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Linear Algebra - Solved Assignments - Fall 2006 Semester
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Solution File Assignment #1 of Linear Algebra
(Fall 2006)
Total marks: 30
Question # 1
Let 1 2 3
1 0 5 2
2 , 1 , 6 , 1
0 2 8 6
ba a a
Determine whether b can be written as a linear combination of1 2 3, ,a a a ?
Solution:
If b can be written as linear combination of 1 2 3, ,a a a
Then
1 1 2 2 3 3x a x a x a b
1 2 3
1 0 5 2
2 1 6 1
0 2 8 6
x x x
Here we have to find the value of 1 2 3, ,x x x
1 3
1 2 3
2 3
5 2
2 6 1
2 8 6
x x
x x x
x x
The argument matrix of this system is as follow
1 0 5 2
2 1 6 1
0 2 8 6
Now doing row operation on this matrix
1 2
1 0 5 2
0 1 4 3 2
0 2 8 6
R R
And on the last row we perform the operation 2 32R R
we get the final matrix as
1 0 5 2
0 1 4 3
0 0 0 0
Now we have solution from this matrix
1 3
1 3
5 2
.
2 5
x x
i e
x x
2 3
2 3
4 3
.
3 4
x x
i e
x x
3x Is free (we can take any value)
There exist a solution of these equation therefore we can write b as linear combination
for example if we take 3 2x then it will give 1 8x and 2 5x . So we can write b as
linear combination as follow
1 2 3
1 0 5 2
2 1 6 1
0 2 8 6
x x x
Putting the values of 1 8x , 2 5x and 3 2x we have
1 0 5 2
8 2 5 1 2 6 1
0 2 8 6
Thus b can be written as a linear combination of 1 2 3,a a a
Question # 2
Describe all the solutions of the following system in parametric vector form
1 2 3
3 5 4x x x
1 2 3
4 8 7x x x
1 2 3
3 7 9 6x x x
Solution: The argument matrix of this system is
1 3 5 4
1 4 8 7
3 7 9 6
After doing row operation 1 21R R and 1 33R R on row 2 and row 3. We get
1 3 5 4
0 1 3 3
0 2 6 6
After doing row operation 2 32R R on row 3 we get
1 3 5 4
0 1 3 3
0 0 0 0
After doing row operation 2 13R R on row 1 we get the final matrix
1 0 4 5
0 1 3 3
0 0 0 0
From matrix we have
1 3
1 3
4 5
.
5 4
x x
i e
x x
2 3
2 3
3 3
.
3 3
x x
i e
x x
3x is free
1 3
2 3 3
3 3
5 4 5 4
3 3 3 3
0 1
x x
x x x
x x
3
5
3
0
4
3
1
here
v
p
so
x v x p
Which is the required parametric vector form.
Question # 3
Let 1 2 3
0 0 4
0 , 3 , 1
2 8 5
v v v
Does { 1 2 3, ,v v v } span
3
R ?
Why or why not?
Solution
1 2 3
3
1 2 3
1 1 2 2 3 3
1 2 3
0 0 4
0 , 3 , 1
2 8 5
have to determine whether arbitrary vector b=(b ,b ,b ) in R can
be expressed as a linear combination
b=k v +k v +k v
of a vectors v , v , v Ex
We
v v v
1 2 3 1 2 3
31 2
1 2 3
1 2 3
1 1 2 3
2 1 2
pressing this terms of components gives
0 0 4
(b ,b ,b )=K 0 3 1
2 8 5
40K 0
(b1,b2,b3)= 0K 3 1
2K 8 5
0 0 4
0 3
K K
KK
K K
K K
b K K K
b K K
3
3 1 2 3
3
1 2 3
2 8 5
0 0 4
0 3 1
2 8 5
It has non zero determinant
So we have det 24 0, tan
{ , , }
K
b K K K
A its maens that system is consis t
Therefore v v v span R
Solution File Of Assignment #2 of Linear Algebra
(Fall 2006)
Question # 1
Determine whether the columns of the following matrix are linear Independent or not.
0 8 5
3 7 4
1 5 4
1 3 2
Solution: We have to see that the equation Ax=0 has trival solution or non trival
solution. Take the matrix
0 8 5 0
3 7 4 0
1 5 4 0
1 3 2 0
Interchange row 4 and row 1. We have
1 3 2 0
3 7 4 0
1 5 4 0
0 8 5 0
After doing row operation 1 23R R on row 2 we get
1 3 2 0
0 2 2 0
1 5 4 0
0 8 5 0
After doing row operation 1 3R R on row 3 we get
1 3 2 0
0 2 2 0
0 2 2 0
0 8 5 0
After doing row operation 2 31R R on row 3 we get
1 3 2 0
0 2 2 0
0 0 0 0
0 8 5 0
Interchange row 3 ad row 4
1 3 2 0
0 2 2 0
0 8 5 0
0 0 0 0
After doing row operation 2
1
2R and 2 38R R respectively we get following matrix
1 3 2 0 1 3 2 0
0 1 1 0 0 1 1 0
0 8 5 0 0 0 3 0
0 0 0 0 0 0 0 0
As we can see from last matrix that there is no free variable and there are three basic
variables 1 2 3, ,x x x . So the equation Ax = 0 has only the trivial solution and column are
linearly independent.
Question # 2
A linear transformation T is defined as T(x)=Ax. Find a vector x whose Image under T is ‘b’ and determine whether x is unique. Where
A =
1 0 2
2 1 6
3 2 5
, b =
1
7
3
Solution: For this we have to solve Ax = b or
1
2
3
1 0 2 1
2 1 6 7
3 2 5 3
x
x
x
The augmented matrix is
1 0 2 1
2 1 6 7
3 2 5 3
We have to do row operations. First 1 22R R on row2 and then 1 33R R on row 3 we
get following matrix respectively.
1 0 2 1 1 0 2 1
0 1 2 5 0 1 2 5
3 2 5 3 0 2 1 0
After doing row operations 2 32R R and
3
1
5R respectively on row3 we get following two
matrixes respectively
1 0 2 1 1 0 2 1
0 1 2 5 0 1 2 5
0 0 5 10 0 0 1 2
At last we do the row operation 3 12R R on row 1 and 3 22R R on row2 we get the
following matrix
1 0 0 3
0 1 0 1
0 0 1 2
Here we have
1
2
3
3
1
2
x
x
x
Vector x is
3
1
2
whose Image under T is ‘b’
Yes x is unique because there exist a unique solution.
Question # 3
Let T : 2 2
R R be a linear transformation such that
T( 1 2,x x ) = (
1 2 1 2,4 5x x x x ) .Find x such that T(x)=(3,8)
Solution:
1 2
1 2
( )4 5
x xT x
x x
Here 3
( )8
T x
after putting we get
1 2
1 2
3
8 4 5
x x
x x
After equating both sides we get
1 2
1 2
3
4 5 8
x x
x x
The augmented matrix of this system is
1 1 3
4 5 8
First applying 1 24R R on row 2 and then
2 11R R on row 1 we get the following two
matrixes respectively.
1 1 3 1 0 7
0 1 4 0 1 4
From last matrix we have
1
2
7
4
x
x
So x is
1
2
7
4
xx
x
Solution File Of Assignment # 3 of Linear Algebra (Fall 2006)
Total marks: 25
Question # 1
Solve the equation Ax=b by using LU-Decomposition for the given matrix
A =
2 3 4
4 5 10
4 8 2
, b =
6
16
2
Solution:
Marks 10
1
1 2 1 3
2
2 3 4 1 0 0
4 5 10 , * 1 0
4 8 2 * * 1
1 3/ 2 2 2 0 01
4 5 10 , * 1 02
4 8 2 * * 1
1 3/ 2 2 2 0 0
0 1 2 4 4 , 4 1 0
0 2 6 4 * 1
1 3/ 2 2 2 0 0
0 1 2 1 , 4 1 0
0 2 6 4 * 1
U L
U R L
U R R and R R L
U R L
2 3
3
1
2
3
1
2
1 3/ 2 2 2 0 0
0 1 2 2 , 4 1 0
0 0 2 4 2 1
1 3/ 2 2 2 0 01
0 1 2 , 4 1 02
0 0 1 4 2 2
,
:
2 0 0 6
4 1 0 16
4 2 2 2
3
U R R L
U
R L
we see that A LU let
Ly b
y
y
y
y
y
3
1
2
3
1
2
3
4
1
1 3/ 2 2 3
0 1 2 4
0 0 1 1
4
2
1
y
and Ux y
x
x
x
x
x
x
Question # 2
Solve the equation Ax=b by taking inverse of the matrix of the following
system of equations.
1 3
1 2 3
1 2 3
2 5
3 4 2
2 3 4 1
x x
x x x
x x x
Marks 10
Solution:
1
2
3
-1
3
1 2
1 3
1 0 2 5
A= 3 1 4 , , 2
2 3 4 1
by [A I ] finding A :
1 0 2 1 0 0
3 1 4 0 1 0
2 3 4 0 0 1
1 0 2 1 0 0
0 1 2 3 1 0 3
2 3 4 0 0 1
1 0 2 1 0 0
0 1 2 3 1 0 2
0 3 8 2 0 1
1 0
x
x x b
x
R R
R R
2 3
3 2 3 1
3
1 1
1
1
2 1 0 0
0 1 2 3 1 0 3
0 0 2 7 3 1
1 0 0 8 3 1
0 1 0 10 4 1
0 0 2 7 3 1
1 0 0 8 3 1
0 1 0 10 4 1 / 2
0 0 1 7 / 2 3/ 2 1/ 2
8 3 1
10 4 1
7 / 2 3/ 2 1/ 2
8 3 1
10 4 1
7 / 2 3
R R
R R and R R
R
Henc
we can easily check that A A I
A
x A b
A A
1
2
3
5
2
/ 2 1/ 2 1
35
43
15
x
x
x
Question # 3
Let A =
1 1 1
0 2 3
5 5 1
.Find the third column of 1
A
without computing
the other columns. Marks 5
Solution
3 1
3
2 3
After performing hte row operations we get,
1 1 1 * * 0
0 2 3 * * 0 R 5R
5 5 1 * * 1
1 1 1 * * 01
0 2 3 * * 0 R4
0 0 4 * * 1
1 1 1 * * 0
0 2 3 * * 0 R 3R
0 0 1 * * 1/ 4
1 1 1 * * 01
0 2 0 * * 3/ 42
0 0 1 * * 1/ 4
2
1 3
1 2
1
R
1 1 1 * * 0
0 1 0 * * 3/ 8 R R
0 0 1 * * 1/ 4
1 1 0 * * 1/ 4
0 1 0 * * 3/ 8 R R
0 0 1 * * 1/ 4
1 0 0 * * 1/ 8
0 1 0 * * 3/ 8
0 0 1 * * 1/ 4
Third column of without computing the other columns is given by,
* * 1/ 8
A * * 3/ 8
*
* 1/ 4
Solution File OF Assignment no.4 Fall 2006 (Linear Algebra)
Question # 1
Let A=
0 1 4
2 3 2
5 8 7
and b=
8
1
1
.Determine whether b is in the column
space of A. Marks 10
Solution: To determine the b is in the column space of A, we see that the aug. matrix is
consist ant or not. Now
Row reducing the augmented matrix [A b]
0 1 4 8
~ 2 3 2 1
5 8 7 1
2 3 2 1
~ 0 1 4 8
5 8 7 1
2 3 2 1
~ 0 1 4 8
0 1/ 2 2 3/ 2
2 3 2 1
~ 0 1 4 8
0 0 0 5 / 2
We conclude that Ax = b is inconsistent and So b is not in the Col of A.
Question # 2
Find the rank of the following matrix,
1 2 4 3 3
5 10 9 7 8
4 8 9 2 7
2 4 5 0 6
Marks 10
Solution:
1 2 4 3 3
5 10 9 7 8A=
4 8 9 2 7
2 4 5 0 6
We have to find the reduced row echelon form
1 2 1 3 1 4Applying -5R +R ,-4R +R ,2R +R
1 2 4 3 3
0 0 11 22 7
0 0 7 14 5
0 0 3 6 0
3 4 42Applying 3R +7R ,3 +11R
1 2 4 3 3
0 0 11 22 7
0 0 0 0 15
0 0 0 0 21
R
3 4Applying -21R +15R
1 2 4 3 3
0 0 11 22 7
0 0 0 0 15
0 0 0 0 0
Since The Matrix has 3 pivot columns ,so the rank A = 3
Rank(A) = 3
Question # 3
By using Cramer’s Rule, solve the following system of equations,
1 2 3
1 3
1 2 3
2 4
2 2
3 3 2
x x x
x x
x x x
Marks 10
Solution:
1 2 3
1 3
1 2 3
2 4
2 2
3 3 2
x x x
x x
x x x
Take determinant,
2 1 10 2 1 2 1 0
1 0 2 2 1 11 3 3 3 3 1
3 1 3
D
= 2(-2) - 1(-3-6) + 1(-1 )
= 2(-2) - 1(-9) - 1
= -4 +9 - 1 = 4
D = 4
1
11
2
22
3
3
4 1 1
2 0 2 16
2 1 3
164
4
2 52
5213
4
4
D
Dx
D
D
Dx
D
D
Dx
3 41
4D
So x1 = -4
x2 = 13
x3 = -1
Solution File of Assignment No. 5
LINEAR ALGEBRA (Fall 2006)
Question # 1 Find the dimension of Null Space and Column Space for the matrix
A=
4 1 2 2
3 2 0 1
1 1 2 1
SOLUTION: In order to find the dimension of the column Space we have to Row
Reduced the given matrix
A=
4 1 2 2
3 2 0 1
1 1 2 1
To Echelon Form:
4 1 2 2
3 2 0 1
1 1 2 1
1 1 2 1
~ 3 2 0 11 3
4 1 2 2
1 1 2 1
~ 0 5 6 2 3 , 41 2 1 3
0 5 6 2
1 1 2 1
~ 0 5 6 22 3
0 0 0 0
A
change R with R
R R R R
R R
1 1 2 1
1~ 0 1 6/5 2/5
250 0 0 0
R
Thus A has two pivot column so the dimension of ColA=2.
For Null space, we need the reduced echelon form.
Further row operations on A yields
1 0 4/5 3/5
0 1 6/5 2/5
0 0 0 0
~
A has two free variables x3 and x4.
Dimensions of Null Space of A, NullA=2
So dim of ColA + dim of NullA=n(the no. of Columns of A)
Which is true. Hence Dimension Theorem is true.
Question # 2
Let A=
4 2 3
1 1 3
2 4 9
. An Eigenvalue of A is 3.Find a basis for
the corresponding eigenspace.
Solution:
The scalar 3 is an Eigenvalue of A if and only if the equation
3AX X
Has the nontrivial solution.
3
3 0
AX X
A I X
Now we solve A-3I
4 2 3 3 0 0
3 1 1 3 0 3 0
2 4 9 0 0 3
1 2 3
1 2 3
2 4 6
A I
Row reduced the augmented matrix for 3 0A I X :
1 2 1 3
1 2 3 0
1 2 3 0
2 4 6 0
1 2 3 0
0 0 0 0
0 0 0 0
~ , 2R R R R
A has the free variables 2 3andx x so 3 in Eigenvalue of A.
The general solution is:
1
2 2 3
3
2 3
1 0
0 1
x
x x x
x
The basis is for the eigenspace is
2 3
1 . 0
0 1
.
Question # 3
Is =4 an Eigenvalue of
3 0 1
2 3 1
3 4 5
? If so find the
corresponding eigenvector.
Solution:
Suppose A=
3 0 1
2 3 1
3 4 5
The scalar 4 is an Eigenvalue of A if and only if the equation
4AX X Has the nontrivial solution.
4
4 0
AX X
A I X
Solve the 4A I .
3 0 1 4 0 0
4 2 3 1 0 4 0
3 4 5 0 0 4
1 0 1
2 1 1
3 4 1
A I
Row reduced the augmented matrix for 4A I :
1 2 1 3
1 0 1 0
2 1 1 0
3 4 1 0
1 0 1 0
~ 0 1 1 0 2 , 3
0 4 4 0
R R R R
2 3
1 2
1 0 1 0
~ 0 1 1 0 4
0 0 0 0
1 0 1 0
~ 0 1 1 0 ,
0 0 0 0
R R
R R
A has free variable 3x so 4 is Eigenvalue of the matrix A.
Let 3x =1
1 3 1
2 3 2
1 2
0 1 0
0 1 0
1 1So and
x x x
x x x
x x
Hence the correspondence eigenvector is
1 1
1 1
1 1
or
Solution File Assignment # 6 ( Linear Algebra)
(Fall 2006)
Total marks: 20
Question # 1
Diagonal the following matrix, if possible
A = 1 4
1 2
Solution:
Step 1: Find the Eigen values of A. The characteristic equation is as follow:
Det |A- I |=0
2
2
2
1 40
1 2
(1 )( 2 ) 4(1) 0
2 2 4 0
6 0
3 2 6 0
( 3) 2( 3) 0
( 3)( 2) 0
3,2
Step 2: Find two linearly independent vectors of A.
Solve the characteristic equation
1
2
1 2 1
1 2
2 2
1 2
1
1
1
2
1
( ) 0
=-3
1-(-3) 4 0
1 -2-(-3) 0
4 4 0 1 1 0 1 1 0~ 1/ 4 ~
1 1 0 1 1 0 0 0 0
0
x as x is a free variable.
0
0
1
1
A I x
For
x
x
R R R
x x
Take t
x x
x t
x t
x tt
x t
V
1
2
1 2 1
1 2
2 2
1 2
1
1
1
2
2
1
1
=2
1-2 4 0
1 -2-2 0
1 4 0 1 4 0 1 4 0~ ~
1 4 0 1 4 0 0 0 0
4 0
x as x is a free variable.
4 0
4 0
4
4 4
1
4
1
For
x
x
R R R
x x
Take s
x x
x s
x s
x ss
x s
V
1 2 and V are linearly independent.V
Step 3: Construct p from the vectors V1 and V2.
1 2
1 4
1 1
4:
Construct D from the corresponding eigen values according to the
arrangement in the above step.
-3 0D=
0 2
will varify AP=PD which is the condition for a diagonalizable m
p V V
p
Step
We
atrix.
1 4 1 4 3 8AP=
1 -2 1 1 3 2
1 4 -3 0 3 8
1 1 0 2 3 2
A is diagonaslizable.
PD
Hence
Question # 2
(a) Classify the following matrices as an attractor,repellor or a saddle
Point of the Dynamical System 1K K
Ax x
A = 1.7 .3
1.2 .8
B = .5 .6
.3 1.4
Solution:
a) A=1.7 .3
1.2 .8
The characteristic equation is as follow:
2
2
2
22
| | 0
1.7 .30
1.2 .8
(1.7 )(.8 ) ( .3)( 1.2) 0
1.36 1.7 .8 0.36 0
1 2.5 0
2.5 1 0
the quardetic equation to find the value of .
( 2.5) ( 2.5) 4(1)(1)-b b 4 2.5 6=
2 2(1)
A I
Applying
ac
a
1
.25 4
2
2.5 2.25 2.5 1.5
2 2
2.5 1.5 2.5 1.5 4 1, ,
2 2 2 2
2,0.5
=2>1 and =0.5<1, hence it a saddle point of the dynamic system K K
As Ax x
b) B=.5 .6
.3 1.4
Solution: The characteristic equation is as follow:
2
2
22
| | 0
.5 .60
.3 1.4
(.5 )(1.4 ) (.6)( .3) 0
0.7 .5 1.4 .18 0
1.9 0.88 0
the quardetic equation,
( 1.9) ( 1.9) 4(1)(0.88)-b b 4 1.9 3.61 3.52=
2 2(1) 2
1.9 0.09 1.9 0.3
2 2
1.9
B I
Applying
ac
a
1
0.3 1.9 0.3,
2 2
2.2 1.6,
2 2
1.1,0.8
=1.1>1 and =0.8<1. Hence it is a saddle point for the dynamic System .K K
As Ax x
(b) Solve the initial value problem /( ) ( )t Ax tx for t 0 with
0
3
2x
and A = 1 2
3 4
Solution:
2
2
2
1
2
|A- I|=0
1- -20
3 -4-
(1 )( 4 ) ( 2)(3) 0
4 4 6 0
3 2 0
2 2 0
( 2) 1( 2) 0
( 2)( 1) 0
2, 1
eigen vectors corresponding to 2
1 ( 2) 2 0
3 4 ( 2) 0
3 2 0
3 2 0
As
The
x
x
2 1
1 2
2 2
1
1
1
1
2
3 2 0~
0 0 0
3 2 0
x as x is a free variable.
3 2 0
3 2
2
3
2 22
3 33
1
corresponding to the eigen value =-1
1-(-1) -2
3 -3
R R
x x
Take p
x p
x p
x p
x pp p
xp
Eigenvectors
1
2
x
x
1 2
2 1
1 2
2 2
1 2
1
1
1
2
1 1 2 2
1
2 2 0 1 1 0 1 1~ ,
3 3 0 1 1 0 2 3
1 1 0~
0 0 0
0
x as x is a free variable.
0
0
1
1
,
( )
t=0
3 2
2 3
t t
R R
R R
x x
Take t
x x
x t
x t
x tt
x t
So
x t a e x a e x
Initially
a
2
1 2
1 2
1 2
1 2
1
1
2
2
2
1
1
23
2 3
2 3......(1)
3 2......(2)
subtrating 2 from 1, we get a 1
this value of a in equation 1.
2(-1)+a 3
3 2 5
2 1( ) 5
3 1
t t
a
a a
a a
a a
a a
By
Put
a
x t e e
Solution File of Assignment No. 7 (Linear Algebra) FALL SEMESTER 2006 Question # 1
Apply the Power Method to A = 2 1
4 5
with 0
1
0x
to estimate
the dominant eigenvalue.Stop when K=5
Solution:
Question # 2
(a) Determine whether the set S= {1 2 3, ,u u u } is an orthogonal set?
0
0 0
1 0
0
1 1
2 1
1
2
Ax
2 1 1 2, 4
4 5 0 4
2 .51 1x
4 14
2 1 .5 1 1 2, 7
4 5 1 2 5 7
2 0.28571 1
7 17
2 1 0.2857
4 5 1
First Compute
Ax
Ax
Ax
x Ax
Ax
2
3 2
2
3 3
0.5714 1 1.5714, 6.1428
1.1428 5 6.1428
1.5714 0.25581 1
6.1428 16.1428
2 1 0.2558 0.5116 1 1.5116, 6.
4 5 1 1.0232 5 6.0232
x Ax
Ax
4 3
3
4 4
5 4
4
0232
1.5116 0.25091 1
6.0232 16.0232
2 1 0.2509 0.5018 1 1.5018, 6.0036
4 5 1 1.0036 5 6.0036
1.5018 0.25011 1
6.0036 16.0036
x Ax
Ax
x Ax
Ax
5 5
2 1 0.2501 0.5002 1 1.5002, 6.0004
4 5 1 1.0004 5 6.0004
Where 1 2 3
1 0 5
2 , 1 , 2
1 2 1
u u u
Solution:
1 2 3
1 2
1 3
1 0 5
2 , 1 , 2
1 2 1
we calculate the products pairs of distinct vectors to check whether the set is orthagonal
or not.
1 0
u . 2 . 1 0 2 2 0
1 2
1
u .
u
u
u u u
2 3
5
2 . 2 5 4 1 0
1 1
0 5
. 1 . 2 0 2 2 0
2 1
, the given set is orthagonal.
u u
So
(b) Show that the set S= {1 2,u u } is an orthogonal basis for
2
R ?
Express the vector y=6
3
as a linear combination of the vectors in S
Where 1 2
3 2
1 6andu u
Solution:
1 2
1 2
1
2
1 1
2 2
The set s will be orthagonal if u .u 0.
3 2. . 6 6 0
1 6
, is orthagonal set.
Now,
6 3y.u . 18 3 15
3 1
6 2. . 12 18 30
3 6
3 3. . 9 1 10
1 1
.
u u
So it
y u
u u
u u
1 21 2 1 2
1 1 2 2
1 2
2 2. 4 36 40.
6 6
,
. . 15 30. .
. . 10 40
3 3
2 4
Now
y u y uy u u u u
u u u u
y u u
Question # 3
Let y=1 2
5 3 3
9 , 5 , 2
5 1 1
u u
.Find the distance from y to the plane
in 3
R spanned by1 2andu u .
Solution:
1 2
1 21 2
1 1 2 2
1
1 1
2
5 3 3
9 , 5 , 2
5 1 1
. .ˆ . .
. .
5 3
. 9 . 5 15 45 5 35
5 1
3 3
. 5 . 5 9 25 1 35
1 1
5 3
. 9 . 2
5 1
y u y uy u u
u u u u
y u
u u
y u
u u
2 2
1 21 2
1 1 2 2
15 18 5 28
3 3
. 2 . 2 9 4 1 14
1 1
putting these values in the formula below:
. .ˆ . .
. .
3 3 335 28
ˆ 5 2 5 ( 235 14
1 1 1
u u
Now
y u y uy u u
u u u u
y
2
3 3 6 3
) 2 5 4 9
1 1 2 1
5 3 2
ˆ 9 9 0
5 1 6
ˆ 4 0 36 40
ˆ 40 2 2 10 2 10
the distance of y from is 2 10.
y y
y y
y y x x
Hence W
Solution File Assignment No.8
Linear Algebra (Fall Semester 2006)
Question # 1
Consider the basis S= {1 2 3, ,u u u } for
3
R .Where
1 2 3
1 0 1
1 , 1 , 2
1 1 3
u u u
Use the Gram-Schmidt Process to transfer S to an orthonormal basis
for 3
R
Solution:
Step 1
1 1
1
1
1
v u
Step 2
1
2 12 2 2 2 12
1
.w
u vv u proj u u v
v
2 1
0 1
. 1 . 1 0(1) 1(1) 1(1) 2
1 1
u v
2
1 1 1 1 3v
2
0 1 0 2 / 32
1 1 1 2 / 33
1 1 1 2 / 3
v
2
2 / 3
1/ 3
1/ 3
v
Step 3
2
3 1 3 2
3 3 3 3 1 22 2
1 2
. .w
u v u vv u proj u u v v
v v
3 1
3 2
3 2
2
2
1 1
. 2 . 1 1(1) 2(1) 3(1) 6
3 1
1 2 / 32 1 1
. 2 . 1/ 3 1( ) 2( ) 3( )3 3 3
3 1/ 3
2 2 3 2 2 3. 1
3 3 3 3
4 1 1 4 1 1 6 2
9 9 9 9 9 3
u v
u v
u v
v
3
3
3
3
1 1 2 / 36 1
2 1 1/ 33 2 / 3
3 1 1/ 3
1 1 2 / 33
2 2 1 1/ 32
3 1 1/ 3
1 2 6 / 6
2 2 3/ 6
3 2 3/ 6
1 6 / 6
0 3/ 6
1 3/ 6
v
v
v
v
3
3
1 6 / 6
0 3/ 6
1 3/ 6
0
1/ 2
1/ 2
v
v
Thus
1
1
1
1
v
, v2 =
2 / 3
1/ 3
1/ 3
, v3 =
0
1/ 2
1/ 2
These are the orthogonal basis for 3R .
Now we have to find the orthonormal basis 3R
2 2 2
1
2 2 2
2
2 2
3
(1) (1) (1) 1 1 1 3
2 1 1 6
3 3 3 3
1 1 20
2 2 2
v
v
v
1
1
1
2
2
2
3
3
3
1/ 3
1/ 3
1/ 3
2 / 6
1/ 6
1/ 6
0
1
2
1
2
vq
v
vq
v
vq
v
Question # 2
The orthogonal basis for the columns space of the matrix
A =
3 5 1
1 1 1
1 5 2
3 7 8
is given by
3 1 3
1 3 1, ,
1 3 1
3 1 3
.Find a QR-factorization of the above matrix
A. Solution:
Let
1 2 3
3 1 3
1 3 1, ,
1 3 1
3 1 3
v v v
Normalize the three vectors to obtain u1, u2, u3.
1 1
1
320
31
11 1 20
1 12020
33
20
u vv
2 2
2
120
13
31 1 20
3 32020
11
20
u vv
3 3
3
320
31
11 1 20
1 12020
33
20
u vv
QR Factorization:
3 3120 20 20
31 120 20 20
31 120 20 20
3 3120 20 20
Q
As R=T
Q A
So take the transport of the above matrix ,
3 31 120 20 20 20
3 31 120 20 20 20
3 31 120 20 20 20
TQ
3 31 1 3 5 120 20 20 20
1 1 13 31 1
20 20 20 20 1 5 2
3 31 1 3 7 820 20 20 20
3 5 13 1 1 3
1 1 111 3 3 1
1 5 2203 1 1 3
3 7 8
9+1+1+9 15 1 5 21 3 1 2 241
3 3 3 3 -5+3+5+7 1 3 6 820
9
R
R
1 1 9 15+1+5-21 3 1 2 24
20 40 261
0 20 220
0 0 24
R
Question # 3
Find the Least Square solution and its error, where
A =
2 1
2 0
2 3
and b=
5
8
1
Solution:
2 12 -2 2 4 4 4 2 0 6 12 8
2 01 0 3 2 0 6 1 0 9 8 10
2 3
TA A
52 -2 2 10 16 2 26 2 24
81 0 3 5 0 3 5 3 2
1
TA b
Then the equation T TA A A b becomes
1
2
12 8 24
8 10 2
x
x
1 10 81
8 1256
TA A
1
ˆ
10 8 241
8 12 256
240 161 =
192 2456
224 41
168 356
4ˆ
3
T Tx A A A b
x
Now
2 1 8 3 54
ˆ 2 0 8 0 83
2 3 8 9 1
Ax
Hence,
5 5 5 5 0
ˆ 8 8 8 8 0
1 1 1 1 0
b Ax
And
ˆ 0
b Ax
The least square error is 0
