Mathematics

Which of the following is not an exact differential?
  • a)
  • b)
  • c)
    xdy - ydx
  • d)
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  5 hours ago


∴ expressions in (a), (b) and (d; are exact.
Since the expression
xdy - ydx
in (c) can not be expressed as du for some function u of x and y, therefore the expression in (c) is not exact.

What is the rank of the linear transformation T : R3 ---> R3 defined by T(x, y, z) = (y, 0, z)?
  • a)
    3    
  • b)
    2
  • c)
    l    
  • d)
    0
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered  •  5 hours ago
We are given that the linear transformation T : R3 ---> R3 defined by
We need to determine the rank of the linear transformation T.
Let (x, y, z) ∈ ker T
ThenT(x, y, z) = (0, 0, 0)
Using the definition of linear transformation, we get,

implies y = 0, z = 0 and x is arbitrary Therefore, ker T = {( x, y, z ) : y = 0, z = 0 and x is arbitrary}
Hence, Nullify of T= 1
Using Rank Nullity theorem, we get Rank T= dim R3 - Nullity T
= 3 - 1 = 2

If the linear transformation T(v) = Av rotates the vectors (-1, 0) and (0,1), x radians clockwise, then:
  • a)
    A =  
  • b)
    A = 
  • c)
    A = 
  • d)
    A = 
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  5 hours ago
We are given that the linear transformation T(v) = Av rotates the vectors (-1,0) and (0,1), π radians clockwise.
We need to find the matrix A.
We know that if a vector (a, b) is rotated through an angle a under T. Then


Therefore, the matrix of T is
A = 

 If X = ,  the rank of X
T
X, where XT denotes the transpose of X, is
  • a)
    0
  • b)
    2
  • c)
    3
  • d)
    4
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  5 hours ago
We are given that the matrix
X=
 
We need to find the rank of XTX. The transpose of X,
XT=
 
Now, 
           
Rank (XTX) = 3

Which one of the following is true?
  • a)
    dim Horn (R3, R4) = 12
  • b)
    dim Horn (R4, R3) = 7
  • c)
    dim Horn (R3, R4) = 7
  • d)
     None of these.
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  5 hours ago
We need to find the dim Horn (R3, R4).
Since dim R3 = 3 and dim R4 = 4
Therefore,dim Horn (R3, R4)
= 3 x 4 = 12
Let U and V be n and m dimensional vector space.
Then dim Hom (U, V) = nm

If A = satisfies the matrix equation A2 - kA + 2I = 0, then what is the value of k?
  • a)
    0
  • b)
    1
  • c)
    2
  • d)
    3
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered  •  5 hours ago
We need to find the value o f k where the matrix
A = satisfies the equations A2 - kA + 2I = 0.
But A = 
So, 
Putting these values in the equation
A2 - kA + 2I= 0, we get
or 
or 1 - 3k + 2 = 0 => k = 1

The system of equations
  • a)
     a unique solutions
  • b)
    finitely many solutions
  • c)
     infinitely many solutions
  • d)
     no solution
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that the system of equation,

It may be written in the matrix form as,

Here, the coefficient of matrix is,

and the augmented matrix is given by,

Reduce the system equation in echelon form using the operations
 R1" and
R3 -> R3 - R1
These operation yield:

and also, R3 --> R3 - 2R2

Here, Rank of A = Rank of aug(A) < number of unknowns.
Hence, the given system of equation has infinite many solutions.

The system of equation 2x + y = 5, x - 3y = -1
3x + 4y = k is consistent, when k is
  • a)
    1
  • b)
    2
  • c)
    5
  • d)
    10
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that the system of equations,
2x + y = 5
x - 3y = - 1
3x + 4y = k is consistent.
We need to find the value of k. The given system of equation may be written as, 
Since, this system of equation is consistent. Therefore

or 2(-3k + 4) -1(k +3) + 5(4 + 9) = 0
-7k + 70 = 0 
k =10

Let :R3 --> R2 be the linear transformation given by
T(x, y, z) = (x, y),  
with respect to standard basis of R3 and the basis {(1,0), (1, 1)} of R3. What is the matrix representation of T?
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
 Let T : R3--> R2 be the linear transformation defined by
T(x, y, z) = (x, y)
we need to determine the matrix of linear transformation T w.r.t. standard basis of R3 and the basis {(1, 0), (1 ,1 } of R2. since {(1 ,0 ,0) , (0,1,0) (0,0,1)} be the standard basis of R3, therefore,

Thus, the matrix of linear transformation T w.r.t. the standard basis R3 and the basis

Consider, the linear transformation
T : R4 ----> R4 given by: 
T(x, y, z, u) = (x, y, 0, 0),
Then, which one of the following is correct?
  • a)
    Rank of T > Nullity of T
  • b)
    Nullity of T > Rank of T
  • c)
    Rank of T = Nullity of T = 3
  • d)
    Rank of T = Nullity of T = 2
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that linear transformation

We need to determine Rank and Nullity of T.
Let (x,y, z, u) ∈ ker T
Then T(x, y, z, u) = (0, 0, 0, 0)
Using the definition of linear transformation we get
(x,y, 0, 0) = (0, 0, 0, 0)
implies x = 0, y = 0, z and u are arbitrary
Therefore,

Hence, Nullity of T = 2
Using Rank Nullity theorem, we get

Let M = . Then, the rank of M is equal to
  • a)
    3
  • b)
    4
  • c)
    2
  • d)
    1
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We need to find the rank of the matrix,

Reduce the matrix to echelon form using the operation.
These operations yield.

and also applying R4 —> 2R4 - R2 we have,

So, rank of M = 2.

If T : R3 ---> R3 is given by
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that a linear tansformation T : R3---> R3 defined by T(x, y, z) = (x - y , y + 3 z, x + 2y).
We need to find T-1.
Let (x, y, z) ∈ ker T. Then
T(x,y, z) = (0, 0, 0)
Using the definition of T, we get
(x -y,y + 3z, x + 2y) = (0, 0, 0) Comparing the components on both sides, we get
x - y =0 , y + 3z = 0, x +2y = 0
Solving for x, y and z, we get
x = 0, y = 0 and z = 0
Hence, ker T= {(0, 0, 0)}, implies T is one-one.
Since T : R3 —> R3 is a linear tansformation and one-one. Hence, T is onto.
Therefore, T-1 exist.
Let (a,b,c) be the image of (x, y, z) under T-1.
Then
T-1(x, y, z) = (a, b,c)
Implies (x, y, z) = T(a, b, c)
Using the definition of T, we get
(x, y, z) = (a - b, b + 3c, a +2b)
Comparing the components on both sides, we get
x =a - b, y = b + 3c, z = a + 2b
Solving for a, b and c, we get

Therefore, T-1(x,y, z)

Solution for the system defined by the set of equations 4y + 3z = 8; 2x - z = 2 and 3x + 2y = 5 is
  • a)
    x = 0; y = 1; z = 4/3
  • b)
    x = 0; y = 1/2; z = 2
  • c)
    x = 1 ;y= 1/2; z = 2
  • d)
    not exist
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We need to determine the solution for the system of equation
4y+3z= 8
2x - z = 2
3x + 2y= 5
This system of equation may be written in matrix form as,

Reduce this system of equation to echelon form using the operations
"R2 ⇔ R1". This operations yields

and also applying "R→ 2R3 - 3R1" which yields

again, "R1→ R3 - R2 " we get

Here the coefficient matrix 
and augmented matrix is given by 
Since, the rank of coefficient matrix ≠ the rank of augmented matrix.
Therefore, the solution does not exist.

The eigen vectors o f the matrix  are written in the form  and .  What is a + b ?
  • a)
    0
  • b)
    1/2
  • c)
    1
  • d)
    2
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that the eigen vectors o f the matrix  are and . We need to find the value o f a + b. The characteristic equation o f the given matrix is, |A - λI| = 0

or (1 - λ)(2 - λ) = 0 or 
λ = 1,2
Put λ = 1 in the equation [ A- λI ] X= 0

or a = 0 and also put X = 2 in the equation [A - λI] X= 0
or 
or 1+2b = 0
or b = 1/2
Now, a + b = 0 + 1/2
a + b = 1/2

For n ≠ m , let T: Rn —> Rm and T2 : Rm —> Rbe linear transformations such that T1T2 is bijective. Then
  • a)
    rank (T1) = n and rank (T2) = m
  • b)
    rank (T1) = m and rank (T2) = n
  • c)
    rank (T1) = n and rank (T2) = n
  • d)
    rank (T1) = m and rank (T2) = m
Correct answer is option 'D'. Can you explain this answer?

Veda Institute answered  •  16 hours ago
We are given that T1 : Rn —> Rand T2 : Rm —> Rn be linear transformation such that T1T2 is bijective, where n ≠ m
Thus, T1T---> Rm ---> Rn is a linear transformation given by
(T1T2) (n) = T1(T2(n))
Since T1T2 is bijective. Therefore, T1Tis non-singular and hence
Rank (T1T2) = m
But Rank (T1T2) < min Rank {T1T2} Therefore,m ≤ min Rank {T1T2}
Case I : If m ≤ n, Then rank of both T1 and T2 will be m.
Thus, we get
Rank (T1) = Rank (T2) = m
Case II : If m ≥ n. Then since T1 is n x m matrix. Therefore, its rank cannot exceed n. Similarly, Rank T2 cannot exceed n. Thus, we get m ≤ n. Therefore m = n.

Let T : R3 —> R3 be defined by T(x1 x2, x3)) = (x1 - x2, x1 - x2,0).
If N(T) and R(T) denote the null space and the range space of T respectively, then
  • a)
    dim N(T) = 2
  • b)
    dim R(T) = 2
  • c)
    R(T) = N(T)
  • d)
    N(T) ⊂ R(T)
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered  •  yesterday
Let T : R3---> R3 be def ined by T (x1, x2, x3) = (x1 - x2, x1 - x2, 0).
and N(T) and R(T) denote the null space and the range space of T respectively.
Let (x1, x2, x3) ∈ ker T.
Then
T(x1 x2, x3) = (0, 0, 0)
using the definition of T, we get
(x1 - x2, x1 - x2, 0) = (0, 0, 0)
Implies x1 = x2, x3 is arbitrary.
Therefore,ker T= {(x1, x2, x3) : x1, x3 arbitrary}
Hence, nullity of T = 2
and Rank of T = 3 - 2 = 1.

A system o f linear equations x + 2 y - z = l l , 3 x + y - 2 z = 10, x - 3y = 5 has
  • a)
    no solution
  • b)
    exactly one solution
  • c)
    exactly 3 solutions
  • d)
    infinitely many solution
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  yesterday
We are given that the system of equation
x + 2y - z =11
3 x + y - 2z = 10
x - 3 y = 5
The given system of equation can be written in the augmented matrix as,

Applying the operations "R2 ---> R2 - 3R1" and R3 —> R3 - Rthese operations yield

and also "R3 ---> R3 - R2 ", which gives,

Here, rank of (A) ≠ rank of aug (A).
Hence, the given system of equation has no solution.

The characteristic polynomial of 3 x 3 matrix A  |λI - A| = λ3 + 3λ2 + 4λ - 3
Let x - trace (A) and y = |A|, the determinant of A.
Then,
  • a)
    x/y = 3/4
  • b)
    x/y = 4/3
  • c)
    x = y = -3
  • d)
    x = 3 and y = -3 
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  yesterday
We are given that the characteristic polynomial o f 3 x 3 matrix, A is
λ3 + 3λ2 + 4λ - 3 = 0
and x = trace (A)
       y = det(A)
The general characteristic polynomial of 3 x 3 matrix is given by,
λ3 - trace (A) λ2 + λ + |A| = 0
Comparing this equation by the given equation, we get
trace = -3 
or x = -3
and |A| = -3
or y = -3
Hence, x = y = -3
 

Let T : R3 --> R3 be defined by T(x1, x2, x3) = (x1 + x2, x2 + x3, x3 + x1) Then T-1 is
  • a)
  • b)
  • c)
  • d)
    None of these
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  yesterday
We are given that a linear transformation T :R3 --> R3 defined by
T(x1, x2, x3) = (x1 + x2, x2 + x3, x3 + x4)
We need  to find T-1
Let (x, y, z) ∈ ker T
then T(x, y, z) =(0, 0, 0)
Using the definition of linear transformation, we get 
(x1 + x2, x2 + x3, x3 + x4) = (0, 0, 0)
Comparing the components of the coordinates, we get
x1 + x2 = 0
x2 + x3 = 0
x3 + x1 = 0
Implies, x1 = 0, x2 = 0 and x3 = 0
Therefore, ker T = {(0,0,0)} , hence, T is one-one.
Since T is a linear transformation from R3 to R3 but R3 is a three dimensional (finite dimensional) vector space.
Hence, T is onto and therefore, T-1 must exist.
Let (x, y, z) be the image of (a,b,c) under T-1,
Therefore T-1(a ,b ,c ) = (x,y,z) Implies, T{x, y, z) = (a, b , c ) Using the definition of linear transformation, we get
(x + y , y + z ,z + x) = (a , b, c) Comparing the components of the co-ordinates, we get
x + y = a
y + z = b
z + x - c
Solving for x, y and z, we get 


and 
Hence,T-1(a, b, c)

That is 
T-1(x,y,z) = Hence,T-1(x1, x2, x3)

 
... more

The system of simultaneous linear equations
x + y + z = 0
x - y - z = 0 has
  • a)
    no solution in R3
  • b)
    a unique solution in R3
  • c)
    infinitely many solution in R3
  • d)
    more than 2 but finitely many solutions in R3 
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  yesterday
We are given that the system of simultaneous linear equations,
x + y + z = 0
x - y - z = 0
The coefficient matrix is given by,

rank of coefficient matrix = 2. Here, the rank of matrix < no. of unknowns therefore, the system of equation has infinitely many solution in R3

Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions of the four functions xj : j = 0, 1, 2, 3 and let D be the differentiation operator. Then the matrix of D in the above ordered basis is
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  yesterday
Let V be the vector space of polynomial functions of degree three or less. Let the ordered basis for V consisting of the functions
xj : j = 0 ,1 ,2 , 3
and let D be the differentiation operator. We need to find the matrix of differentiation operator D in the basis xj, j = 0,1, 2, 3, that is {1, x, x2, x3}.
Therefore, D(1) = d/dx(1) = 0 
= 0.1 + 0.x + 0.x2 + 0.x3
D(x) = d/dx (x) = 1 = 1.1 + 0 . x + 0 .x2 + 0.x3 ax
D(x2) =d/dx(x2) = 2x = 0.1 + 2.x + 0.x2 + 0.x3 ax
D(x3) = d/dx (x3) = 3x2 = 0.1 + 0.x + 3.x2 + 0.x4
Thus, the matrix of differentiation operator is

Let T : C—> Cn be a linear operator of rank n - 2. Then,
  • a)
    0 is not an eigen value of T
  • b)
    0 must be an eigen value of T
  • c)
    1 can never be an eigen value of T
  • d)
    1 must be an eigen value of 
Correct answer is option 'B'. Can you explain this answer?

Veda Institute answered  •  yesterday
We are given that a linear operator T : Cn—> Cn of rank n - 2.
Since T : Cn —> Cis a linear operator of rank n - 2.
Therefore, T is singular operator, Hence 0 must be an eigen value of T.

Let 

Then,
  • a)
    f is continuous at (0, 0) and the partial derivatives fx, fy exists at every point o f R2
  • b)
    f is discontinuous at (0,0) and fx, fy does not exists at every point o f R2
  • c)
    f is discontinuous at (0, 0) a n d fx , fy exists at (0,0)
  • d)
    None of the above
Correct answer is option 'C'. Can you explain this answer?

Veda Institute answered  •  yesterday
Let us suppose (x, y) approaches (0, 0) along the line y = mx. Which is a line through the origin. Put y = mx and allows x —> 0, we get

which depends on m, therefore the limit of f(x, y) at (0, 0) does not exists. Hence, f(x, y) is discontinuous at origin.
Now,
since fy exists at origin.

The number of 6 digit numbers in which all the odd digits and only odd digits appear, is 
  • a)
  • b)
    6!
  • c)
  • d)
    5!/2
Correct answer is option 'A'. Can you explain this answer?

Veda Institute answered  •  yesterday
Clearly, one of the odd digits 1, 3, 5, 7, 9 will repeated. The number of selections of the sixth digit = 5C1 = 5.
Required, number of numbers = 5 x (6!/2!)

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