LINEAR ALGEBRA
SOLVED ASSIGNMENTS
SEMESTER FALL-2004
Solution of Assignment # 1 of MTH501
(Fall2004)
Q1. Consider the system of equations
2
2 3
x y z a
x z b
x y z c
Show that the system to be consistent , the constant a, b and c must satisfy c = a + b.
Answer:
The augmented matrix correspond to the above system is
1 1 2
1 0 1
2 1 3
a
b
c
. We will find out the echelon form of this matrix which can be
obtained by the following steps.
2 1 3 1
2 3 2
1 1 2
1 0 1
2 1 3
1 1 2
0 1 1 , 2
0 1 1 2
1 1 2
0 1 1 ,
0 0 0
a
b
c
a
b a by R R R R
c a
a
a b by R R R
c a b
Now as you know that if augmented matrix of a linear system in echelon form have any row of the form [0 0 0… b] where b is non zero then that system will be inconsistent. Now in the above matrix we have third row as [0, 0, 0, c-a-b] and
in order to system be consistent we must have c –a –b =0 which is equivalent to c = a + b.
Q2. Is there a value of r so that the x = r, y = 2, z =1 is a solution to the following
linear system? If there is a value find it.
3 2 4
4 5
2 3 2 9
x z
x y z
x y z
Answer:
We will try to find out the solution of the above system, so the augmented
matrix for the above system is
3 0 2 4
1 4 1 5
2 3 2 9
and the echelon form of the
augmented matrix can be obtained by the following steps.
1 2
2 1 3 1
2
2
3 0 2 4 1 4 1 5
1 4 1 5 3 0 2 4 Re
2 3 2 9 2 3 2 9
1 4 1 5
0 12 5 19 3 , 2
0 5 4 1
1 4 1 5
5 19 10 1
12 12 12
0 5 4 1
1 4 1 5
5 190 1 5
12 12
23 830 0
12 12
by placing R by R
by R R R R
R
R
3
3
1 4 1 5
5 190 1 12
12 12
0 0 23 83
R
R
23 z = 83
z = 83/23 This doesn’t match with the value of z given in the question. So we can’t find the
value of r from the given system of linear equations. Note:
There are more then one method to solve that question as you can also
put the given values x=r y=2 and z=1 in the system of linear equation and
then you can conclude the same. But concept used in this question is that “If
system of linear equations is consistent then it will have unique solution or
infinite many solutions”. This is not possible that a system of linear equations
have 2 or 3 or 10 solutions.
Q3. An oil refinery produces low-sulfur and high-sulfur fuel. Each ton of low-sulfur requires 5 minutes in the blending plant and 4 minutes in the refining plant; each
ton of high sulfur fuel requires 4 minutes in the blending plant and 2 minutes in the refining plant. If the blending plant is available for 3 hours and the refining
plant is available for 2 hours, how many tons of each type of fuel should be manufactured so that the plants are fully utilized?
Answer:
The data given in the question can be formed into system of linear
equations as Low-sulfur High-sulfur Blending plant 5 4
Refining Plant 4 2 As we are given in the question that we have blending plant and refining plant
available for 3 and 2 hours respectively. Let x tons of low-sulfur and y tons of high sulfur be the amount should be manufactured so that plants are fully utilized. Then from the above data we must have the system,
5x + 4y = 180
4x + 2y = 120 Augmented matrix for the above system
is
1 2
5 4 180 1 2 60
4 2 120 4 2 120
1 2 60
0 6 120
by R R
So we have x + 2y = 60 and -6y = - 120, thus we have y= 20 tons and x = 20tons are the required manufactured tons of each low-sulfur and high-sulfur so that we
can utilize both plants for the given time.
Q4. (a) Find a linear equation in the variables x and y that has the general solution
x = 5 + 2t, y= t.
(b) Show that x=t, 1 5
2 2y t is also the general solution of the equation in
part (a).
Answer:
(a)
We have to find out linear equation for which the given parametric equations x = 5 + 2t, y= t define the coordinates of any point on that line.
And in order to get the required linear equation we will simply find an equation independent of the parameter “t”
Suppose x = 5 + 2t ---------- (1)
y = t--------------- (2) Put t=y in eq. (1) we get
x= 5 + 2y
This is the required linear equation. (b)
We have a linear equation x= 5 + 2y ----------- ( A)
Put x = t in eq.(A) we get
t =5 + 2y --------------- (B)
From (B) y = ( t – 5)/2
Y = t/2 – 5/2
1 5
2 2y t
This shows that x=t, 1 5
2 2y t is also the general solution of the equation in part (a).
Q5. Consider
5 1
2 , 2 2
7 3 3
k
x y and w
k
find the value(s) of k such that the
vector w is in the span of x and y. Answer:
We have to find the value(s) of k (if possible) under the condition that w is in the
span of x and y W is in the span of x and y so we have to check whether x1x + x2y = w have solution?
Here x1 and x2 are constants. To answer this, row reduce the augmented matrix [ x y w]:
12
1
1 2 1 3
2
2 3
5 1
2 2 2
7 3 3
2 2 2
5 1
7 3 3
1 1 11
5 12
7 3 3
1 1 1
0 4 5 5 , 7
0 4 10
1 1 1
5 10 1
4 4
0 4 10
1 1 1
50 1 4
4
0 0 5
k
k
k R
k
k R
k
k R R R R
k
kR
k
kR R
The third equation is 0x2 = -5 this is not possible which shows that the system has no solution.
The vector equation x1x + x2y = w has no solution and so w is not in the span of x and y. Thus we can not able to find the value(s) of k.
Solution of Assignment # 2 of MTH501
(Fall2004) Q1.
If the columns of the matrix
12 7 11 9 5
9 4 8 7 3
6 11 7 3 9
4 6 10 5 12
span R4 then write
the vector
1
4
5
7
as linear combination of the columns of the above matrix.
Solution:
We have to find out the values of the unknowns such that we can write the given
vector
1
4
5
7
as 1 2 3 4 5
12 7 11 9 5 1
9 4 8 7 3 4
6 11 7 3 9 5
4 6 10 5 12 7
c c c c c
which is
correspond to the system of linear equations whose augmented matrix is,
12 7 11 9 5 1
9 4 8 7 3 4
6 11 7 3 9 5
4 6 10 5 12 7
The Augmented Matrix is:
12 7 11 9 5 1
9 4 8 7 3 4
6 11 7 3 9 5
4 6 10 5 12 7
~ 1
1 7 /12 11/12 9 /12 5 /12 1/12
9 4 8 7 3 4 1
6 11 7 3 9 5 12
4 6 10 5 12 7
R
~
2 1
3 1
4 1
1 7 /12 11/12 9 /12 5 /12 1/129
0 5 / 4 1/ 4 1/ 4 3/ 4 19 / 46
0 15 / 2 3/ 2 3/ 2 13/ 2 11/ 24
0 11/ 3 19 / 3 2 31/ 3 20 / 3
R R
R R
R R
2
1 7 /12 11/12 9 /12 5 /12 1/12
0 1 1/ 5 1/ 5 3/ 5 19 / 5 4~
0 15 / 2 3/ 2 3/ 2 13/ 2 11/ 2 5
0 11/ 3 19 / 3 2 31/ 3 20 / 3
R
3 2
4 2
1 7 /12 11/12 9 /12 5 /12 1/12
(15 / 2)0 1 1/ 5 1/ 5 3/ 5 19 / 5~
(11/ 3)0 0 0 0 2 34
0 0 28 / 5 41/15 122 /15 109 /15
R R
R R
1 2 3 4 5
3 4 5
5
3 4 5
12 7 11 9 5 1
3 19
2 34
84 41 122 109
c c c c c
c c c
c
c c c
There are four equations and five unknowns so one variable is taken to be arbitrary Let c2 = s
On solving these four equations we get
5
4
3
1
17
783/ 25
967 / 25
1 10637 7047[1 7 85]
12 25 25
c
c
c
c s
As we are able to find the values of the unknowns so we can write the given
vector
1
4
5
7
as 1 2 3 4 5
12 7 11 9 5 1
9 4 8 7 3 4
6 11 7 3 9 5
4 6 10 5 12 7
c c c c c
Hence the given vector is the linear combination of columns of the given matrix.
Q2. Determine whether the Homogeneous system
1 2 3 5
1 2 3 4 5
1 2 3 5
3 4 5
2 2 0
2 3 0
2 0
0
x x x x
x x x x x
x x x x
x x x
has
non-trivial solution or not. If it has non-trivial solutions then write the solution in parametric form.
Solution:
Matrix of coefficients for the above system is
2 2 1 0 1
1 1 2 3 1
1 1 2 0 1
0 0 1 1 1
then
reduced echelon form of the matrix is
1 1 2 0 1
0 0 0 3 0
0 0 1 0 1
0 0 0 1 0
thus we have,
1 2 3 5 1 2 3 5
4 4
3 5 3 5
4
2 0 2
3 0 0
0
0
x x x x x x x x
x x
x x x x
x
Also by putting the value of 3 5x x in first equation we get,
1 2 5 5 1 2 52x x x x x x x which shows that 2x and 5x are free variables. So
we can take 2 5x t and x s where s and t are arbitrary real numbers, thus the
parametric form of the solution is given by the following equation.
1
2
3
4
5
1 1
1 0
0 1
0 0 0
0 1
x t s
x t
x s t s
x
sx
.
Q3.
(a) Determine whether the given vectors (1, 2, 3, 5), (2, 7, 6, 0), (1, 1, 1, 1) and (3, 5, 7, 9) are linearly independent or not. Justify your answer.
(b) For what values of c the vectors (-1, 0, -1), (2, 1, 2) and (1, 1, c) are linearly dependent in R3.
Solution:
Consider the linear combination of the given vectors as
1 2 3 4
1 2 1 3 0
2 7 1 5 0
3 6 1 7 0
5 0 1 9 0
c c c c
. Now we will solve that homogeneous
system and if we gat the non trivial solution of that system then the given four
vectors will be linear dependent otherwise the four vectors will be linear independent. Matrix of coefficients for the above homogeneous system is,
1 2 1 3
2 7 1 5
3 6 1 7
5 0 1 9
and the echelon form of that matrix is
1 2 1 3
1 10 1
3 3
10 0 1
2
500 0 0
3
.
Since the matrix has pivot position in each row hence the homogeneous system
has only trivial solution. Thus we can conclude that the given vectors are linearly independent. Also note that the homogenous system in Q2 has non trivial solution because second row of the matrix of coefficients in echelon form has no pivot
position. (b)
In order to get the required value or values of c we will solve the homogeneous system of linear equations correspond to the linear combination
1 2 3
1 2 1 0
0 1 1 0
1 2 0
x x x
c
and we will find out the value of c such that this
homogeneous system will have non-trivial solution. So the matrix of coefficients
is
1 2 1
0 1 1
1 2 c
and the echelon form of that matrix is
1 2 1
0 1 1
0 0 1c
.
Now if you choose c such that c -1 = 0 then the above system will have a free
variable namely x3 and thus has non trivial solution, which implies that for this value of c we have three vectors linearly dependent in R3 thus the required value of c is 1.
Q4.
(a) Show that the mapping2 5:T R R defined by
T(x, y) = (x, x-y, 2x-4y, y, x + y) is a linear transformation.
Solution (a):
We will show that the given transformation satisfies the conditions.
(i) T (u + v) = T(u) + T(v) Where u = (a, b) and v = (c, d) are the arbitrary two elements of R2.
(ii) T(cu) = c T(u)
(i) T(u + v) =T(a + c, b + d) since we have (u + v) =(a + c, b + d) now by definition of transformation we can write,
T(a + c, b + d) = (a + c, a + c – b – d, 2a + 2c -4b – 4d, b + d, a + c + b + d)
= (a, a – b, 2a -4b, b, a + b) + (c, c – d, 2c – 4d, d, c + d) = T(a, b) + T(c, d)
Because by definition of transformation we have, T(a, b) = (a, a – b, 2a -4b, b, a + b) T(c, d) = (c, c – d, 2c – 4d, d, c + d)
Thus we have T(a + c, b + d) = T(a, b) +T(b, d) or T (u + v) = T(u) + T(v)
(ii) T(cu) = T(ca, cb) = (ca, ca – cb, 2ca -4cb, cb, ca + cb) = c (a, a – b, 2a -4b, b, a + b) =c T(a, b) T(cu) = c T(u)
Thus both conditions are satisfied by the given transformation hence the transformation is linear.
(b) If we have a linear transformation T:R3 R3 such that
1 1 0 1 0 1
0 2 , 1 0 0 0
0 0 0 1 1 0
T T and T
then find the matrix of
transformation T and find the image of
1
2
3
x
x
x
.
Solution (b):
Since we know that matrix of linear transformation T:Rm Rn is
1 2 3
1 2 3
( ) ( ) ( ) . . . ( )
(1,0,0,...,0), (0,1,0,...,0), (0,0,1,...,0)... (0,0,0,...,1)
m
m
T e T e T e T e
Where e e e e
Thus matrix of the linear transformation is
1 0 0 1 1 1
0 1 0 2 0 0
0 0 1 0 1 0
A T T T
and hence the image of
1
2
3
x
x
x
is given
by
1 2 31 1
2 2 1
3 3 2
1 1 1
2 0 0 2
0 1 0
x x xx x
T x x x
x x x
.
Q5. (a) Find a linear transformation which projects every element of R3 into xz-
plane and then find the image of
10
15
100
.
(c) Three friends play a game in which there are always two winners and one loser. They have the understanding that the loser gives each winner an
amount equal to what the winner already has. After three games, each just lost once and each has 24$. With how much money did each begin?
Solution (a):
Since we have to find out a linear transformation which projects every element of R3 into xz-plane that is the linear transformation when acts on an
element of R3 made the y component of that vector 0. Thus we can say that
formula for that linear transformation is
1 1
2
33
0
x x
T x
xx
and thus the image of
10
15
100
under such a linear transformation is
10
0
100
.
Solution (b):
Let x, y and z be the starting money of the three friends A, B and C
respectively. Now we will consider the three games separately.
1st game:
Suppose that in the first game A having x money is loser then by the given conditions in the question after 1st game A, B and C will have x - y - z, 2y and 2z dollars
respectively.
2nd game:
Suppose that in the second game B having 2y money is loser then by the given conditions in the question after 2nd game A, B and C will have 2x -2 y -2z,
2y - 2z – x + y + z =3y – z – x and 4z dollars respectively.
3rd game:
Suppose that in the third game C having 4z money is loser then by the given conditions in the question after 3rd game A, B and C will have 4x -4 y -4z,
6y – 2z - 2x and 4z -2x + 2y +2z -3y + z+ x=7z – x- y dollars respectively.
Now as we are given in the question that after three games each A, B and C has 24$. So we have, 4x -4 y -4z = 24
6y – 2z - 2x= 24 7z – x - y = 24
Augmented matrix for that system is
4 4 4 24
2 6 2 24
1 1 7 24
and echelon form of that
matrix is
1 1 7 24
0 0 2 3
0 0 1 12
and this gives us z = 12$, y = 21$ and x = 39$.
You can also check that solution satisfies all the given conditions in the question.
Solution of Assignment # 3 of MTH501
(Fall2004)
Q1. Given that the matrix A is symmetric if
6 1 1
1 14 4
1 4 2
tA
and
2 3 1 1
1 2 3 2 4
1 4 3
x y z
A x y z
x y z
then find the values of x, y and z (if
possible).
Solution:
Since the given matrix is symmetric so we have A=At so we can write
2 3 1 1 6 1 1
1 2 3 2 4 1 14 4
1 4 3 1 4 2
x y z
x y z
x y z
As we know that two
matrices will be equal if the order and corresponding entries of the matrices are equal. So by using this fact we can write,
2 3 6
2 3 2 14
3 2
x y z
x y z
x y z
So augmented matrix for the system is
1 2 3 6
2 3 2 14
3 1 1 2
and echelon form of the this
matrix is
1 2 3
0 1 2 1, 2, 3
0 0 1
x y z
, thus we get,
1 4 9 1 1 6 1 1
1 2 6 6 4 1 14 4
1 4 3 2 3 1 4 2
. Hence 1, 2, 3x y z are
the required values.
Q2. If
9 0 1 2 4 1
3 4 4 0 1 6
2 8 0 1 8 0
A B
then find
(i) The element 32a of the matrix AB without calculating the matrix AB.
(ii) The element11a of the matrix BA without calculating the matrix BA
Solution (i):
Since the 32a element of the product AB will be the sum of the
corresponding elements of the third row of A and second column of B. Thus we
have 32
4
1
2 8 0 8 2 4 8 ( 1) 0 8 0
a
Solution (ii):
Since the 11a element of the product BA will be the sum of the
corresponding elements of the first row of B with the first column of A. Thus we have
11
2 4 1 9
3 18 12 2 22
2
a
Q3. Consider the Matrices
3 0 1 1 2 1
0 0 4 0 4 1
1 1 0 1 8 0
A B
then show
that 1 1 1( )AB B A .
Solution:
First of all we will find out the product AB then we will find out the
inverse of AB. So
4 14 3
4 32 0
1 6 2
AB
and
1
0.5333 0.0833 0.8000
0.0667 0.0417 0.1000
0.0667 0.0833 0.6000
AB
--------------- (i)
Now we have
1 1
0.3333 0.0833 0 0.8000 0.8000 0.2000
0.3333 0.0833 1 0.1000 0.1000 0.1000
0 0.2500 0 0.4000 0.6000 0.4000
A and B
1 1
0.5333 0.0833 0.8000
0.0667 0.0417 0.1000
0.0667 0.0833 0.6000
B A
----------------------- (ii)
From (i) and (ii) it is easy to see that1 1 1( )AB B A .
Q4. Consider the Transformation T: R2 R2 defined by T(x) = A x where
cos sin
sin cosA
the show that,
(i) The mapping T is a rotation through the angle .
(ii) A A A
(iii) Find the inverse of A .
Solution (i):
First of all you should note that the transformation defined by the formula
T(x) = A x is a linear transformation. Let a
xb
and we can
write1 0 1 0
0 1 0 1
a ax a b T aT bT
b b
then by definition of
transformation we have,
1 cos sin 1 cos 0 cos sin 0 sin
0 sin cos 0 sin 1 sin cos 1 cosT and T
Now consider the geometric presentation of these two images, we can take the
unit vectors1
0
and 0
1
along the unit circle obviously unit vector1
0
makes an
angle 0 with the x-axis and vector0
1
is along the y-axis which makes an
angle2
with the x-axis as shown in the fig below.
Now as we can easily see that image of1 cos0
0 sin 0
is a rotation through an
angle as1 cos( 0)
0 sin( 0T
similarly we have
cos0 2
1sin
2
and this is also a
rotation of an angle as
cos( )0 2
1sin( )
2
T
. Since from the result of
trigonometry we have cos( ) sin sin( ) cos2 2
and
. Thus in fact
the given matrix of transformation only rotates that point through an angle of .
Now as every element1 0
0 1
ax a b
b
is a linear combination of unit
vectors this rotates through an angle . Hence the transformation is a rotation
through an angle .
Solution (ii):
Since we have
cos sin
sin cosA
thuscos sin
sin cosA
,cos( ) sin( )
sin( ) cos( )A
and we have
cos sin cos sin cos cos sin sin cos sin sin cos
sin cos sin cos sin cos cos sin sin sin cos cosA A
Now as we now that
cos( ) cos cos sin sin sin( ) cos sin sin cos .
Hence we have, cos( ) sin( )
sin( ) cos( )A A A
.
Solution (iii):
We know that if we have
1 1a b a bA then A
c d c dad bc
Thus by using the above formula we
have
1 2 2
2 2
1
cos sin1cos sin 1
sin coscos sin
cos sin
sin cos
A as so
A
Q5. Consider the system of linear equations
6u - 2v - 4w + 4y = 2 3u - 3v - 6w + y = - 4 -12u + 8v + 21w - 8y = 8
-6u - 10w + 7y = - 43 Find the solution of the above system using LU decomposition.
Solution:
First of all we will find out the LU factorization of the matrix of
coefficients of the above system which is
6 2 4 4
3 3 6 1
12 8 21 8
6 0 10 7
A
Step1: Zero out below the first diagonal entry of A. We get
2 1 3 1 4 1
1 0 0 06 2 4 4
10 2 4 1 1 0 01
, 2 , 20 4 13 0 2
2 * 1 00 2 14 11
1 * * 1
R R R R R R
Where
in the second matrix entries below first diagonal entry: 1 are the negative of the multipliers of the row operations which we did in the first matrix.
Step2: Zero out the entries below the second entry of the diagonal.
3 2 4 2
1 0 0 06 2 4 4
10 2 4 1 1 0 0
2 , 20 0 5 2
2 2 1 00 0 10 12
1 1 * 1
R R R R
Step3: Zero out the entries below the third entry of the diagonal.
4 3
1 0 0 06 2 4 4
10 2 4 1 1 0 0
2 20 0 5 2
2 2 1 00 0 0 8
1 1 2 1
R R
.
Thus we have
1 0 0 06 2 4 4
10 2 4 1 1 0 0
20 0 5 2
2 2 1 00 0 0 8
1 1 2 1
U L
.
Now our system becomes LUx = b where,
1 0 0 06 2 4 4 2
10 2 4 1 41 0 0
, ,20 0 5 2 8
2 2 1 00 0 0 8 43
1 1 2 1
u
vL U x and b
w
y
Let Ux = z where
1
2
3
4
z
zz
z
z
and by solving Lz = b we get,
11
2 2 1
33 1 2
44 1 2 3
21 0 0 02
114 4 51 0 0
228
8 2 2 22 2 1 043
1 1 2 1 43 2 32
zz
z z z
zz z z
zz z z z
Now we will solve the system Ux = z which will give the solution of the above system.
6 2 4 4 2 4
0 2 4 1 5 1.2
0 0 5 2 2 6.9
0 0 0 8 32 4.5
u y
v w
w v
y u
Solution of Assignment # 4 of MTH501
(Fall2004)
Q1. Let
1 2 0 0 0
3 4 0 0 0
0 0 5 1 2
0 0 3 4 1
U
and
3 2 0 0
2 4 0 0
0 0 1 2
0 0 2 3
0 0 4 1
V
Determine whether U and V are block diagonal matrices and find UV using block
matrix multiplication. Solution
U =
1 2 0 0 0
3 4 0 0 0
0 0 5 1 2
0 0 3 4 1
and V =
3 2 0 0
2 4 0 0
0 0 1 2
0 0 2 3
0 0 4 1
Here A22 = 1 2
3 4
O23 =0 0 0
0 0 0
O22 = 0 0
0 0
B23 = 5 1 4
3 4 1
C22 =
3 2
2 4
O22 = 0 0
0 0
O32 =
0 0
0 0
0 0
D32 =
1 2
2 3
4 1
Then U and V becomes
22 23 22 22
32 3222 23
22 22 23 32 22 22 23 32
22 22 32 23 22 22 23 32
22 22
23 32
1 2 3 2
3 4 2 4
7 6
17 10
1 25 1 2
2 33 4 1
4 1
1 9
7 5
7 6
A O C OU and V
O DO B
A C O O A O O DUV
O C O B O O B D
A C
B D
UV
0 0
17 10 0 0
0 0 1 9
0 0 7 5
is the required result
Q2. Check whether the given system is diagonally dominant?
-x1 + 4x2 – x3 = 3 4x1 – x2 = 10 - x2 + 4x3 = 6
Then solve the system after making appropriate changes by Jacobi’s Method. Also solve the above system by Gauss Seidal method. (Only three iterations and
show your calculations) Solution:
From the given system we see that,
1 4 1 .
1 4 .
4 1 .
is not staisfied
is not staisfied
staisfied
Thus the given system is not diagonally dominant. But by interchanging first two rows
we get the system 4x1 – x2 = 10
-x1 + 4x2 – x3 = 3 - x2 + 4x3 = 6
Now you can check that
4 1 .
4 1 1 .
4 1 .
staisfied
staisfied
staisfied
hence we made the system diagonally
dominant.
Jacobi’s Method:
4x1 – x2 = 10 -x1 + 4x2 – x3 = 3
- x2 + 4x3 = 6
1 21
1 1 32
1 23
0 0 0
1 2 3
1
1
1
2
1
3
1 1 1
1 2 3
2
1
2
2
10
4
3
4
6
4
( , , ) (0,0,0)
0
10
4
3
4
6
4
10 3 6( , , ) ( , , )
4 4 4
sec 1
10 3/ 4 43
4 16
3 10 / 4
kk
k kk
kk
xx
x xx
xx
Take initial iteration x x x
For first iteration put k
x
x
x
x x x
For ond iteration put k
x
x
2
3
2 2 2
1 2 3
3
1
3
2
3
3
3 3 3
1 2 3
6 / 4 28
4 16
6 3/ 4 27
4 16
43 28 27( , , ) ( , , )
16 16 16
2
10 28 /16 188
4 64
3 43/16 27 /16 118
4 64
6 27 /16 123
4 64
188 118 123( , , ) ( , , )
64 64 64
x
x x x
For third iteration put k
x
x
x
x x x
Seidal method
4x1 – x2 = 10
-x1 + 4x2 – x3 = 3 - x2 + 4x3 = 6
1 21
11 1 3
2
11 2
3
0 0 0
1 2 3
1
1
1
2
1
3
1 1 1
1 2 3
10
4
3
4
6
4
int ( , , ) (0,0,0)
0
10 0 10
4 4
3 10 / 4 0 22
4 16
6 22 /16 118
4 64
10 22 118( , , ) ( , , )
4 16 64
sec
kk
k kK
Kk
xx
x xx
xx
In ial iteration x x x
For first iteration put k
x
x
x
x x x
For ond iterat
2
1
2
2
2
3
2 2 2
1 2 3
3
1
3
2
3
3
1
10 22 /16 182
4 64
3 182 / 64 118 / 64 492
4 256
6 492 / 256 2028
4 1024
182 492 2028( , , ) ( , , )
64 256 1024
2
10 492 / 256 3052
4 1024
3 3052 /1024 2028 /1024 8152
4 4096
ion put k
x
x
x
x x x
For third iteration put k
x
x
x
3 3 3
1 2 3
6 8152 / 4096 32728
4 16384
3052 8152 32728( , , ) ( , , )
1024 4096 16384x x x
Q3. Consider the matrix
4 0 7 3 5
0 0 2 0 0
7 3 6 4 8
5 0 5 2 3
0 0 9 1 2
then find
(i) (3, 1)th Cofactor of the above matrix. (ii) Minor correspond to the element a55. (iii) Determinant of the matrix using such a row or column which involves the
least amount of computations. Solution:
(i) (3, 1)th Cofactor of the above matrix is 3 1
0 7 3 5
0 2 0 0( 1)
0 5 2 3
0 9 1 2
.
(ii) Minor correspond to the element a55 is
4 0 7 3
0 0 2 0
7 3 6 4
5 0 5 2
(iii) From the given matrix you can see easily that second row and second column have only one non-zero entries, so if we expand the determinant by taking second row or column then we will involve fewer calculations.
So expanding the given matrix by second
row2 3
4 0 7 3 54 0 3 5
0 0 2 0 07 3 4 8
7 3 6 4 8 ( 1)5 0 2 3
5 0 5 2 30 0 1 2
0 0 9 1 2
Now again
expanding the determinant by second column will reduce the calculations.
So
2 3
2 2
4 0 7 3 54 0 3 5
0 0 2 0 07 3 4 8
7 3 6 4 8 ( 1) 25 0 2 3
5 0 5 2 30 0 1 2
0 0 9 1 2
4 3 5
2 ( 1) 3 5 2 3
0 1 2
Again
expanding by the third row we
get
2 3
2 2 3 2 3 3
4 0 7 3 54 0 3 5
0 0 2 0 07 3 4 8
7 3 6 4 8 ( 1) 25 0 2 3
5 0 5 2 30 0 1 2
0 0 9 1 2
4 3 54 5 4 3
2 ( 1) 3 5 2 3 6 ( 1) ( 1) ( 1)5 3 5 2
0 1 2
( 6) 12 15 8 15 ( 6){(3) ( 7)} 24
Q4. (i) Determine the volume of the parallelepiped which has the following
vectors as its adjacent sides,
3 2 , 10 5 2 3a i j k b i j k and c i j k .
(ii) Consider the linear transformation T: R3 R3 defined as
1 1 2 1
2 1 2 3
3 1 2 3
3 2
0 4 5
0 2 2
x x x x
T x x x x
x x x x
If S be the volume of the parallelepiped in R3 formed by the vectors (1, 1, 1) ,(2, 3, 7) and (3, 5, 7) find the volume of T(S).
Solution: (i) Volume of the parallelepiped is equal to the determinant of the
given vectors which
is
3 2 110 5 1 5 1 10
1 10 5 3 2 1 120 16 8 1282 3 1 3 1 2
1 2 3
(ii) First of all we will write the matrix of transformation which
is
3 2 1
0 4 5
0 2 2
Also we know that
Volume of T(S) = (Determinant of matrix of transformation of T) * (Volume of S)
Volume of S =
1 1 13 7 2 7 2 3
2 3 7 1 1 1 14 7 1 65 7 3 7 3 5
3 5 7
but volume will be
positive so we have Volume of S = 6.
3 2 14 5
0 4 5 3 3( 18) 542 2
0 2 2
. Thus we have,
Volume of T(S) = (-54) * 6 = -324, but we will take the positive value so, Volume of T(S) = 324
Q5.
(i) Let 1n . Then Rn consists of (a) n real numbers. (b) n-tuples of real numbers
(c) n-tuples of vectors Solution:
(b) is the correct answer as every element of Rn has the form (x1, x2,…, xn) where x1, x2,.., xn are real numbers, also (x1, x2,…, xn) =(y1, y2,…, yn) if and only if x1,= y1, x2,= y2…, xn,= yn .
(ii) In a vector space V over a field F scalar multiplication is given by a map
(a) V V F
(b) F V V
(c) F F F Solution:
The correct answer is (b) as we have scalar multiplications from
F V V to vector space V.
(iii) Which of the following is statement is not an axiom for the real vector space?
(a) For all ,x y V we have x + y = y + x
(b) For all , ,x y z V we have (x + y) + z = x + (y + z)
(c) For all , ,x y z V we have (xy)z = x(yz)
Solution:
As we know that a Vector space V along with a field is a group under addition and satisfy the scalar multiplication properties but a vector space is associate
with respect to multiplication is not necessary. Thus option (c) is incorrect.
(iv) Which of the following is true? If V is a vector space over the field F
(a) / ,x y x V y V V
(b) / ,x y x V y V V V
(c) / ,v v V F F V
Solution:
The option (a) is true as we know that for a vector space V we
have , / ,x y V then x y V which means that x y x V y V V .
(v) Check whether the given subsets form subspaces of Rn
(a) 1 2/, ...n
nU x R x x x
(b) 2 2
1 2/,nU x R x x
(c) 1/, 1nU x R x
Solution:
In order to check whether the given subsets are subspaces or not we will check the two conditions.
(i) If ,x y U x y U .
(ii) If c F and v U then cv U
(a) 1 2/, ...n
nU x R x x x is a subspace of Rn as for,
1 2 1 2 1 2 1 2, ( , ,... ) ( , ,... ) ... , ...n n n nx y U where x x x x and y y y y also x x x y y y
Then we have,
1 2 1 2 1 1 2 2
1 1 2 2 1 2 1 2
( , ,... ) ( , ,... ) ( ,. ,.. )
... ... , ...
n n n n
n n n n
x y x x x y y y x y x y x y U
because x y x y x y as x x x y y y
So first condition of subspace is satisfied by U.
Now take1 2 1 2
1 2 1 2 1 2
( , ,.., ) ..
( , ,.., ) .. ..
n n
n n n
c F and v U where v x x x suchthat x x x
cv cx cx cx U because x x x cx cx cx
Hence the set given in the part (a) is a subspace of the vector space Rn.
(b) 2 2
1 2/,nU x R x x
U is not a vector space by providing a counter example
U contains all those vectors in which the square of first two elements must be equal Let s,wU S= (1, -1, 2, …n)
W = (2, 2, 3, …n) S+w = (3, 1, 5,….2n)
But in s + w the square of first two elements are not equal so s+w does not belongs to U so addition is not defined in U. For this reason U is not a vector space by the definition of vector space.
(c) 1/, 1nU x R x
U contains all those vectors in which the first element is 1
Let s and w belong to U Then s = (1,2,3,…n) and w = (1,3,4…n)
s + w = (2, 5, 7, ….2n) s + w does not belong to U because the first element is not 1
So addition is not defined in U So U is not a Vector space under the given condition.
Solution of Assignment # 5 of MTH501
(Fall2004) Question # 1
a. For the bases 3{( 4,3,7), (2, 1,0), (1,0,13)}of R find the
coordinate of the element (-7, 5, 1) relative to the bases .
b. For the bases 2 2 2
2{2 ,3 1, }x x x x of find the coordinate of the
216 5 9x x relative to .
Solution:
(i)
In order to get the coordinate of the element (-7, 5, 1) relative to the given
bases 3{( 4,3,7), (2, 1,0), (1,0,13)}of R we will solve the following system.
4 2 1 7
3 1 0 5
7 0 13 1
x y z
The values of x, y and z will be the required coordinates.
Augmented matrix of that system is
4 2 1 7
3 1 0 5
7 0 13 1
and by solving it we get the
values of unknowns as x = 2, y = 1, z = -1. (ii)
In order to get the coordinate of the element 216 5 9x x relative to the given
bases 2 2 2
2{2 ,3 1, }x x x x of we will try to find out the values of a, b and c such
that
2 2 2 2 0
2 0 2 0
2 3 1 16 5 9
2 3 16 5 9
a x x b x c x x x x
a b c x a x bx x x x
comparing the coefficients of
2 0,x x and x we get the following system
2 3 16
5
9
a b c
a
b
by solving it we get the values
of unknowns as a = -5, b = 9, c = -1. Thus the coordinate of the given polynomial
is ( 5,9, 1)
Question # 2
(i) Find the change of coordinate matrix that change coordinates to
coordinates where 2 2 2 2 2{ 1, 1, 1}, { 4,4 3 2,2 3}x x x x x x x x x are
bases for 2 .
(ii) Let b1 =1
3
, b2 = 2
4
,c1 = 7
9
, c2 = 5
7
and consider the bases
for R2 given by B = {b1, b2} and C = {c1, c2} a. Find the change of coordinates matrix from C to B
b. Find the change of coordinates matrix from B to C. Solution:
(i) In that question we will find out the coordinates of elements of using the bases . As we did in Q1, thus
2 2 24 1 1 1x x a x x b x c x And solving as we did in above question we
get2
2
2, 3 1 4 3
1
a b and c thus coordinate of x x
. Similarly we
have2
1
4 3 2 2
3
coordinate of x x
, 2
1
2 3 1
1
coordinate of x
.
Now as we know that
2 2 24 4 3 2 2 3
2 1 1
3 2 1 .
1 3 1
P x x x x x
P is the required matrix
1 2 1 2
1
. ,
1 2 | 7 5 1 0 |5 3|
3 4 |9 7 0 1 |6 4
5 3
6 4
. ( ) sin ( int )
4 3 21( )
6 52
B C C B
B C
C B B C
a Notice that P is needed rather than P and compute
b b c c
So P
b By part a and u g the property with B and C erchanged
P P
3/ 2
3 5 / 2
Question # 3
For dimension of Column space; Dimension of Row space and Rank of the following
matrix
1 1 1 2
1 0 2 3
2 4 8 5
.
Solution:
In order to get the dimension of column space, dimension of row space and rank of the given matrix first of all we will find out the echelon form of the given
matrix which is
1 1 1 2
0 1 3 1
0 0 0 3
. From the reduced echelon form it is quite
clear that the columns 1, 2 and four have the pivot positions thus the
corresponding columns in the given matrix are linearly independent and form a bases for the column space of the given matrix. Hence dimension of column space
of the given matrix is 3. Also from the echelon form it is clear that the non-zero rows of the given matrix are three so forms a bases for the row space thus dimension of row space as well
as rank of the matrix is 3. Defective matrix:
Let A be an nn matrix. If there is an eigenvalue of A such that
the geometric multiplicity of is less than the algebraic multiplicity of , then A is called a defective matrix.
Question # 4
Find all the eigenvalues and eigenvectors of the matrix A:
4 1 1 1
16 3 4 4
7 2 2 1
11 1 3 4
A
Also, show that the matrix A is defective. Solution:
First of all we will get the eigenvalues of the given matrix for which we will solve the equation determined by the determinant det( ) 0A tI now we have,
4 3 2 3det( ) 5 9 7 2 ( 1) ( 2)A tI t t t t t t
Thus det( ) 0A tI gives us t = 1, 1, 1, 2 as roots of the characteristic polynomials which
are the eigenvalues of the given matrix. As the eigenvalue 1 repeats three times so it has
Algebraic multiplicity of eigenvalue 1 is 3 and Algebraic multiplicity of eigenvalue 2 is 1.
In order to get the geometric multiplicity we will find out the dimension of the eigenspaces correspond to these eigenvalues, for this we will solve the homogeneous
system.
1 0 ,
6 1 1 1 0 6 1 1 1 0
16 1 4 4 0 48 3 12 12 0
7 2 0 1 0 42 12 0 6 0
11 1 3 2 0 66 6 18 12 0
6 1 1 1 0 6 1 1 1 0
0 5 4 4 0 0 5 4 4 0
0 5 7 1 0 0 0 3 3 0
0 5 7 1 0 0 0 0 0 0
A I and the augmented matrix is
4
1
2 4
4
3
44
4
3 3
5 5
8 8
5 5
1
1
x
x
x xx x
x
xx
x
Thus the vector
3
5
8
5
1
1
form bases for the eigenspaces correspond
to the eigenvalue 1. Thus the geometric multiplicity of the eigenvalue 1 is 1 which is less then the algebraic multiplicity of 1. Thus the matrix is by definition defective.
5 1 1 1 0 5 1 1 1 0
16 2 4 4 0 80 10 20 20 02
7 2 1 1 0 35 10 5 5 0
11 1 3 3 0 55 5 15 15 0
5 1 1 1 0 6 1 1 1 0
0 6 4 4 0 0 3 2 2 0
0 3 2 2 0 0 0 0 0 0
0 6 4 4 0 0 0 0 0 0
A I
3 4
1
2 3 4 3 4
3
34
4
1 13
2 2 2
3 03 3 3
0 3
x x
x
x x x x xx
x
xx
x
Shows that dimension of eigenspaces is 2 which is the geometric multiplicity of the eigenvalue 2 thus the algebraic multiplicity of eigenvalue 2 is less then Geometric multiplicity of eigenvalue 2.
But eigenvalue 1 has geometric multiplicity less than algebraic multiplicity thus the given matrix is defective.
Question # 5
(i) Find all 2 2 matrices for which 1
2
is an eigenvector correspond to the
eigenvalue 5.
Solution:
We have to find out the matrix a b
c d
such
that
1 1 2 55
2 2 2 10
2 5 5 2
2 10 10 2
a b a b
c d c d
a b a b
c d c d
thus the required matrix
is5 2
10 2
b b
d d
where b and d are any real numbers.
(ii) True or false; A square matrix A is invertible if and only if 0 is not an eigenvalue of A.
Solution:
The above statement is true. (iii) If 5 is an eigenvalue of matrix A then find the eigenvalue of A5 without
any calculations and justify your answer. Solution:
As we are given 5 is the eigenvalue of A then the eigenvalue of A5 will be
55. Justification:
Suppose is an eigenvalue of A then we will show that A2 will have
eigenvalue 2 . Let x be the eigenvector correspond to that eigenvalue then by
definition we have,
2
2
2 2 2 2
,
.
Ax x
A Ax A x A x Ax
but Ax x so
A x x
A x x is an eigenvalue of A
By using principle of Mathematical induction we can show that k will be the eigenvalue
of 2,3,4...kA for k .
(iv) Find eigenvalues of the matrix
1 2 3 5 0
0 9 4 2 1
0 0 3 1 2
0 0 0 4 3
0 0 0 0 1
A
without any
calculation also find the eigenvalues of the matrix A-1. If you are given that A is invertible.
Solution:
Since we know that a triangular matrix has its eigenvalues as diagonal elements, so the eigenvalues of the given matrix are 1, 9, 3, 4. Also we are given A is
invertible so the eigenvalues of A-1 will be 1,1 1 1
, ,9 3 4
. Where we use the fact that if is an
eigenvalue of an invertible matrix A then1
will be an eigenvalue of A-1.
(v) Consider the matrix1
1 1
kA
, where k is any arbitrary constant. For
what values of k the matrix A will have two distinct eigenvalues? When is
there no real eigenvalue? Solution:
In order to get the answer of the asked questions we will solve the
equation
210 0 1
1 1
1
kA I k
From this you can easily decide that for k = 0 you will get the repeated eigenvalues and all other values give distinct eigenvalues. Also if the value of k is less then zero then you will get the complex eigenvalues so there will be no real eigenvalue if k is less then 0.
Question # 6
Show that A is diagonalizable, where
25 8 30
24 7 30
12 4 14
A
.
Solution:
First of all we will find out the eigenvalues of the given matrix which
are 1 2 31 2and . A will be diagonalizable if it will have three linearly
independent eigenvectors correspond to these eigenvalues. So now we will calculate the
eigenvectors correspond to these eigenvalues. Eigenvector correspond to eigenvalue 1 are
1 2
1 4
3 3
0 4
u and u
and eigenvector correspond to the eigenvalue 2 is 3
4
4
2
u
.
Take the matrix
1 4 4 1 0 0
3 3 4 0 1 0
0 4 2 0 0 2
P and D
our claim is that A = PDP-1 so
we will not calculate the inverse of the matrix P but equivalently we will show that AP = PD.
Question # 7
Determine whether the signals 2 , 4 , ( 5)k k k are solution of the difference
equation 3 2 122 40 0k k k ky y y y . Also determine whether these signals
forma basis for the solution space of the same equation. Solution:
Take 3
32 2 .k k
k ky then y and soon
the difference equation becomes,
3 2 1
3 2
2 2 222 402 0
2 2 2 22(2) 40 0
0 0
k k k k
k
Hence 2k is a solution of the given difference equation.
Take 3
34 4 .k k
k ky then y and soon
the difference equation becomes,
3 2 1
3 2
4 4 224 404 0
4 4 4 22(4) 40 0
0 0
k k k k
k
Hence 4k is a solution of the given difference equation.
Take 3
35 5 .k k
k ky then y and soon
the difference equation becomes,
3 2 15 5 22 5 40 5 0
5 125 25 22( 5) 40 0
0 0
k k k k
k
Hence 5k
is a solution of the given
difference equation.
Now take the Casorati matrix of the given signals, and if the signals are linearly independent then these signals will form the bases for the solution space.
2 (4) ( 5)
2 (4) 5 0
2 (4) 5
1 1 1 1 1 1
0 2 7 0 2 7
0 4 21 0 0 35
Take k
k k k
k+1k+1 k+1
k+2k+2 k+2
1 1 1
2 4 -5
4 8 25
Since the echelon form of the matrix has pivot element thus the signals are linearly independent and form bases for the solution space of given difference equation.
Solution of Assignment # 6 of MTH501 (Fall2004)
Q1. Define 4
3
( 3)
( 1): ( )
(1)
(3)
p
pT P by T p
p
p
then,
c. Show that T is a linear transformation.
d. Find the matrix for T relative to the basis {1, t, t2, t
3}
for 3P and the standard basis for 4 .
Solution: First of all we will show that the given transformation is linear
for this we will show that (i) T( p + q) = T(p) + T(q) (ii) T(cp) = cT(p)
Where p and q are arbitrary elements of polynomial space P3 and c is scalar.
(i) Let
2 3 2 3
0 1 2 3 0 1 2 3
2 3
0 0 1 1 2 2 3 3
p a a x a x a x and q b b x b x b x
p q a b a b x a b x a b x
then we have,
0 0 1 1 2 2 3 3
0 0 1 1 2 2 3 3
0 0 1 1 2 2 3 3
0 0 1 1 2 2 3 3
0 1 2 3 0 1 2
0 1 2 3
0 1 2 3
0 1 2 3
3 9 27
3( )
3 9 27
3 9 27 3 9 2
3 9 27
a b a b a b a b
a b a b a b a bT p q
a b a b a b a b
a b a b a b a b
a a a a b b b
a a a a
a a a a
a a a a
3
0 1 2 3
0 1 2 3
0 1 2 3
7
3 9 27
( ) ( )
b
b b b b
b b b b
b b b b
T p T q
Similarly you can show that the second condition satisfied by T.
(ii) First of all we will find out the image of bases of P3 and then we will find out the coordinate of these images in terms of standard
basis of R4.
4
4
4
1 1
1 1(1) (1) tan
1 1
1 1
3 3
1 1( ) (1) tan
1 1
3 3
9
1( ) (1)
1
9
T Coordinate of T in R with s dard basis
T t Coordinate of T in R with s dard basis
T t Coordinate of T in R wi
4
9
1tan
1
9
27 27
1 1( ) (1) tan
1 1
27 27
th s dard basis
T t Coordinate of T in R with s dard basis
So the required matrix is
1 3 9 27
1 1 1 1
1 1 1 1
1 3 9 27
.
Q2. Find the B-matrix of the
transformation
14 4 14
33 9 31
11 4 11
x Ax where A
, when B= {b1, b2,
b3} where 1 2 3
1 1 1
2 , 1 , 2
1 1 0
b b b
.
Solution: Let the transformation be T defined
by
14 4 14
33 9 31
11 4 11
x Ax where A
then the required B-matrix will
be
1 2 3( ) ( ) ( )T b T b T b .
Now
1 14 4 14 1 14 8 14 8
( 2 ) 33 9 31 2 33 18 31 15
1 11 4 11 1 11 8 11 8
1 14 4 14 1 14 4 14
( 1 ) 33 9 31 1 33 9 31
1 11 4 11 1 11 4
T
T
4
7
11 4
1 14 4 14 1 14 8 0 6
( 2 ) 33 9 31 2 33 18 0 15
0 11 4 11 0 11 8 0 3
T
So the required matrix is
8 4 6
15 7 15
8 4 3
.
Q3. Let T be the transformation whose standard matrix is given
by
5 0 0 0
0 3 0 0
1 4 3 0
1 2 0 3
. Find a basis for 4 with the property that B
T is
diagonal.
Solution: First of all we will find out the Eigenvalues of the given matrix,
also we know that eigenvalues of triangular matrix are the elements in its diagonal so eigenvalues of the given matrix are 5,-3 and 3. Now we
will find out the basis for eigenspaces of these eigenvalues. For eigenvalue 5 we will solve the homogeneous
system
1
5 0
0 0 0 0 0 0 0 0 0 0 0 0
0 2 0 0 0 2 0 0 1 4 8 0
1 4 8 0 1 4 8 0 0 1 0 0
1 2 0 8 0 2 8 8 0 0 8 8
8
0
1
1
A I augmented matrix is
and basis for eigenspace is b
basis
Basis of the eigenspaces for eigenvalue 3 is2
0
3
2
1
b
, basis for -3,
1 2
0 0
0 0,
1 0
0 1
c c
Now taking
8 0 0 0 5 0 0 0
0 3 0 0 0 3 0 0
1 2 1 0 0 0 3 0
1 1 0 1 0 0 0 3
P and D
.
Now using the Diagonal Matrix Representation Theorem on page # 325 the
required B-matrix of the transformation is
5 0 0 0
0 3 0 0
0 0 3 0
0 0 0 3
D
.
Q4.
For the matrix1 2
1 3
, find an invertible matrix P and a matrix C
such that
A = PCP-1.
Solution:
First of all we will find out the eigenvalues of the given
matrix.
2
2
1 20 4 3 2 0
1 3
4 5 0
1, 4 5 sin .
4 16 202
2
Here we have a b and c then u g quadratic formula
i
Now we will find out the eigenvector correspond to the eigenvalue 2 - i, for
which we will solve
1
2
1 2
1 2
1 2 0
1 1 0
1 2 0......(1)
1 0...........(2)
xi
i x
i x x
x i x
In order to get the nontrivial solution we should know that both equation 1
and 2 must determine the same relationship between x1 and x2.And we can use either equation to get the relationship between these two variables.
Take equation 2 we get 1 1x i by taking x2 = 1 then the eigenvector will
be1
1
iv
.
Now taking Re Im 2, 1a b
P v v and C where a bb a
.So we have
1 1 2 1,
1 0 1 2P C
And you can check now that A = PCP-1
.
Q5. Solve the initial value problem
0, (0)x Ax x x where 0
3 1 1 7
12 0 5 3
4 2 1 16
A and x
Solution:
First of all we will find out the eigenvalues of the matrix A now the eigenvalues of the matrix A are -1, 1, 2 and corresponding eigenvectors
are 1 2 3
2
1 1 2 2 3 3
1 3 1
2 , 1 , 1
2 7 2
( ) t t t
u u u
and thus general solution is x t a e u a e u a e u
.Apply the initial
condition we get the system 1 2 3
7 1 3 1
3 2 1 1
16 2 7 2
a a a
by solving that
system you will get the values of constants. Q6.
Consider the vectors
7 1
6 , 5
13 3
w x
e. Find distance between w and x. f. Compute a vector in the direction of w having unit length.
g. Compute. .w x w w
x and wx x
.
Let 1 2, ,..., nw span v v v . Show that if x is orthogonal to each vj,
for1 j n then x is orthogonal to every vector in w.
Solution: Since we know that distance between w and x
is
2 2 27 1 6 5 13 3 64 1 100
165
w x
w x
We will find out the unit vector in the direction of w which is
71
6254
13
w
w
.
.
. 7 30 39 62, 1 25 9 35
w xx
x
w x x
So we have
1. 62
535
3
w xx
x
.
Similarly we have
3
2254. 254254
35 35
w ww
x
.
Since each element y in w will be of the
form 1 1 2 2 1 2... , ,..., n .n n ny c v c v c v where c c c are real umbers . Now take
1 1 2 2 1 1 2 2
1 2
. . ... . . ... .
. 0
. (0) (0) ... (0) 0
n n n n
i j
n
x y x c v c v c v c x v c x v c x v
Since we have x v for i j
x y c c c
Hence proved as we take arbitrary element of w.
Q7. Determine whether the QR factorization of the matrix
1 2 5
1 1 4
1 4 3
1 4 7
1 2 1
is
possible or not? Justify! If it is possible then find Q and R. Solution:
First of all we will show that columns of the given matrix are linearly independent, which will show that the matrix can be factored
as QR.
1 2 5 1 2 5
1 1 4 0 3 1
1 4 3 0 6 2
1 4 7 0 6 2
1 2 1 0 0 4
From this it is quite clear that every column
has a pivot position hence columns of the given matrix are linearly independent. So QR factorization is possible.
Now we will find out the orthogonal basis for the column space of the given
matrix note that the vectors 1 2 2
1 2 5
1 1 4
1 , 4 , 3
1 4 7
1 2 1
x x x
form basis
for the column space. We will use Gram Schmidt process to find out the orthogonal basis for the column space.
2 11 2 2 1
1 1
2 1 1 1 2
3 1 3 23 3 1 2 3 1 3 2 2 2
1 1 2 2
3
1
1.
1.
1
1
3
0
. 5 . 5 3
3
3
. ., . 20, . 12, . 36
. .
2
0
2
2
2
x vv v x v
v v
x v and v v v
x v x vv x v v x v x v v v
v v v v
v
Normalize these vectors we get, 1 2 3
1 1
2 21 30 0
1 01 1 1 1
1 , 3 ,6 2 25
1 31 1
1 32 2
1 1
2 2
v v v
Now form the matrix Q which has columns as normalize basis which we
calculate above. So we have
1 1 1
2 25
10 0
5
1 1 1
2 25
1 1 1
2 25
1 1 1
2 25
Q
Find the matrix R by using the equation Qt A = R.
Q8.
(i) Let3 8
1 6y and u
. Compute the distance from y to the line
through u and the origin. (ii) Find the closest point to y in the subspace W spanned by v1 and
v2 where 1 2
3 3 1
1 1 1, ,
5 1 1
1 1 1
y v v
Solution: (i) Since we know that the distance from a vector y to a line
through the line from u and origin is y projection of y on u also we know
that
.
.
83. 30, . 100
610
y uprojection of y on u u
u u
y u u u projection of y on u
3 8 5.43
1 6 2.810
29.16 7.84 37
y projection of y on u
y projection of y on u
(ii) Since we know the closest point to y in the subspace W spanned by v1
and v2 is
1 21 2
1 1 2 2
1 1 1
11
1 1
2 2 2
11
1 1
. .
. .
. 9 1 5 1 6, . 9 1 1 1 12
3
1. 1
1. 2
1
. 3 1 5 1 6, . 4
1
1. 3
1. 2
1
3 1 6
1 1 21 3 1
1 1 22 2 2
1 1 2
y v y vy v v
v v v v
y v v v
y vv
v v
y v v v
y vv
v v
y
3
1
1
1
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