(This is written in Java script; it works on the computer I'm using now; I don't
know if it works on all computers. Maybe if you are better at Java script than me
you can help me write the code in a better way.)
Can you get them all turned on at once, and then all off again. Does it make any
difference which order you make your clicks? Are there different ways to get to
all on or all off?
What if we start from a different position?
What if we add more connections?
What if we start from a different graph?
What about this?
Or this:
Or this:
Can you solve the puzzle? How is linear algebra involved?
The answer is that you need to find the inverse of the matrix of the connections.
How this worked was explained in class. We also had to use "mod 2" arithmetic, that
is 1+1=0 when we're thinking about 1 meaning switch once; switching twice is the
same as switching zero times.
For example, for the first of the above graphs, we can lable the lights a,b,c,d:
We start with them all off, so if s
_{a} is the state of a, s
_{b}
the state of b, etc, and the state can be either 0 for off or 1 for on, then we
start with:
(s_{a},s_{b},s_{c},s_{d})=(0,0,0,0)
We want to end up with
(s_{a},s_{b},s_{c},s_{d})=(1,1,1,1)
So the change we want to make is (1,1,1,1) (meaning change everything). (If we'd
started with (1,1,0,0), we'd only want to make the change (0,0,1,1), meaning only
change the last two).
If A is the matrix of the graph, which describes how clicking one button affects
all the buttons, then if v is a vector describing which buttons you click, eg (1,0,0,1)
would mean click a and d, but not b and c, then the change you get will be the vector
Av. We want
Av=(1,1,1,1)
So v will be given by
v=A^{1}(1,1,1,1)
(note, the vectors should have been drawn vertically).
In our case, A is given by
/

1

1

1

0

\



1

1

0

1





1

0

1

1



\

0

1

1

1

/

So to find how to make whatever change in the lights you want, you just need to
be able to find the inverse of A, using mod 2 arithmetic. (Which is like ordinary
arithmetic, except whenever you get an even number, you write 0 instead, and odd
numbers can always be replaced by 1.)
This is a very brief outline of the solution. For more details, see the following
references:
 Problem solving using geometry games, by Ole Bjorkqvist. (This is probably the best
reference for an introductory approach; unfortunately I'm not sure where I got it
from.)
 The sigma game and Cellular Automata, by Klaus Sutner, American Math Montly, Jan
1990, Vol 97, #1
 Linear cellular automata and the Garden of Eden, by Klaus Sutner, The Mathemaitcal
Intelligencer Vol. 11, No. 2, 1989
Assignment
To be handed in on Tuesday 3rd June.
 section 6.1: 6, 16, 20, 22, 24, 26, 32
 section 6.2: 8, 10, 16
 section 6.3: 2, 4, 6, 12, 22
What will be on the quiz
On Thursday 29th of May there will be a quiz, with a question like the following:
Do the vectors (1,0,5), (1,2,5) and (1,3,1) span R
^{3}?
Find the inverse of an appropriate matrix to find the solutions to the following
problems:
a_{1}(1,0,5)+a_{2}(1,2,5)+a_{3}(1,3,1)=(1,0,0)
b_{1}(1,0,5)+b_{2}(1,2,5)+b_{3}(1,3,1)=(0,1,1)
c_{1}(1,0,5)+c_{2}(1,2,5)+c_{3}(1,3,1)=(0,0,2)
Solution:
To ask if we can span R
^{3} with these vectors means can we get to any point
in R
^{3} using them, eg, how can we get to (x,y,z) for any choice of x,y,z.
This means asking if the following vector equation has a solution:

/

1

\



/

1

\



/

1

\



/

x

\


a



0



+

b



2



+

c



3




=



y





\

5

/



\

5

/



\

1

/



\

z

/


Solving this is the same as solving the matrix vector equation:
/

1

1

1

\

/

a

\


/

x

\



0

2

3





b



=



y



\

5

5

1

/

\

c

/


\

z

/

if
a=(a,b,c), and
x=(x,y,z) (strictly speaking, I should say
a
is the transpose of (a,b,c), to make it vertical) (note,
a is not the same
as a  one is a vector, the other is a number). Suppose the matrix is called A.
The the above equation is
Aa=x
This equation tells us how to get from knowing (a,b,c) to knowing (x,y,z). But we
want to know the reverse. We are told (x,y,z), and we want to find (a,b,c). The
transformation that will do that for us, if it exists, will be the inverse of A,
and we'll have
a=A^{1}x
So, if we can find an inverse for A, then the vectors do span R
^{3}, and
then we can use A
^{1} to solve the problems of how to get to the three
given vectors, (1,0,0), (0,1,1), and (0,0,2).
The inverse is found as follows, using row reduction, as described in class:
/

1

1

1



1

0

0

\


/

1

1

1



1

0

0

\


/

1

1

1



1

0

0

\




0

2

3



0

1

0






0

2

3



0

1

0






0

2

3



0

1

0




\

5

5

1



0

0

1

/

R_{3}5R_{1}

\

0

0

4



5

0

1

/

R_{3}/ 4

\

0

0

1



5/4

0

1/4

/


/

1

1

1



1

0

0

\


/

1

1

1



1

0

0

\

R_{1}R_{3}

/

1

1

0



1/4

0

1/4

\


R_{2}3R_{3}



0

2

0



15/4

1

3/4



R_{2}/2



0

1

0



15/8

1/2

3/8






0

1

0



15/8

1/2

3/8





\

0

0

1



5/4

0

1/4

/


\

0

0

1



5/4

0

1/4

/


\

0

0

1



5/4

0

1/4

/

R_{1}R_{2}

/

1

0

0



13/8

1/2

1/8

\




0

1

0



15/8

1/2

3/8




\

0

0

1



5/4

0

1/4

/

so the inverse matrix is:
/

13/8

1/2

1/8

\



15/8

1/2

3/8



\

5/4

0

1/4

/

(Check with MatLab!)
Now we can solve the problems:
/

13/8

1/2

1/8

\

/

1

\


/

13/8

\



15/8

1/2

3/8





0



=



15/8



\

5/4

0

1/4

/

\

0

/


\

5/4

/

/

13/8

1/2

1/8

\

/

0

\


/

3/8

\



15/8

1/2

3/8





1



=



1/8



\

5/4

0

1/4

/

\

1

/


\

1/4

/

/

13/8

1/2

1/8

\

/

0

\


/

1/4

\



15/8

1/2

3/8





0



=



3/4



\

5/4

0

1/4

/

\

2

/


\

1/2

/

The solutions are
a
_{1}=13/8, a
_{2}=15/8, a
_{3}=5/4, b
_{1}=3/8,
b
_{2}=1/8, b
_{3}=1/4, c
_{1}=1/4, c
_{2}=3/4, c
_{3}=1/2,
Lecture 6 Back to the
Linear Algebra Notes Index.
Lecture 8