Linear Algebra - Solved Assignments - Fall 2004 Semester

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. Ans wer:

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           Ans wer:

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 1 0 1 12 12 12 0 5 4 1 1 4 1 5 5 19 0 1 5 12 12 23 83 0 0 12 12 by placing R by R by R R R R R R         _{ }   _                   _  _{ }              _       _{ }           _    _           3 3 1 4 1 5 5 19 0 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?

Ans wer:

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 2

y t is also the general solution of the equation in part (a).

Ans wer: (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”

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 2 y t

This shows that x=t, 1 5

2 2

y 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.

Ans wer:

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.

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 1 1 5 1 2 7 3 3 1 1 1 0 4 5 5 , 7 0 4 10 1 1 1 5 1 0 1 4 4 0 4 10 1 1 1 5 0 1 4 4 0 0 5 k k k R k k R k k R R R R k k R k k R 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.

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 1/12 9 0 5 / 4 1/ 4 1/ 4 3 / 4 19 / 4 6 0 15 / 2 3 / 2 3 / 2 13 / 2 11/ 2 4 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                _    _      _  _  _  _  _    _                _    _                

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 ofx3  x5in first equation we get,

1 2 2 5 5 1 2 5

x   x x x x   x x which shows thatx2andx5are free variables. So we can take x2 t and x5 swhere 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 s x                                                        _ _  _{ }  _{ }   . 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:

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 1 0 1 3 3 1 0 0 1 2 50 0 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 c 1              .

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 mapping 2 5 :

T R R defined by

T(x, y) = (x, x-y, 2x-4y, y, x + y) is a linear transformation. Solution (a):

(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 3 1 1 2 2 1 3 3 2 1 1 1 2 0 0 2 0 1 0 x x x x 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 3 3 0 x x T x x x        _            

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 t A       _{ }      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          _{  }  _{ }                 . Hencex1,y 2,z3are 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

(ii) The elementa₁₁of the matrix BA without calculating the matrix BA Solution (i):

Since the a₃₂ 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 a11 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(AB)1 B A1 1. 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)

Q4. Consider the Transformation T: R2  R2 defined by T(x) = A_x where cos sin sin cos A_          

 the show that,

(i) The mapping T is a rotation through the angle. (ii) A A_{ }  A_{ }

(iii) Find the inverse ofA_. Solution (i):

First of all you should note that the transformation defined by the formula T(x) = A_x is a linear transformation. Let x a

b        and we can write 1 0 1 0 0 1 0 1 a a x 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 cos

T    and T               _    _    _    _                                 

Now consider the geometric presentation of these two images, we can take the unit vectors 1 0      and 0 1    

 along the unit circle obviously unit vector 1 0    

 makes an angle 0 with the x-axis and vector 0

1    

 is along the y-axis which makes an angle

2 

with the x-axis as shown in the fig below.

Now as we can easily see that image of 1 cos 0 0 sin 0             is a rotation through an angleas 1 cos( 0) 0 sin( 0 T            _      similarly we have cos 0 ₂ 1 sin 2                     

and this is also a

rotation of an angleas cos( ) 0 ₂ 1 sin( ) 2 T      _              _     

. Since from the result of

trigonometry we havecos( ) sin sin( ) cos

2 and 2

 

       . Thus in fact

the given matrix of transformation only rotates that point through an angle of.

Now as every element 1 0

0 1 a x 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 cos A_            thus cos sin sin cos A_            , 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 cos

A 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 1 a b a b A then A c d ad bc c d       _{ }  _{ }      

Thus by using the above formula we

have 1 2 2 2 2 1 cos sin 1 cos sin 1 sin cos cos 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 0 6 2 4 4 1 0 2 4 1 1 1 0 0 , 2 , 2 0 4 13 0 2 2 * 1 0 0 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 firs t matrix.

Step2:

Zero out the entries below the second entry of the diagonal.

3 2 4 2 1 0 0 0 6 2 4 4 1 0 2 4 1 1 0 0 2 , 2 0 0 5 2 2 2 1 0 0 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 0 6 2 4 4 1 0 2 4 1 1 0 0 2 2 0 0 5 2 2 2 1 0 0 0 0 8 1 1 2 1 R R       _{ }  _{  }  _{ }   _{  }    _{  }   _{ }   _{  } .

Thus we have 1 0 0 0 6 2 4 4 1 0 2 4 1 1 0 0 2 0 0 5 2 2 2 1 0 0 0 0 8 1 1 2 1 U L       _{ }  _{  }  _{ }   _{  }    _{ }    _{ }   _{  } .

Now our system becomes LUx = b where,

1 0 0 0 6 2 4 4 2 1 0 2 4 1 4 1 0 0 , , 2 0 0 5 2 8 2 2 1 0 0 0 0 8 43 1 1 2 1 u v L U x and b w y   _{       }   _{     }   _    _{   } _ _{ }           _{ }  _{     }   _{     }       Let Ux = z where 1 2 3 4 z z z z z             

and by solving Lz = b we get,

1 1 2 2 1 3 3 1 2 4 4 1 2 3 2 1 0 0 0 2 1 1 4 4 5 1 0 0 2 2 8 8 2 2 2 2 2 1 0 43 1 1 2 1 43 2 32 z z z z z z z z z z z 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    _       

22 23 22 22 32 32 22 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 2 5 1 2 2 3 3 4 1 4 1 1 9 7 5 7 6 A O C O U and V O D O B A C O O A O O D UV 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

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

1 2 1 1 1 3 2 1 2 3 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 k k k k k k k x x x x x x x

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 2 1 1 1 1 3 2 1 1 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 k k k k K K k x x x x x x x 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 16384 x 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

row 2 3 4 0 7 3 5 4 0 3 5 0 0 2 0 0 7 3 4 8 7 3 6 4 8 ( 1) 5 0 2 3 5 0 5 2 3 0 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 5 4 0 3 5 0 0 2 0 0 7 3 4 8 7 3 6 4 8 ( 1) 2 5 0 2 3 5 0 5 2 3 0 0 1 2 0 0 9 1 2 4 3 5 2 ( 1) 3 5 2 3 0 1 2                      _   _  _    Again

get

 



 







2 3 2 2 3 2 3 3 4 0 7 3 5 4 0 3 5 0 0 2 0 0 7 3 4 8 7 3 6 4 8 ( 1) 2 5 0 2 3 5 0 5 2 3 0 0 1 2 0 0 9 1 2 4 3 5 4 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 3

a 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 1 10 5 1 5 1 10 1 10 5 3 2 1 120 16 8 128 2 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 1 3 7 2 7 2 3 2 3 7 1 1 1 14 7 1 6 5 7 3 7 3 5 3 5 7

         but volume will be

3 2 1 4 5 0 4 5 3 3( 18) 54 2 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) Letn1. 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 allx y V,  we have x + y = y + x

(b) For allx y z, , Vwe have (x + y) + z = x + (y + z) (c) For allx y z, , Vwe 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 havex 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) U



xRn/,x₁x₂  ... x_n



(b)



2 2



1 2 /, n U xR x x (c) U



xRn/,x₁ 1



Solution:

In order to check whether the given subsets are subspaces or not we will check the two conditions.

(i) Ifx y, U   x y U.

(ii) Ifc F  and vU then cv U

(a) U



xRn/,x₁x₂  ... x_n



is a subspace of Rn as for,

1 2 1 2 1 2 1 2

, ( , ,... _n) ( , ,... _n) ... _n, ... _n

x 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 take 1 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 /, n U 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,wU

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) U



xRn/,x₁ 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 {2x x, 3x 1,x }of

     find the coordinate of the

2

16x 5x9relative 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 16x2 5x9relative to the given

bases 2 2 2

2 {2x x, 3x 1,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

{x x 1,x 1,x 1}, {x x 4, 4x 3x 2, 2x 3}

            are

(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 2

4 1 1 1

x   x a x   x b x c x  And solving as we did in above question we

get 2

2

2, 3 1 4 3

1 a b and c thus coordinate of x x _

          _   _ _{ }     . Similarly we have 2 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 2 4 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 2 1 ( ) 6 5 2 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 nn 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 determinantdet(A tI )0now we have,

4 3 2 3

det(A tI ) t 5t 9t    7t 2 (t 1) (t2)

Thusdet(A tI )0gives 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 4 4 4 3 3 5 5 8 8 5 5 1 1 x x x x x x x x x 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 0 2 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 3 4 4 1 1 3 2 2 2 3 0 3 3 3 0 3 x x x x x x x x x x x x 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 5 5 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

is 5 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kwill be the eigenvalue ofAk for k2,3, 4... .

(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 matrix 1

1 1

k A  _

 , 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 q uestions we will solve the

equation





2 1 0 0 1 1 1 1 k A 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 1 and 3 2. 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 signals2 , 4 , ( 5)k k  kare solution of the difference equationyk3yk222yk140yk 0. Also determine whether these signals

forma basis for the solution space of the same equation. Solution:

Take 3

3

2k 2k .

k k

y  then y    and so on 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 y_k 4kthen y_k34k3and so on.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 4kis a solution of the given difference equation.

Take y_k  

 

5 kthen y_k3  

 

5 k3and so on.the difference equation becomes,

 

 

 

 

  



3 2 1 5 5 22 5 40 5 0 5 125 25 22( 5) 40 0 0 0 k k k k k                   

Hence

 

5 kis 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 + 1 k + 1 k + 1 k + 2 k + 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 p T 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, t

2

, t

3

}

for

P3

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 P

3

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 b T 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 P

3

and then we

will find out the coordinate of these images in terms of standard

basis of R

4

.

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 1 tan 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= {b

1

, b

2

,

b

3

} 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