Vishwakarma Institute of Technology

Pune

Linear Algebra and Differential equations

(HS 1071)

Unit IV
Linear Transformation
 Introduction
In this chapter we shall study functions from and standard (arbitrary)
vector space to another standard (arbitrary) vector space and its various
properties. The aim of such study is to show how a linear transformation
(mapping or function) can be represented by a matrix. The matrix of a
linear transformation is uniquely determined for a standard (particular)
basis.

Definition and Examples of Linear Transformation

Definition : A mapping T : V  W from a vector space
V to a vector space W is called a linear transformation
if it satisfies following conditions.
For Every u , v  V and   R
(i ) T  u  v   T(u )  T(v)
(ii ) T  u   T(u ).
Definition Domain, Codomain and range of linear
transformation
If function T : V → W is linear transformation then V is called
the domain of T and W is called co-domain of T, while T(V ) is
range of linear transformation T.

Example. Identity transformation

The transformation I : V → V defined as I(u) = u, for every u in
V is a linear transformation from V to V . This transformation
is called as identity transformation on V .

Example. Zero(Null) transformation

The transformation T : V → W defined as T(u) = 0,
for every u in V is a linear transformation from V to W. This
transformation is called as zero(null) transformation on V .
Note: If V=W , then T is called as a linear operator

Theorem : If 𝑇: 𝑉 → 𝑊 is a linear transformation then T(0)=0.

But converse is not true i.e. If T(0)=0 then 𝑇: 𝑉 → 𝑊 may or may not
be a linear transformation.

Remark: (i) Zero transformation is a linear transformation.

(ii) Identity transformation is a linear transformation
Example:
Let T : R  R defined as T ( x  y , x ). Deter min e whether T is linear transformation.
2 2

Solution : Let u  ( x1 , y1 ) v  ( x2 , y2 )  R 2
(i ) Here u  v  ( x1  x2 , y1  y2 )
Consider T (u  v)  T (( x1  x2 ), ( y1  y2 ))
 ( x1  x2  y1  y2 , x1  x2 )
and T (u )  T (v)  T ( x1 , y1 )  T ( x2 , y2 )
 ( x1  y1 , x1 )  ( x2  y2 , x2 )
 ( x1  y1  x2  y2 , x1  x2 )
 ( x1  x2  y1  y2 , x1  x2 )
Therefore T (u  v)  T (u )  T (v)
(ii ) Here k .u  k ( x1 , y1 )  (kx1 , ky1 )
Consider T (k .u )  T (kx1 , ky1 )  (kx1  ky1 , kx1 )
and kT (u )  kT ( x1 , y1 )  k ( x1  y1 , x1 )  (kx1  ky1 , kx1 )
T (k .u )  kT (u )
Hence T is linear transformation.
Examples:
(i) Let 𝑇: 𝑅 2 → 𝑅 2 be defined as T(x, y) = (x + 1, y + 2) Determine whether T is
a linear transformation.

(ii) Let 𝑇: 𝑅 2 → 𝑅 3 be defined as T(x, y) = (𝑥 2 , 𝑦 2 , xy) Determine whether T

is a linear transformation.

(iii) Let 𝑇: 𝑅 3 → 𝑅 2 be defined as T(x, y, z) = (x+y+z, 2x−3y +4z) Determine

whether T is a linear transformation.

(iv) Let 𝑇: 𝑅 2 → 𝑅 3 be defined as T(x, y) = (0, 0, 0) Determine whether T is a

linear transformation.
Example : Every matrix transformation is
linear transformation. Consider A mn matrix
n
which transforms a vector in R into a vector
of R . The transformation T : R  R
m n m
is
defind as T( x)  Ax.
Solution : Let x, y  R n
 T( x  y )  A( x  y )  Ax  Ay  T( x )  T( y ).
T(x)  A(x)  A(x)  T(x).
Example : The tranformation T : C1  F ,
( where C1  set of all differentiable functions, F  set of all functions.)
df
defind by T( f ( x))   f ( x) is a linear transformation .
dx

Example : The tranformation T : C  a, b   R,

b
defined by T  f ( x)    f ( x) d x is a linear transformation .
a
 1 
 1 2 3   1 
Ex : Let A    , u  3  , b   
 1 2 1  1  1

the Transformation T : R 3  R 2 , is defined as T( x)  Ax

(a) Find T(u )
(b) Find x in R 3 , such that T( x)  b.
(c) Is there more than one x,such that T( x)  b.

1 
 1 2 3    4 
 1 2 1 3    4 
    
 1
  x1 
 1 2 3    1 
Ax  b    
   1
x

2
 1 2 1    
 x3 

1 2 3 1 1 2 3 1
 A | B    R2  R1   .
 1 2 1 1 0 4 4 0
This gives 4x2  4x3  0, x1  2x2  3x3  1.
Let x3  t  R, then x2  t , x1  1  t.
1  t 
 
Thus x   t  , t  R.
 t 
Example : Let A 68 matrix.Find the values of a and b,
such that T : R a  R b defind by T( x)  Ax is well defined.

Example : What should be the order of matrix A such that

A represents the matrix of the transformation T : R 5  R 7 .
Example : Suppose T : R 2  P2 be a linear transformation such that
1 2 1 
T    2  x  2 x , T    2  x.Find T   .
2

1 1   2

1  1  2
Let    a    b    a  3, b  1
2 1 1 
1  1  2 
   3      .But T is a linear transformation
2 1 1 
1  1  2
 T    3T    T  
2 1 1 
 3(2  x  2 x )  (2  x)  4  2 x  6 x .
2 2
Matrix of a Linear Transformation

Consider the linear transformation T : R n  R m such that

let e1 , e2 , ..., en  be a standard basis for R n .
 a11   a12   a1n 
a  a  a 
 21   22   2n 
 .   .   . 
let T(e1 )    , T(e 2 )    , ..., T(e n )   
 .   .   . 
 .   .   . 
     
 am1   am 2   amn 
Then the m  n matrix whose n columns correpond to
T(ei ), i  1, 2,..., n .
  a11 a12 ... a1n 
a a 
 21 22 ... a2 n 
 . . . . 
A   T(e1 ) T(e 2 ) ... T(e n )    
 . . . . 
 . . . . 
 
 am1 am 2 ... amn 
is such that T(v)  Av for every v in R n .
A is called the standard matrix for T.
 Example : Find the standard matrix of the linear
transformation T : R 3  R 2 defined by T( x, y, z )   x  3 y, 2 x  y  .

 1    0   0 
    1     3      0
T(e1 )  T  0      , T(e2 )  T  1      , T(e1 )  T  0      .
 0   2  0    1  1    0 
     

1 3 0 
  T  A   T(e1 ) T(e 2 ) T(e3 )     .
 2 1 0 
 dp
Example : T : P3  P2 be a linear transformation defined by T ( p( x))   p( x).
dx
Find the matrix of the linear transformation.

Solution : The standard basis for P3  1, x, x 2 , x 3 

T(1)  0, T( x)  1, T( x 2 )  2 x, T( x 3 )  3 x 2

0 1  0 0

T(1)  0  0  , T( x)  1  0  , T( x 2 )  2 x   2  , T( x 3 )  3 x 2  0 
0  0   0   3

0 1 0 0 
 A   0 0 2 0  ( th e m a trix o f tra n s fo rm a tio n )
 0 0 0 3 
  a b  
T : M 2 (R)  P3 , T      ax  bx  cx  d ,
3 2

 c d  
find the matrix of the linear transformation.

1 0 0 0
0 1 0 
0
T   0 0 1 0
 
0 0 0 1
 Kernel and Range of a Linear Transformation
Definition: Kernel of a linear transformation (Null space): Let T : V → W be
a linear transformation, then the set of all vectors in V which maps into 0
is called the Kernel of T. It is denoted by ker(T).
i.e., ker(T) = {u ∈ V : T(u) = 0}.

Definition: Range of a linear transformation: Let T : V → W be a linear

transformation, then the set of all vectors in W those are images under T
of at least on vector in V is called the range of T. It is denoted by: Range(T).
i.e., Range(T) = {w ∈ W : T(u) = w for some u ∈ V }.
Note:
(i) If T : V → W is zero linear transformation then Ker(T) = V
and R(T) = {0}
(ii) If I : V → W is Identity transformation then Ker(I) = {0} and
R(I) = V .

Example: (i) Find kernel and range of zero transformation.

(ii) Find the kernel and range of identity
transformation

Theorem
Let T : V → W be linear transformation then
(a) The kernel of T is a subspace of V.
(b) The range of T is a subspace of W.
Definition: Rank: Let T : V → W be a linear transformation then the
dimension of the range of T is called the rank of T. Notation: Rank(T).

Definition: Nullity: Let T : V → W be a linear transformation then the

dimension of kernel of T is called the nullity of T. Notation: nullity(T)

Rank-Nullity(Dimension) Theorem

Statement : Let T : V → W be a linear transformation with dim(V ) = n

then,
rank(T) + nullity(T) = n

i.e., for any linear transformation the rank plus nullity is equal to the
dimension of domain vector space.
Example: Let A be a 6×7 matrix with rank 4. What is the dimension of
solution space of AX = 0.

Example: If 𝑇: 𝑅4 → 𝑅3 is defined as,

𝑇 𝑥1 , 𝑥2 , 𝑥3 , 𝑥4
= (𝑥1 − 𝑥2 + 2𝑥3 + 𝑥4 , 𝑥1 − 𝑥2 + 3𝑥3 + 2𝑥4 , 2𝑥1 − 2𝑥2
+ 4𝑥3 + 2𝑥4 )
then find bases for range and kernel of T. Hence determine rank and
nullity of T and verify the dimension theorem.
One-to-One
Onto

W  T(V)

T (u )  T (v)  u  v OR
u  v  T (u )  T (v)

One-One + Onto = Bijective

Definition : One-to-one Linear Transformation
A transformation T : V → W is one-to-one(injective) if distinct
elements of V are transformed into distinct elements of W.
That is if u ≠ v ⇒ T(u) ≠T(v).

Definition : Onto Linear Transformation

A transformation T : V → W is said to be onto(surjective) if for
given w ∈ W there exist v ∈ V such that T(v) = w i.e. if range of
T is the entire co-domain W.

A linear transformation which is both one-one and onto is

called bijective.
Theorem: A linear transformation T : V → W is one-one iff
ker(T) = {0}.

Theorem: Let T : V → W be a linear transformation where W

is finite dimensional then T is onto iff rank T is equal to
dimension of W. 𝜌 𝑇 = dim(𝑤)

Theorem: Let T : V → W be a linear transformation with

vector spaces V and W both of dimension n then T is one-one
iff it is onto.
 x 
 x  y  2 z 
Example: Consider the linear transformation 𝑇: 𝑅2 → 𝑅2 defined as T  y   
  x  2 y  3z 
 z 

i) Find the Kernel 𝐾𝑒𝑟 𝑇 and the Range 𝑅 𝑇 of T.

ii) Find the basis for the Kernel 𝐾𝑒𝑟 𝑇 and the Range 𝑅 𝑇 of T.
iii) State the dimension of the Kernel 𝐾𝑒𝑟 𝑇 and the Range 𝑅 𝑇 of T.
Solution: 𝑇: 𝑅2 → 𝑅2  x   x  y  2 z 
T  y  
  x  2 y  3z 
 z 

 x  y  2 z  1   1  2
  x  2 y  3z    1 x   2  y  3  z
       
i) Therefore Range, R(T)=span1   1  2  
 ,   ,  
  1  2  3  

ii) To find basis of R(T)

  1 1 2 
[T ]  
 1 2 3 
1 1 2 
R1  R2
 0
 1 1 

Rank of [T]=2
Dimension of R(T)=2
Therefore Basis of R(T) is  1  ,  1  or  1  ,  2   or  1 ,  2  
              
 1  2    1  3    2   3  

Ker (T )   X  R 3 | TX  0
 x  0
1 1 2     
TX  0   y  0
0 1 1     
 z   0 

Rank=2
Therefore system has one parametric solution.
y+z=0 ⟹ y=-z and x-y-2z=0 ⟹ x=z.
Let z=t, t∈ R,
Therefore y=-t and x=t.
  1  
 
 ker(T )  span   
 
1
 1  
  

 1  
Basis of ker(T)=   
  1 
 1  
  
And dimension of Ker(T)=1
Example: Let T : R 2
R
2
be a linear transformation such that
1 0  0
   2    1    2
T 0    ,T  1    3 ,T  0   2
0  1  0   1  
     

 x
  2
Find T  y. Hence find the image of   under T. Write the
 z  1
  0
 

matrix of T. Is the transformation one-one and onto? Justify.

 x  1  0  0
        
T  y   T  x 0  y 1  z  0
 z   0  0 1
        
1 0  0
     
 xT  0   yT  1   zT  0 
0 0 1
     
 2  1  2
 x  y   z 
 1   3  2
 2x  y  2z 
  
 x  3y  2z 
 2 
   4 1  0   3 
T  1       
 0   2  3  0   5 
 
 Matrix
 2 1 2 
[T ]   
 1 3 2 
 1 3 2   1 3 2 
[T ]  
R 12
 
R 2 2 R 1
 
 2 1 2  0 7 6

Rank=2, Nullity=1
Rank=2=Dimension of Co-domain
Nullity=1≠0
Therefore T is onto but not one-one.
Invertible/Regular Linear Transformation

A linear transformation T : R n  R n defined by T  X   AX,is said to be

invertible or non singular or regular if the matrix of transformation A is
non singular matrix, i.e., invertible.

The corresponding inverse transformation T 1 : R n  R n is given by

T 1  Y   A 1Y.
  x   2x  4 y 
Consider, T:R  R defined as T    
2 2
.
 y   3x  5 y 
Is T a regular transformation? If yes, find the image of
 2  1
  under T .
 3

2 4 1  5 4   5 / 22 4 / 22 
T   3 5  22  0. T    3 2   3 / 22 2 / 22  .
1

22    
 5 / 22 4 / 22   2   1 / 11 
 3 / 22 2 / 22   3    6 / 11
    
 2x  y
Example: Show that the map T  y    x  y  is a linear transformation.
x
   
Write the matrix of T. Is the transformation regular or non-singular?
1
Justify. If yes find the image of  2  under T
1

 x   2x  y 
Solution: Given T  
 y  x  y 
 x   x 
u  ,v    ,  R
1 2

 y   y 
 1  2

   x1  x    2 x1  2 x 2   y 1  y 2 
T  u  v   T      
2

  y    
 1
y 2  x1  x 2   y 1  y 2 
  (2 x  y )  (2 x  y )   2x  y   2x  y 
    1

1 1 2 2 1 2 2

 (
 x1  y 1)  ( x 2  y 2)       
 x1 y 1   x 2 y 2 
  T (u )  T (v )

2 −1
Matrix of T is [T]=
1 1
[T ]  2  1  3  0
Therefore T is regular

1 1
1  1 1  3 3
 1
 
T  3  1 2   1 
 2
 3 3 

 1 1
 3 3   2  1 
Image of 2 under T
1
     
1   1 2  1  0 
   3 3 
Composite Linear Transformation
T:V  U
S: U  W
S T:V  W
(S T)( v)  S(T( v))

S T   ST 
  x
   x  2y  z 
T : R 3  R 2 defined by T  y     and S : R 2
 R 4
defined by
 z   x  5y  z 
 
 2y 
 
 
x x  y
S   .Which is well defined T Sor S T ? Find the matrix of the same.
 y   2x 
 
 x  y 

As T : R  R and S : R  R , S T is a
3 2 2 4

well defined map. S T : R  R . 3 4

0 2   2 10 2 
1 1  1 2 1  0 7 2 
  
 S T    S T   
 
   
2 0   1 5 1  1 4 2 
   
1 0   1 2 1
𝐎𝐫𝐭𝐡𝐨𝐠𝐨𝐧𝐚𝐥 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧:
A linear transformation 𝑇: R𝑛 → R𝑛 defined as T(X)=AX
is said to orthogonal if A is an orthogonal matrix.

𝐎𝐫𝐭𝐡𝐨𝐠𝐨𝐧𝐚𝐥 𝐌𝐚𝐭𝐫𝐢𝐱 ∶
Matrix A is said to be orthogonal matrix if
AA𝑇 = A𝑇 A = I,where I is a 𝑛 × 𝑛 identity matrix.
 a  b a  b 
Find the condition on a and b so that the matrix  
 a  b a  b 
is orthogonal.

 a  b a  b   a  b a  b  1 0
AA  I  
t
    
 a  b a  b  a  b a  b   0 1 
        a  b  a  b    b  a  a  b  1 0
2 2
a b b a
  
 a  b  a  b    a  b  b  a  a  b  a  b  
2 2
0 1 
 2a 2  2b 2  1 and a 2  b 2  0. This gives

1
a=b=  .
2
 Geometric Linear Transformations in 𝑹𝟐

Reflection

Scaling

Transformation
in 𝑅2 Shearing

Rotation

Projection
Reflection:

The reflection of a geometric object through a line produces

the mirror image of the object across the line.

Type 1 : Reflection about X-axis

Type 2 : Reflection about Y-axis
Type 3 : Reflection about the line y=x
Type 4 : Reflection about the line y=-x
Type 5 : Reflection about origin
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
1 0 
Consider A   
0 1
1 0   0   0  1 0   1   1 
0 1  0    0  , 0 1  0    0  ,
         
 1 0   0   0   1 0   1  1 
0 1  1    1 , 0 1 1   1
         
1 0   x   x 
0 1  y     y 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
 1 0 
Consider A   
 0 1
1 0   0   0   1 0   1   1
0 1  0    0  ,  0 1   0    0  ,
         
 1 0   0   0   1 0   1  1
 0 1   1    1 ,  0 1  1   1 
         
 1 0  x    x 
 0 1  y    y 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
0 1 
Consider A   
1 0 
0 1   0   0  0 1   1   0 
1 0   0    0  , 1 0   0    1  ,
         
 0 1   0   1   0 1   1  1
 1 0   1    0  , 1 0   1    1 
         
0 1   x   y 
1 0   y    x 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
 0 1
Consider A   
 1 0 
 0 1  0   0   0 1  1   0 
 1 0   0    0  ,  1 0   0    1 ,
         
 0 1  0   1  0 1 1  1
 1 0   1    0  ,  1 0  1   1
         
 0 1  x    y 
 1 0   y     x 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
 1 0 
Consider A   
 0 1
 1 0   0   0   1 0   1   1
 0 1  0    0  ,  0 1  0    0  ,
         
 1 0   0   0   1 0  1  1
 0 1  1    1 ,  0 1 1   1
         
 Reflection through origin

 1 0   x    x 
 0 1  y     y 
    
Scaling:

A transformation on an object that results in contraction or

dilation (stretching) is called a scaling.

Type 1: Horizontal Scaling

Type 2 : Vertical Scaling
Type 3 : Scaling in both directions
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
k 0
Consider A   
0 1
 k 0  0   0   k 0  1   k 
0 1  0    0  , 0 1  0    0  ,
         
 k 0   0   0   k 0   1  k 
 0 1   1    1  ,  0 1   1   1 
         
 k 0  x   kx 
0 1  y    y 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
1 0 
Consider A   
0 k 
1 0   0   0  1 0   1   1 
0 k   0    0  , 0 k   0    0  ,
         
 1 0   0   0   1 0   1  1 
 0 k   1    k  ,  0 k   1   k 
         
1 0   x   x 
0 k   y    ky 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
k 0
Consider A   
0 1
 k 0  0   0   k 0  1   k 
0 1  0    0  , 0 1  0    0  ,
         
 k 0   0   0   k 0   1  k 
 0 1   1    1  ,  0 1   1   1 
         
 k 0  x   kx 
0 1  y    y 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
1 0 
Consider A   
0 k 
1 0   0   0  1 0   1   1 
0 k   0    0  , 0 k   0    0  ,
         
 1 0   0   0   1 0   1  1 
 0 k   1    k  ,  0 k   1   k 
         
1 0   x   x 
0 k   y    ky 
    
Scaling

 k 0   x   kx 
 0 k   y    ky 
    
Shearing :

The transformation which produces the visual effect of

Slanting is called shearing

Type 1 : shear in X-direction

Type 2 : shear in Y-direction
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
1 k 
Consider A   
0 1 
1 k   0   0  1 k   1   1 
0 1   0    0  , 0 1   0    0  ,
         
 1 k   0   k   1 k   1  1  k 
 0 1   1    1  ,  0 1   1   1 
         
1 k   x   x  ky 
0 1   y    y 
    
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
1 0
Consider A   
k 1
 1 0  0   0   1 0  1   1 
k 1  0    0  , k 1  0    k  ,
         
 1 0   0   0   1 0   1  1 
 k 1   1    1  ,  k 1   1   1  k 
         
 1 0  x   x 
 k 1   y    kx  y 
    
Rotation :

The transformation which rotates given a point in 𝑅2 through

an angle 𝜃 is called Rotation.

Type 1 : counterclockwise rotation(θ > 0)

Type 2 : clockwise rotation (θ < 0)
  0   1   0   1
S :square with vertices   ,   ,   ,  
 0   0   1   1
cos   sin  
Consider A   
 sin  cos  
cos   sin    0   0   cos   sin    1   cos  
 sin       ,     ,
 cos    0   0   sin  cos    0   sin  
cos   sin    0    sin  
 sin     ,
 cos    1   cos  
cos   sin   1  cos   sin  
 sin     
 cos   1  cos   sin  
  0   1   0  1
S :square with vertices   ,   ,   ,  
 0   0   1   1

In particular if   then the images are
2
0 1  0   0  0 1  1   0 
1 0   0    0  , 1 0   0    1  ,
         
0 1  0   1 0 1 1  1
1 0   1    0  , 1 0  1   1 
         
 Rotation

 1 0
1 1 
   

 1
0
 
 Rotation
cos   sin    x   x cos   y sin  
 sin     
 cos    y   x sin   y cos  
 Example:
Find the transformation which produces shear in
posive X- direction of factor 3, followed by reflection through
the line y=x.
 1 3
:- Shear in posive X- direction of factor 3 : A=  
 0 1 
0 1 
reflection through the line y=x : B   
 1 0 
 0 1   1 3  0 1 
The matrix of transformation is BA=       .
 1 0   0 1   1 3
 y1  0 1  x1   x2 
Hence the required transformation is Y=        
 2 
y 1 3  2   1
x x  3 x2
Example: Give a geometric description of the linear transformation
defined by the matrix product

1 0 
First action is by marix   , which is expansion in
0 3
0 1 
positive Y -direction of factor 3.   is reflection about
1 0 
y  x. Hence the effect is scaling in positive Y -direction of
factor 3 followed by reflection about y  x.
Example:

 x  x  y 1 1
The matrix of transformation T   =   is A=   .
 y  y   0 1
0  0 1 1
We will first find images of vertices   ,   ,   and   .
0  2  2 0
1 1  0   0  1 1  0   2  1 1  1   3 
0 1  0    0  , 0 1  2    2  , 0 1  2    2 
              
1 1  1   1 
and       .
0 1  0   0 
0 1
 
   2
 2

0 1
     2
0 0  
 2  3
 
 2

0 1
   
0 0
Example: Find the matrix of the transformation T : R  R which produces 2 2

the effect of shear of factor 3 along positive Y-direction followed by reflection

0
about y=-x followed by clockwise rotation through an angle 30 . Hence find
0
the image of u   1  under the transformation T.
Solution : T : R 2  R 2
1 0 
Shear factor 3 along positive Y  axis A   
3 1 
0 1
Re flection about y   x is B   
 1 0 
 3 1 
cos(30)  sin(30)   2 2 
Clockwise rotation through 30 is C  
0
 
 sin( 30)  cos( 30)   1 3 
 2 2
 3 1   3 1   1  3 3  3 
 2 2  0 1 1 0   2 2   3 1  2 2 2
Re quired transformation is CBA     1 0  3 1      1 0    
  1 2 3     1 3    3  3
2 
1
2  2 2  2 2
0
Im age of u    is
1
 1  3 3  3  0    3 
 2 2 2  2
CBAu    1    
  3 2  3 2 1 2     2 
1
Example: Find the combined transformation matrix for the following
sequence of transformations: Rotation by 450 in clockwise direction,
followed by scaling in x and y directions by factors 3 and ½ respectively,
followed by reflection through the xy-plane.
Apply it on the point [ 1 3].

Example: Find the concatenated transformation matrix for successive

transformations
i) Scaling in x,y co-ordinate by factor 4,2 units respectively.
ii) Rotation by in clockwise direction by 900
iii) Reflection about y=x
Apply the resulting concatenated transformation matrix on the
position vector [2 2 1].
Example: Obtain the concatenated matrix representation of the
following transformations
i) Scaling in x and y co-ordinate by factor by -1,2 units respectively.
ii) Rotation about z-axis by 900 .
iii) Shear of factor 4 along positive Y-direction.
Hence find the transformed position vector of the point A[3 2 ].

Example: Find combine transformation matrix for the following

sequence of transformation
i)Rotation in clockwise direction by an angle −300
ii)Scaling in X and Y axis by 2,3units respectively
Apply this combine transformation matrix on the point P[3,1]
Example: Obtain the concatenated transformation matrix for the
following sequence of transformation:
1) Scaling in x and z co-ordinate by factor 2 and 3 respectively.
2) Rotation in anticlockwise direction by an angle 600 .
Apply the resulting transformation matrix on the position vector
P[2 11].