Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Row Reduction, Finished
Recall that we had three row operations
• Row Interchange/Exchange: take two rows and switch their locations
...

• Row Addition: take two rows, add a real multiple of one to the other
...

Those three operations do not change the solution set of a system of linear equations
...
Find the current matrix’s leftmost non-zero column
...

2
...
Use Row Interchange to make that row the
top row
...
Use Row Multiplication to make the top row’s leading value equal to one
...
Use Row Addition to make all values directly below that leading one equal to zero
...
Go back to step 1 using ONLY the rows below the top row
...
Using
the topmost non-zero value in the first column will mean taking that 3, exchanging the top
two rows and dividing by 3
...
There’s
no reason not too, it’s just that the algorithm has to be exact
...

=⇒
 0
= R1  0 1 −2 2
3 
1 −2 2
3 
R4 − 2R1
2 1
4 0 −1
1
0 −1
2 0
Now we ignore the first row, it’s got a leading one, now we just need to put the remainder
of the matrix in order
...
First: divide the third row by -3
for




1 1
1 0 −1
1 1
1 0 −1
 0 1 −2 2
 0 1 −2 2
3 
3 




 0 0
 0 0
2 
0 1
0 1
2 
R4 = R4 + 2R3
0 0
0 1
2
0 0
0 0
0
This is the actual essence of the method:
1
...

2
...

3
...

The official algorithm is just more precise
...

Categorizing Answers
The REF is not optimal to write out the general solution, but it IS good for figuring out
how many solutions you have
...


 0 0
0 1
2 
0 0
0 0
0
These are the key to categorizing the solution
...
The pivots are at the x1 , x2 and x4 columns, meaning we have written it up with
x3 as a free variable and the others as dependent variables
...

• If there is a pivot on the Right Hand Side (the constant column, with no variables
associated with it) then you have a contradiction and no solution
...

• If there is no contradiction and every variable has a pivot then every variable is dependent and you get one, unique, solution
...

Question: did it have to be x3 as a free variable in the previous example? The answer is no,
in fact
...
Notice that x1 and x2 are both related to
x3 ’s value
...


2

Reduced Row Echelon Form:
This is the upgraded version, remember, with zeros above as well as below the pivots
...
Here’s how you get something into RREF:
1
...

2
...

3
...

4
...
Go back to step 2 (selecting the row directly
above)
...
Now to get the general solution (the set of all solutions to the
system)
...
We get

 
 
 

x1
−3t
0
−3
 x2   −1 + 2t   −1   2 

 
=
 

t ∈ R
...

Theorem: a matrix has a single (i
...
, unique) row equivalent RREF
...

As was stated, some questions are about whether there’s a solution (or an infinite number) rather than getting the general solution
...



0
2
4


−1 1
3 x
 1 0 −1 y 
0 2
4 z
is the system to be solved
...

R3 − 2R2
0 0
0 z − 2y − 2x
Notice, this means that the system has a solution if and ONLY if z − 2y − 2x = 0
...

There’s an additional thing we can do here
...
Notice, however, that for the arrangement


 

  
−1
1
3
x
 1  + b  0  + c  −1  =  y 
a
0
2
4
z
the solution set will be a = y + c and b = y + x − 2c, c free (and z − 2y − 2x = 0)
...
For ANY vector in the span of these vectors, there is a way of expressing
it with c = 0, so the vector corresponding to c does not contribute to the span
...

Span




0
2
4
0
2
Example:



 
 
 
1 
−2
−1
 1
Reduce the spanning set for R3 :  −1  ,  1  ,  1  ,  −3  to a basis
...


4

Matrix Operations
Matrix addition and scalar multiplication we have covered (in the Mn×m space)
...

Definition: The Transpose of a matrix, written AT , simply reverses the rows and columns
...

0 −1 −2
3 −2
Definition: Matrix multiplication: it is somewhat complicated
...
The
values of AB are as follows: the ith row, jth column (element i, j, or (AB)i,j ) will have a
value equal to the dot product of the ith row of A and the jth column of B
...



[
]
2 −1
0 −2 −3
3 , B =

...

AB =  (0)(0) + (3)(1)
(4)(0) + (1)(1)
(4)(−2) + (1)(3)
(4)(−3) + (1)(0)
1 −5 −12
[
] [
]
0 + 0 − 12
0−6−3
−12 −9
BA =
=

...

Here are two basic matrices:


1 0 0 ··· 0
 0 1 0 ··· 0


...



...

In =  0 0
...


...


...
Additionally, you have
matrices 0n×m or just 0, which are simply the zero vector from Mn×m
...
Assume that all multiplications,
addition, etc, are defined
...

• (AT )T = A
• (A + B)T = AT + B T
• (AB)T = B T AT
Finally, we get to some exercises
...
2: 1
...
b), 4
...
4: 1
...
bdfg), 4

---