TA: Geoff Williams
Tutorials: Thursday 1:30-2:20 in TSH-120
E-mail:williagg@math.mcmaster.ca
Office hours: Wednesday 1:30-3:30 at the Math Help Centre
Tutorial #3
Transformations
An Example of a (Linear) Transformation:
We have a machine that transforms Lego blocks. That is to say, we put Lego blocks into the
machine and the machine outputs Lego blocks.
If we put in:
If we put in:
1 yellow
1 yellow
1 black
block →
blocks →
it outputs:
1 red
1 blue
blocks
it outputs:
1 red
1 white
2 blue
blocks
Now here’s a question, what will the output be if we put it 3 yellow and 2 black blocks?
Before we answer this question, let’s first improve our notation. So far we have been putting
two types of blocks into the machine (yellow blocks and black blocks) and getting out three
types of blocks (red, white and blue blocks). So this is a transformation from 2-dimensions to
3-dimensions. Let’s call our transformation T. In symbols we write that T is a ransformation
from 2-dimensions to 3-dimensions as:
T : R2 → R3
or for this particular example we can write:
T:
#yellow
# black

#red
→  # white 
#blue

From the first piece of information given above about how our machine works we have that:


1
1
T
= 0 
0
1
and from the second piece of information about our machine we get that:


1
1
T
= 1 
1
2
So if we now try to answer the question of what the output will be if we input it 3 yellow
and 2 black blocks, we may guess that since:
1
3
2
1
0
=3
1
1
+2
we’ll get that
3
2
T


3
= 2 
5
This would be the correct output if in fact our transformation was linear. (As a counterexample if # (white blocks) = # (black blocks)2 , this transformation would not be linear.).
1
But, assuming our transformation is linear, what is T
=?
4
0
is.
To answer this question it would be nice to know what T
1
1
0
Then we will know what both T
and T
are.
0
1
T
0
1
=T
1
1
−
1
0
1
0
1
1
=T
−T
1
0
 
 

1
0
1
= 1 − 0 = 1 
2
1
1

This gives us that:
T
1
4
= 1T
+ 4T
0
1
 



1
1
0
= 1 0  + 4 1  =  4 
1
1
5

Infact using these two basic transformations we can figure out any transformation:
a
1
0
T
= T a
+b
b
0
1




1
0
= a 0  + b 1 
1
1
or written as a matrix
1
T
0

T
0
1

1 0  0 1  a
b
1 1
2
a
b
Section 2.2
Exercise 11



 
2
2
1
Assume that A  −1  = 0 = A  0  and that AX = B has a solution X0 =  −1 .
3
3
2
Find a two parameter family of solutions to AX = B.
Solution:

 

2
1
Let Y1 =  −1  and Y2 =  0 . Then tY1 and sY2 are also solutions to the homogeneous
3
2
system AX = 0, since
A(tY1 ) = t(AY1 ) = t0 = 0
and
A(sY2 ) = s(AY2 ) = s0 = 0
Moreover linear combination of the solutions Y1 and Y2 (i.e. tY1 + sY2 ) will be a two-parameter
family of solutions to AX = 0.
A(tY1 + sY2 ) = A(tY1 ) + A(sY2 ) = 0
But we have to find a solution to the system of equations AX = B. Notice that
A(tY1 + sY2 ) + AX0 = 0 + B = B ⇒ A(tY1 + sY2 + X0 ) = B


  

1
2
2
Hence tY1 + sY2 + X0 = t  −1  + s  0  +  −1  is a two-parameter family of solutions
2
3
3
to the system of equations AX = B.
3
Section 2.3
Exercise 9
In each case either prove an assertion or give an example showing that it is false.
a). If A 6= 0 is a square matrix, then A is invertible.
Solution:
False. Counterexample is:
1 0
0 0
b). If A and B are both invertible, then A + B is invertible.
Solution:
False. Counterexample
is:
2 0
−1
0
A=
and B =
. Then
0 1
0 −1
2 0
−1
0
1 0
A+B =
+
=
0 1
0 −1
0 0
c). If A and B are both invertible, then (A−1 B)T is invertible.
Solution:
True.
[(A−1 B)T ]−1 = [B T (A−1 )T ]−1 = ((A−1 )T )−1 (B T )−1 = AT (B −1 )T = (B −1 A)T
4
Section 2.4
Exercise 8
a). Factor A as a product of elementary matrices. (see also Example 3 p.65 in the book)
1 1
A=
2 1
Solution:
The reduction of A → I is as follows:
1 1 R2 −2R1
1
1
−−−−→
2 1
0 −1
1 0
E1 =
−2 1
1 1
−R2
−−→
0 1
1
0
E2 =
0 −1
1 0
R1 −R2
−−−−→
0 1
1 −1
E3 =
0
1
Here E3 E2 E1 A = I ⇒ A−1 = E3 E2 E1 ⇒ A = (E3 E2 E1 )−1 = E1−1 E2−1 E3−1 =
1 0
1
0
1 1
=
2 1
0 −1
0 1
Notes typed by Olga Krylova and Geoff Williams.
5