solution is
called independent.
Thus for any system of linear equations there are three possibilities for the answer:
or
inconsistent
(no solution) = 
independent

usually just called inconsistent.
consistent
(1 unique solution) = 
independent

usually just called independent
consistent
(infinitely many solutions) = 
dependent

usually just called dependent
Part IV.
There is another viable use for Gaussian Elimination.
Recall that the identity matrix for the n x n square matrices is the matrix with ones in the main
diagonal and zeroes everywhere else.
For example the 3 x 3 identity matrix is
1 0 0
0 1 0


0 0 1
The matrix B is called the inverse of A if AB = BA = I (the identity matrix) and it is denoted by
A-1.
Let us explore how we can use Gaussian Elimination to derive the inverse of a 3x3 (or any
square) matrix.
LinearThinking 10.6
Rev. S06
Mathematical Investigations II
Name
Consider the matrix of the coefficients from problem (b) above:
1  3 2 


Let A = 4  11 6 
 3 2  5
Now we want to augment matrix A with the identity matrix to get:
1  3 2 1 0 0


[A|I] = 4  11 6 0 1 0
 3 2  5 0 0 1


Now row reduce the left side (matrix A) of the above 3x6 matrix:
1  3 2 1 0 0


4  11 6 0 1 0 
 3 2  5 0 0 1


Locate A–1.
(You can check this with your calculator.)
LinearThinking 10.7
Rev. S06
Mathematical Investigations II
Name
Use this same process to find the inverse of the following matrices:
b.
1
 1

2

1
1 2 1
1 0 1

1 1 0

3 1 0
d.
2 1  1


4 3 2 
 6 5 5 
a.
1
1 1


  1 0 2 .
 1  1 3
c.
 2 1  1


4 3 2 
 6 4 1 
e.
What happens in the last two cases? How does this relate to your work above?
LinearThinking 10.8
Rev. S06