(New page: << Back to Homework 3 Homework 3 Ben Horst: A :: B :: C ----)
 
m (3)
 
(5 intermediate revisions by the same user not shown)
Line 3: Line 3:
 
Homework 3 Ben Horst:  [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]]  ::  [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]]  ::  [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]]
 
Homework 3 Ben Horst:  [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]]  ::  [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]]  ::  [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]]
 
----
 
----
 +
==1==
 +
Bob can find the inverse matrix, and multiply the encrypted message by it to find the intended message.
 +
 +
==2==
 +
Since Eve has intercepted several pairs on in/out, and conveniently enough, the output forms a basis, she can decode the message easily.
 +
 +
The intercepted vectors <2,0,0>, <0,1,0>, <0,0,3> form a basis for the decoding (as a linear combination, they can form any other vector, thus any other vector may be written as a sum of these three).
 +
 +
==3==
 +
Using this method, <2,23,3> must be a combination of <2,0,0>, <0,1,0>, and <0,0,3>.
 +
 +
It is simple to find that combination: 1*<2,0,0> + 23*<0,1,0> + 1*<0,0,3> = <2,23,3>.
 +
 +
We use those coefficients on the input (linearity) to find the output:
 +
 +
1*<1,0,4> + 23*<0,1,0> + 1*<1,0,1> = <2,23,5>  which corresponds to [b,w,e] .

Latest revision as of 12:46, 19 September 2008

<< Back to Homework 3

Homework 3 Ben Horst: A  :: B  :: C


1

Bob can find the inverse matrix, and multiply the encrypted message by it to find the intended message.

2

Since Eve has intercepted several pairs on in/out, and conveniently enough, the output forms a basis, she can decode the message easily.

The intercepted vectors <2,0,0>, <0,1,0>, <0,0,3> form a basis for the decoding (as a linear combination, they can form any other vector, thus any other vector may be written as a sum of these three).

3

Using this method, <2,23,3> must be a combination of <2,0,0>, <0,1,0>, and <0,0,3>.

It is simple to find that combination: 1*<2,0,0> + 23*<0,1,0> + 1*<0,0,3> = <2,23,3>.

We use those coefficients on the input (linearity) to find the output:

1*<1,0,4> + 23*<0,1,0> + 1*<1,0,1> = <2,23,5> which corresponds to [b,w,e] .

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett