Further pure mathematics 3 matrix algebra

Philosophy of the course
• The course does not aim to present a teaching plan for any of
the FP3 topics. This is left to your professional judgement as a
teacher.
• The course is more about presenting each topic in a way that
provides insights into the topic beyond the syllabus.
• In general the course will present the necessary knowledge and
skills to teach the Edexcel FP3 module effectively.
• Relevant examination problem solving techniques will be
demonstrated.

Session structure
• Section 1: Introduction to matrices.
• Section 2: Linear transformations represented by
matrices.
• Section 3: The transpose and inverse of a matrix.
• Section 4: Eigenvalues and eigenvectors.
• Section 5. Symmetric matrices and their
reduction to diagonal form.
• Section 6. Typical examination questions.

•Section 1: Introduction to
matrices.

In mathematics a matrix is a rectangular array of
numbers which represents information (such as routes in
a network or rules to effect a transformation in space).
The array is presented within brackets.
Examples of matrices are:
What is a matrix?

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
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
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










0
1
4
1
2
1
3
0
2
,
1
3
,
5
4
3
2
This has 2 rows and 2
columns. It is called a
2  2 matrix.
This has 2 rows and 1
column: a 2  1 matrix
3 rows and 3 columns: a
3  3 matrix.
The entries of a matrix
are referred to by their
(row, column) position.
This is the (1, 2) entry.

Consider the simultaneous equations:
Another (matrix) way of representing these equations is:
And similarly for the second row:
Such a matrix representation simplifies the study of simultaneous equations
(especially when the number of variables is larger than 2) and, as we’ll see
shortly, also helps us to represent and study transformations of the plane.
How can matrices arise in the classroom?

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

32
18
5
4
3
2
y
x
32
5
4
18
3
2




y
x
y
x
  y
x
y
x
y
x




















3
2
3
2
3
2
:
follows
as
column
with the
matrix
the
of
)
(
row
first
he
multiply t
we
LHS
on the
product
he
multiply t
To
  y
x
y
x












5
4
5
4
This following slides were presented in the session on FP1 matrices.

Let T a 90o anticlockwise turn in the x – y plane about
the origin.
How can matrices arise in the classroom?
1
0
to
by
mapped
gets
0
1
point
The 
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
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

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

T
0
1
to
by
mapped
gets
1
0
point
The 
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


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
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


T
to
by
mapped
is
at
vertex
one
with
rectangle
blue
the
Now
T
y
x




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


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







y
x









x
y
at
vertex
one
with
rectangle
red
..the 







x
y




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


























y
x
y
x
x
y
y
x
T
0
1
1
0
:
means
This  
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
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

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
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

 

y
x
0
1
1
0
T
under
0
1
of
Image 







T
under
1
0
of
Image 








• The earliest evidence of the use of matrices to study simultaneous equations occurs in
the Jiuzhang suanshu or Nine Chapters on the Mathematical Art . This Chinese text
dates from between 206 BC to 50 AD.
• This set of simultaneous equation problem appears in the Jiuzhang suanshu:
• “There are three types of corn, of which three bundles of the first, two of the second, and
one of the third make 39 measures. Two of the first, three of the second and one of the
third make 34 measures. And one of the first, two of the second and three of the third
make 26 measures. How many measures of corn are contained of one bundle of each
type?”
• In algebraic form this is:
• In the Jiuzhang suanshu this is represented in a equivalent form as:
• The text then proceeds to solve the equations using a method identical to the
Gaussian elimination of the late 17 century.
History of the use of matrices to represent simultaneous equations.
26
3
2
1
34
1
3
2
39
1
2
3









z
y
x
z
y
x
z
y
x
39
34
26
1
1
3
2
3
2
3
2
1
The use of matrices to represent transformations is relatively new. Jan de Witt, a
Dutch mathematician, used them in his 1659 text Elements of Curves.

•Section 2: Linear
transformations represented by
matrices.

•The study of linear transformation is
essentially an extension of the FP1 topic
now extended to 3  3 real matrices.
However it will be instructive to review
the earlier exposition in the following
slides.

The transformation: a 90o anticlockwise rotation about the origin.

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






y
x
vector
a
Consider 
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
y
x
    )
1
.(
..........
:
is
rule
then the
,
write
we
If u
kT
u
k
T
y
x
u 




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



:
to
maps 
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x
y
y
x
T

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x
y
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


y
x
k
vector
enlarged
he
consider t
Now 
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
y
x
k
:
to
maps
evidently 
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x
y
k
y
x
k
T

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

x
y
k
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

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

y
x
kT
x
y
k
y
x
k
T
Thus

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

y
x
1
1
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y
x
y
x
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
1
1
y
x
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
y
x
T
1
1

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y
x
T
1
1
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y
x
y
x
T
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
1
1
and
vectors
wo
Consider t
y
x
y
x
:
shown
ram
parallelog
blue
the
of
vertex
4
on the
lies
sum
Their th
1
1
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
y
x
y
x
:
under
and
,
of
images
the
are
ram
parallelog
red
of
vertices
The
1
1
1
1
T
y
x
y
x
y
x
y
x
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.
dimensions
3
with
space
real
to
applies
also
above
analysis
The
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
1
1
1
1
:
have
must
we
s
coordinate
of
addition
for
rule
ram
parallelog
By the
y
x
T
y
x
T
y
x
y
x
T
      )
2
.(
..........
is
rule
then the
,
write
we
If
1
1
v
T
u
T
v
u
T
y
x
v
and
y
x
u





















The transformation: a 90o anticlockwise rotation about the origin.
:
ram
parallelog
red
the
onto
ram
parallelog
blue
the
maps
T

Linear transformations of real 3 dimensional space.
numbers.
real
are
and
i.e.
short)
for
space
3
or
(
dimensions
3
with
space
real
in the
;
vectors
any two
and
scalar
any
Consider
1
1
1
3
1
1
1
z
, y
z, x
y,
x,
z
y
x
v
z
y
x
u
k























   
     
v
T
u
T
v
u
T
ii
u
kT
u
k
T
i
T




)
)
:
conditions
two
the
satisfies
that
spaces
3
or
2
of
mation
A transfor
n.
nsformatio
linear tra
a
called
is
n.
nsformatio
linear tra
a
is
origin
about the
space
l
dimensiona
2
of
rotation
ise
anticlockw
90
the
Thus o

Representing a linear transformation of 3 space by a 3  3 matrix.
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

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











z
y
x
T
z
y
x
T 0
0
0
space.
3
in
any vector
be
Let
plane.
the
of
n
nsformatio
linear tra
a
be
Let










z
y
x
T

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

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







z
T
y
x
T 0
0
0
0
0
0
.
1
0
0
,
0
1
0
,
0
0
1
suppose
And
3
3
3
2
2
2
1
1
1






















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









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



























c
b
a
T
c
b
a
T
c
b
a
T
......
3
2
1
3
2
1
3
2
1
































z
y
x
c
c
c
b
b
b
a
a
a
z
y
x
T

































z
T
y
T
x
T 0
0
0
0
0
0
:
imply
2)
and
1)
Rules

































1
0
0
0
1
0
0
0
1
zT
yT
xT

































3
3
3
2
2
2
1
1
1
c
b
a
z
c
b
a
y
c
b
a
x

















3
2
1
3
2
1
3
2
1
zc
yc
xc
zb
yb
xb
za
ya
xa
.
of
matrix
the
is
3
2
1
3
2
1
3
2
1
T
c
c
c
b
b
b
a
a
a











Linear transformation terminology
space.
3
in
any vector
be
Let
plane.
the
of
n
nsformatio
linear tra
a
be
Let










z
y
x
T
T
z
y
x
c
b
a
c
b
a
z
y
x
T under
of
the
is
say that
we
If









































image


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.
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are
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vectors
the
of
images
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i.e.
of
matrix
the
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Linear transformations. Exercise.
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i
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2
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of
ations
transform
following
that the
Show
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Exercise solutions
space.
3
in
vectors
any two
be
and
let
and
scalar
any
be
Let
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z
y
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T

The matrix of a linear transformation. Example.

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x
y
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T 
:
n
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linear tra
the
of
matrix
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Find
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1
1
1
0
1
0
0
0
1
:
c
b
a
T
Solution
;
0
0
1
0
1
0
2
2
2
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c
b
a
T
.
1
0
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1
0
0
3
3
3
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b
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T
.
of
matrix
the
is
1
0
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2
1
3
2
1
3
2
1
T
c
c
c
b
b
b
a
a
a
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The matrix of a linear transformation. Exercise.


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z
y
x
z
y
x
T
3
2
:
n
nsformatio
linear tra
the
of
matrix
the
Find 

The matrix of a linear transformation. Exercise solution

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z
y
x
z
y
x
T
3
2
:
n
nsformatio
linear tra
the
of
matrix
the
Find 
;
0
0
1
0
0
1
:
1
1
1
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c
b
a
T
Solution
;
0
2
0
0
1
0
2
2
2
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c
b
a
T
.
3
0
0
1
0
0
3
3
3
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c
b
a
T
.
of
matrix
the
is
3
0
0
0
2
0
0
0
1
3
2
1
3
2
1
3
2
1
T
c
c
c
b
b
b
a
a
a



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
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












•Section 3: The determinant,
transpose and inverse of a
matrix.

• This section is again essentially an extension
of the FP1 topic on now the inverse of a 3  3
real matrix. However a new concept of the
transpose of a matrix will be needed to
compute the inverse matrix.
• While the exposition for determinant in the
2  2 is easy to understand that of the 3  3
case is beyond the scope of this course
(essentially being a second year undergraduate
pure mathematics module).
• One may therefore ask the question as to why
this particular section is included in the FP3
module.

What is a determinant? The 2  2 case.
T2


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d
b
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a
M
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case.
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to
go
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this
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1
1
and
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0
,
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,
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ices
with vert
square
the
:
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to
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1
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1
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1
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By
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c
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b
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a
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square
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of
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ed
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Note that if a matrix A has det A = 0 then it cannot have an inverse
and is called singular. If det A ≠ 0 then A is called non-singular.
The rule for computing the inverse of a 2  2 matrix.

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providing
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1
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:
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3
a
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The 

The inverse of a 3  3 matrix. Example 1.
.
2
5
1
1
1
2
2
3
1
matrix
the
of
inverse
the
Determine
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A
:
are
row
by
entries
nine
The
:
adj
find
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: A
Solution
      11
5
1
1
2
det
1
entry
3)
(1,
;
5
2
1
1
2
det
1
entry
2)
(1,
;
3
2
5
1
1
det
1
entry
1)
(1,
4
3
2

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
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
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
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


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
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
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






      8
5
1
3
1
det
1
entry
3)
(2,
;
4
2
1
2
1
det
1
entry
2)
(2,
;
4
2
5
2
3
det
1
entry
1)
(2,
5
4
3



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




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

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




      5
1
2
3
1
det
1
entry
3)
(3,
;
3
1
2
2
1
det
1
entry
2)
(3,
;
1
1
1
2
3
det
1
entry
1)
(3,
6
5
4


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
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
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
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.
5
3
1
8
4
4
11
5
3
adj















 A
4
22
15
3
)
11
,
5
,
3
(
)
2
,
3
,
1
(
adj
row1
row1
det
Then








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
 A
A
A
 


















5
8
11
3
4
5
1
4
3
4
1
adj
det
1
Finally 1 T
A
A
A

The inverse of a 3  3 matrix. Example 2.
.
4
1
3
1
2
1
4
0
2
matrix
the
of
inverse
the
Determine











A
:
are
row
by
entries
nine
The
:
adj
find
First we
: A
Solution
      5
1
3
2
1
det
1
entry
3)
(1,
;
1
4
3
1
1
det
1
entry
2)
(1,
;
7
4
1
1
2
det
1
entry
1)
(1,
4
3
2



































      2
1
3
0
2
det
1
entry
3)
(2,
;
4
4
3
4
2
det
1
entry
2)
(2,
;
4
4
1
4
0
det
1
entry
1)
(2,
5
4
3



































      4
2
1
0
2
det
1
entry
3)
(3,
;
2
1
1
4
2
det
1
entry
2)
(3,
;
8
1
2
4
0
det
1
entry
1)
(3,
6
5
4


































.
4
2
8
2
4
4
5
1
7
adj
















 A
6
20
14
)
5
,
1
,
7
(
)
4
,
0
,
2
(
adj
row1
row1
det
Then

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A
A
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4
2
5
2
4
1
8
4
7
6
1
adj
det
1
Finally 1 T
A
A
A

The inverse of a 3  3 matrix. Exercise.
;
1
0
1
1
2
3
5
4
1
;
2
1
3
2
2
2
4
0
3
matrices
following
the
of
inverses
the
Determine
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A

The inverse of a 3  3 matrix. Exercise answers.
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75
.
1
5
.
0
25
.
0
2
5
.
0
5
.
0
75
.
0
5
.
0
25
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14
4
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6
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6
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2
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2
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10
1
;
1
0
1
1
2
3
5
4
1
;
2
1
3
2
2
2
4
0
3
1
1
B
A
B
A

The inverse of a transformation in 3 space.


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y
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T
A
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z
y
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2
1
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3
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1
0
0
0
0
1
:
is
ation
transform
inverse
Then the
0
0
0
0
1
is
this
And
.
of
inverse
the
find
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0
0
0
0
1
is
This
.
for
matrix
the
need
First we
:
:
ation
transform
the
of
inverse
the
find
want to
we
Suppose



•Section 4: Eigenvalues and
eigenvectors.

Introduction to eigenvectors and eigenvalues.
 
6
2
3
6
2
3
6
6
2
3
6
2
3
6


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the
ing
correspond
an
is
and
the
ing
correspond
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is
is,
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.
the
to
ing
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are
lines
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on the
origin
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from
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points
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And
.
of
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called
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and
factors
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12
18
2
3
0
4
9
0
to
ed
transform
is
2
3
And
.
2
3
form
the
of
are
line
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3
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line
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3
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9
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ed
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3
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4
9
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ed
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0
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left
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through t
lines
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find
to
aim
We
.
0
4
9
0
is
matrix
whose
space
2
of
n
nsformatio
linear tra
he
Consider t
2
eigenvalue
r
eigenvecto
eigenvalue
r
eigenvecto
s
eigenvalue
rs
eigenvecto
A
s
eigenvalue
a
a
a
a
a
a
a
a
x
y
m
a
a
a
a
a
a
a
a
x
y
m
m
m
m
ma
a
mx
y
a
ma
a
ma
ma
a
a
ma
a
mx
y
A
a
a
a
a
a
a
a
a

Algorithm to find to eigenvalues and eigenvectors.
6.
eigenvalue
with
rs
eigenvecto
are
,
0
,
2
3
So
2
3
6
6
4
9
6
0
4
9
0
have
we
6
case
the
For
6.
eigenvalue
with
rs
eigenvecto
are
,
0
,
2
3
So
2
3
6
6
4
9
6
0
4
9
0
have
we
6
case
the
For
.
6
and
6
are
s
eigenvalue
The
0
36
zero
be
must
t
determinan
its
t,
determinan
of
meaning
the
mind
bearing
ly,
Consequent
area.
destroys
so
and
origin
h the
linethroug
particular
a
in
vectors
all
s
annihilate
0
4
9
0
that
means
This
0
0
0
4
9
0
0
0
0
0
0
4
9
0
0
0
0
4
9
0
Then
eigenvalue
with
0
4
9
0
of
r
eigenvecto
an
be
Let
:
follows
as
derived
is
alogrithm
efficient
An
case.
3
3
in the
especially
efficient
not
is
slide
previous
the
of
method
The
2

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
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





a
a
a
x
y
y
x
x
y
y
x
y
x
a
a
a
x
y
y
x
x
y
y
x
y
x
y
x
y
x
y
x
y
x
y
x
A
y
x













0000
00000
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00000
00000
00000
00000
00000
00000

Eigenvalues and eigenvectors: Example 1 in the 3  3 case.
       
   







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
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
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
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
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

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



continued
3.
eigenvalue
with
rs
eigenvecto
are
,
0
,
0
0
value
zero
-
non
any
can take
and
0
0
0
0
2
4
4
3
8
3
3
3
3
2
4
7
3
5
3
3
2
4
0
7
3
0
0
5
have
we
3
case
the
For
.
eigenvalue
each
to
ing
correspond
rs
eigenvecto
the
Find
2.
Step
.
5
and
7
3,
are
rs
eigenvecto
0
3
7
5
0
2
4
7
3
det
1
0
3
4
0
3
det
1
0
3
2
0
7
det
1
5
.
the
is
this
:
0
3
2
4
0
7
3
0
0
5
det
Solve
1.
Step
3
2
4
0
7
3
0
0
5
of
rs
eigenvecto
and
s
eigenvalue
the
Find
4
3
2
a
a
z
y
x
y
x
y
x
x
z
y
x
z
y
x
y
x
x
z
y
x
z
y
x
equation
stic
characteri
A












0 0
0 0
0

Eigenvalues and eigenvectors: Example 1 in the 3  3 case (contd)
.
a
a
a
a
y
y
y
z
y
x
y
x
z
y
x
z
y
x
y
x
z
y
x
z
y
x
y
x
x
z
y
x
z
y
x
a
a
a
z
y
x
z
y
x
x
x
z
y
x
z
y
x
y
x
x
z
y
x
z
y
x
5
eigenvalue
with
rs
eigenvecto
are
,
0
,
25
.
2
4
25
.
2
4
1
2
and
4
4
1
2
1
and
4
0
0
0
8
2
4
12
3
0
5
5
5
3
2
4
7
3
5
5
3
2
4
0
7
3
0
0
5
have
we
5
case
the
For
7.
eigenvalue
with
rs
eigenvecto
are
,
0
,
2
0
2
and
0
0
0
0
4
2
4
3
12
7
7
7
3
2
4
7
3
5
7
3
2
4
0
7
3
0
0
5
have
we
7
case
the
For


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


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

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
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
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
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






































































































































































































Eigenvalues and eigenvectors: Example 2 in the 3  3 case
       
      
      

































































































































































 






















































continued
3.
eigenvalue
with
rs
eigenvecto
are
,
0
,
,
0
5
5
)
3
(
)
2
(
2
0
5
5
)
2
(
3
)
1
(
)
3
........(
)
2
.......(
)
1
.......(
0
0
0
2
3
2
3
2
3
3
3
3
5
3
3
1
1
3
1
5
1
3
1
1
have
we
3
For
.
eigenvalue
each
to
ing
correspond
rs
eigenvecto
the
Find
2.
Step
.
2
and
6
3,
are
rs
eigenvecto
:
0
6
3
2
0
18
9
2
0
36
7
0
42
9
2
4
10
7
0
14
3
3
2
4
6
1
0
1
3
5
1
det
1
3
1
3
1
1
det
1
1
1
1
1
5
det
1
1
0
1
1
3
1
5
1
3
1
1
det
Solve
1.
Step
1
1
3
1
5
1
3
1
1
of
rs
eigenvecto
and
s
eigenvalue
the
Find
2
2
3
2
3
2
4
3
2
a
a
a
a
x
z
x
y
y
x
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
equation
stic
characteri
A



























0
0
0
0
0

Eigenvalues and eigenvectors: Example 2 in the 3  3 case (contd)










































































































































































































































continued
2.
eigenvalue
with
rs
eigenvecto
are
,
0
,
0
,
0
0
20
)
3
(
)
2
(
3
0
20
)
2
(
3
)
1
(
)
3
........(
)
2
.......(
)
1
.......(
0
0
0
3
3
7
3
3
2
2
2
3
5
3
2
1
1
3
1
5
1
3
1
1
have
we
2
For
.
6.
eigenvalue
with
rs
eigenvecto
are
,
0
,
2
2
,
0
4
4
)
3
(
)
2
(
0
4
4
)
2
(
)
1
(
)
3
........(
)
2
.......(
)
1
.......(
0
0
0
5
3
3
5
6
6
6
3
5
3
6
1
1
3
1
5
1
3
1
1
have
we
6
For
.
a
a
a
x
z
y
y
y
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
a
a
a
a
x
y
x
z
z
x
z
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x


0 0
0

Particular eigenvectors.





















 

























































9
4
16
nominate
we
:
5
eigenvalue
with
of
rs
eigenvecto
are
,
0
,
25
.
2
4
1
2
0
nominate
we
:
7
eigenvalue
with
of
rs
eigenvecto
are
,
0
,
2
0
1
0
0
nominate
we
:
3
eigenvalue
with
of
rs
eigenvecto
are
,
0
,
0
0
3
2
4
0
7
3
0
0
5
:
examples
are
Here
case.
each
in
r
eigenvecto
particular
a
nominate
can
However we
.
eigenvalue
each
to
ing
correspond
rs
eigenvecto
of
number
infinite
an
are
There
A
a
a
a
a
A
a
a
a
A
a
a
A 0
0 0
0 0
0 0

Eigenvalues and eigenvectors . Exercise.
.
5
4
3
2
1
3
2
3
1
;
0
3
1
1
2
1
3
2
3
.
eigenvalue
each
to
ing
correspond
r
eigenvecto
particular
a
and
s
eigenvalue
the
determine
matrices
following
For the























 
 B
A

Eigenvalues and eigenvectors. Exercise solutions
       
             
   
 
ely)
(respectiv
1
1
1
;
1
1
1
;
14
1
11
is
4
2;
1;
eigenvalue
r with
eigenvecto
an
:
get
we
or
for
values
e
appropriat
choosing
By
.
;
14
11
0
0
0
2
4
4
;
5
5
3
11
0
0
0
4
3
2
3
2
;
2
3
3
2
;
3
3
3
2
4
;
4
4
4
;
2
2
2
;
3
2
3
2
3
4
;
2
;
1
0
3
1
1
2
1
3
2
3
consider
rs
eigenvecto
compute
To
.
4
and
2
1,
are
rs
eigenvecto
0
3
1
1
0
1
3
1
2
3
)(
1
3
0
1
3
1
2
3
2
3
0
3
1
2
1
det
1
3
1
1
1
det
1
2
3
1
2
det
1
3
0
3
1
1
2
1
3
2
3
det
0
3
1
1
2
1
3
2
3
2
2
4
3
2

































































































































































































































































































































 


































 






















































 

z
x, y
z
x
z
y
z
x
z
y
y
z
y
x
z
y
x
z
y
z
x
z
y
z
y
x
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
equation
stic
characteri
A





















Eigenvalues and eigenvectors. Exercise solutions (contd)
       
            
  
(resp.)
1
1
1
;
3
3
1
;
7
2
2
is
(resp.)
2
;
2
3;
eigenvalue
r with
eigenvecto
an
:
get
we
or
for
values
e
appropriat
choosing
By
.
;
3
7
2
0
0
0
5
5
4
4
;
2
3
3
2
4
3
0
0
0
7
4
3
2
3
2
3
;
3
4
3
2
3
3
2
3
3
;
2
4
3
2
4
3
2
3
4
;
2
2
2
;
2
2
2
;
3
3
3
5
4
3
2
3
2
3
2
;
2
;
3
5
4
3
2
1
3
2
3
1
consider
rs
eigenvecto
compute
To
.
2
and
2
3,
are
rs
eigenvecto
0
3
1
3
0
0
3
6
3
9
)
1
)(
3
(
1
0
9
3
2
9
3
3
3
4
1
0
4
3
1
3
det
1
2
5
3
2
3
det
1
3
5
4
2
1
det
1
1
0
5
4
3
2
1
3
2
3
1
det
5
4
3
2
1
3
2
3
1
2
2
4
3
2
















































































































































































































































































































































































 

































































z
x, y
z
y
y
x
z
y
z
x
y
z
y
x
z
y
y
x
z
y
z
y
x
z
y
x
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
z
y
x
equation
stic
characteri
B





















•Section 5: Symmetric matrices
and their reduction to diagonal
form.

Definition of a symmetric matrix.
matrices.
3
3
symmetric
are
5
4
2
4
1
3
2
3
1
and
0
2
3
2
2
1
3
1
3
So
.
if
symmetric
is
matrix
square
A



























B
A
A
A
A T

Eigenvalues and eigenvectors of a symmetric matrix.
.)
resp
(
1
0
1
2
;
1
2
1
6
;
1
1
1
3
1
0
1
;
1
2
1
;
1
1
1
1
1
3
1
5
1
3
1
1
2)
products).
scalar
take
:
(proof
other
each
lar to
perpendicu
are
rs
eigenvecto
three
these
1)
:
Note
.)
resp
2
and
6
3,
s
eigenvalue
to
ding
(correspon
1
0
1
;
1
2
1
:
1
1
1
are
rs
eigenvecto
Particular
2.
eigenvalue
with
rs
eigenvecto
are
,
0
,
0
6;
eigenvalue
with
rs
eigenvecto
are
,
0
,
2
3;
eigenvalue
with
rs
eigenvecto
are
,
0
,
:
are
These
rs.
eigenvecto
and
s
eigenvalue
its
computed
slide
previous
a
in
have,
we
and
symmetric
evidently
is
1
1
3
1
5
1
3
1
1




























































































































































a
a
a
a
a
a
a
a
a
a
a
A
0
0
0
0

Eigenvalues and eigenvectors of a symmetric matrix. (contd)
     
D
AP
P
DP
AP
P
P
DP
AP
D
P
A










































































































































































1
1
So
exists.
,
0
3
1
1
1
2
1
det
As
then
s
eigenvalue
of
matrix
diagonal
the
as
2
0
0
0
6
0
0
0
3
and
rs
eigenvecto
lar
perpendicu
of
matrix
the
as
1
1
1
0
2
1
1
1
1
write
we
if
So
......(*)
..........
2
0
0
0
6
0
0
0
3
1
1
1
0
2
1
1
1
1
1
1
1
0
2
1
1
1
1
1
1
3
1
5
1
3
1
1
:
writing
as
thesame
is
that this
show
will
arithmetic
Matrix
.)
resp
(
1
0
1
2
;
1
2
1
6
;
1
1
1
3
1
0
1
;
1
2
1
;
1
1
1
1
1
3
1
5
1
3
1
1
satisfy
rs
eigenvecto
and
s
eigenvalue
its
and
symmetric
evidently
is
1
1
3
1
5
1
3
1
1
0
0
0
0
0
0

Diagonalising a symmetric matrix 1.
.
determine
to
required
not
is
one
always,
not
but
Generally,
.
called
is
equation
the
writing
and
,
rs,
eigenvecto
s,
eigenvalue
finding
of
process
the
matrix
symmetric
a
Given
.
s,
eigenvalue
of
matrix
diagonal
the
as
2
0
0
0
6
0
0
0
3
rs,
eigenvecto
orthogonal
of
matrix
the
is
1
1
1
0
2
1
1
1
1
where
,
,
1
1
3
1
5
1
3
1
1
for
that
found
We
1
1
1









































P
A
D
AP
P
D
P
A
D
P
D
AP
P
A
ing
diagonalis
matrix.
symmetric
every
to
applies
This
in
s
eigenvalue
of
order
the
to
correspond
in
rs
eigenvecto
the
of
order
the
and
D
P
0
0
0
0
0
0

Diagonalising a symmetric matrix 2.
matrix.
orthogonal
1.
length
of
rs
eigenvecto
an
called
is
then
1
length
have
rs
eigenvecto
the
If
find
to
need
that we
is
process
ing
diagonalis
in this
difference
The
now
is
equation
ing
diagonalis
the
and
so
1
0
0
0
1
0
0
0
1
2
1
0
2
1
6
1
6
2
6
1
3
1
3
1
3
1
2
1
6
1
3
1
0
6
2
3
1
2
1
6
1
3
1
now
However
s
eigenvalue
of
matrix
diagonal
the
as
2
0
0
0
6
0
0
0
3
where
,
have
still
we
2
1
6
1
3
1
0
6
2
3
1
2
1
6
1
3
1
then
1
length
of
rs
eigenvecto
consider
now
we
If
.
1
1
3
1
5
1
3
1
1
for
ion
considerat
Another
1
1
P
D
AP
P
P
P
PP
D
D
AP
P
P
A
T
T
T

























































































0
0
0
0
0
0
0

•Section 6. Typical examination
questions.












k
A
2
4
2
0
2
4
2
3 (a) Show that det A = 20 – 4k.
(b) Find A– 1
(c) Given that k = 3 find the eigenvalues A.
Example 1.
     
 
 




















































































































































































4
2
4
2
16
3
2
8
4
2
8
4
4
20
1
adj
det
1
4
2
4
2
16
3
2
8
4
2
8
4
0
2
2
3
det
2
2
4
3
det
2
0
4
2
det
2
4
2
3
det
4
4
3
det
2
4
2
det
2
4
0
2
det
4
2
2
det
2
2
0
det
adj
(b)
4
20
16
16
4
12
4
4
8
2
2
4
3
2
4
0
2
det
1
4
4
2
2
det
1
2
2
2
0
det
1
3
det
(a)
:
1
4
3
2
k
k
k
k
A
A
A
k
k
k
k
k
k
k
A
k
k
k
k
k
A
Solution
T
0
0
0 0
0












k
A
2
4
2
0
2
4
2
3 (a) Show that det A = 20 – 4k.
(b) Find A– 1
(c) Given that k = 3 find the eigenvalues of A.
Example 1 (contd)
      
       
    
 
  
   
8.
and
1
are
s
Eigenvalue
0
1
8
1
0
8
7
1
0
20
4
3
1
0
1
16
1
4
4
1
3
0
4
4
4
2
2
2
4
3
3
0
3
2
4
2
2
4
2
3
det
(c)
:
2
2











































































is
equation
stic
Characteri
Solution 0
0
0
0
0
0
0














0
1
9
1
0
2
1
k
k
A
(a) Find values of k for which A is singular.
Given that A is non-singular
(b) Find, in terms of k, A–1
Example 2.
  
 
   










































































 



























 







 








 







































 






















k
k
k
k
k
k
k
k
A
A
A
k
k
k
k
k
k
k
k
k
k
k
k
k
k
A
k
k
A
k
k
A
k
k
k
k
k
k
k
A
Solution
T
9
9
18
9
2
2
6
3
1
adj
det
1
2
9
18
2
9
9
1
0
1
det
0
2
det
1
2
1
det
1
9
1
det
0
9
2
det
0
1
2
1
det
1
9
1
0
det
0
9
0
det
0
1
1
det
adj
(b)
6.
or
3
hen
singular w
is
6
or
3
when
0
det
6
3
18
9
1
9
1
0
det
2
0
9
0
det
0
1
1
det
det
(a)
2
1
2
2
0
0
0
0 0
0












3
4
4
4
5
0
4
0
1
A
(a) Verify that is an eigenvector of A and find the corresponding
eigenvalue.
(b) Show that 9 is another eigenvalue of A and find the corresponding
eigenvector.
(c) Find another eigenvalue and corresponding eigenvector and write down a
matrix P and a diagonal matrix D such that P–1 AP = D.
Example 3.











1
2
2
    
 
       .
9
and
3
,
3
are
rs
eigenvecto
0
9
3
3
0
27
6
3
3
factor
a
is
there
(a)
part
:
0
81
9
9
0
16
80
1
7
9
0
5
4
4
0
1
8
1
0
3
4
4
4
5
0
4
0
1
det
is
The
(b)
3.
eigenvalue
with
of
r
eigenvecto
an
is
1
2
2
1
2
2
3
3
6
6
1
2
2
3
4
4
4
5
0
4
0
1
1
2
2
(a)
2
2
3
2
3
2













































































































































ion
stic equat
characteri
A
A
Soln. 0 0
0
0
0
0












3
4
4
4
5
0
4
0
1
A
(b) Show that 9 is another eigenvalue of A and find the corresponding
eigenvector.
(c) Find another eigenvalue and corresponding eigenvector and write down a
matrix P and a diagonal matrix D such that P–1 AP = D.
Example 3 (contd)
























































































































































































































3
0
0
0
3
0
0
0
9
1
1
1
2
1
1
2
1
2
so
3,
eigenvalue
with
of
r
eigenvecto
an
is
1
2
2
Also
.
3
eigenvalue
with
of
r
eigenvecto
an
is
1
1
1
9;
eigenvalue
with
of
r
eigenvecto
an
is
1
1
2
.
;
2
0
0
0
4
4
4
2
4
2
;
6
4
4
4
4
4
8
3
3
3
;
9
9
9
3
4
4
4
5
4
3
;
9
3
4
4
4
5
0
4
0
1
.
3
and
9
to
ing
correspond
rs
eigenvecto
of
n
Calculatio
(c)
,
(b)
1
AP
P
P
A
A
A
z
y
z
x
z
y
x
z
y
x
z
y
z
x
z
y
x
z
y
z
x
z
y
x
z
y
x
z
y
x
z
y
z
x
z
y
x
z
y
x
z
y
x
Soln.
0
0 0
0
0

0.
>
and
constants
are
and
where
.
0
3
1
4
1
a
c
p, a, b
c
b
a
p
A









 

Given AAT = kI , for some constant k.
(a) Find the values of p, k, a, b and c.
(b) Find det A
Example 4.
      2
57
2
6
1
2
12
4
2
3
1
det
2
2
2
2
2
3
0
3
1
4
1
So
2
,
2
2
,
2
2
8
,
,
2
,
18
25
.
0
,
2
,
18
,
4
,
0
3
3
,
3
,
18
0
0
0
0
0
0
3
4
3
9
3
4
3
18
0
0
0
0
0
0
3
4
3
9
3
4
3
18
.
1
0
4
3
1
0
3
1
4
1
:
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2












































































































































 

A
A
b
c
a
a
a
b
c
a
a
a
a
a
b
c
a
c
b
a
c
b
a
c
a
p
k
k
k
k
c
b
a
pc
a
c
b
a
pc
a
p
p
c
b
a
p
k
k
k
AA
c
b
a
pc
a
c
b
a
pc
a
p
p
c
b
a
p
c
p
b
a
c
b
a
p
AA
Solution
T
T 0
0 0
0
0 0
0
0 0

Further pure mathematics 3 matrix algebra

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Further pure mathematics 3 matrix algebra