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IIT JAM Mathematics Practice Test- 6 - Mathematics MCQ


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30 Questions MCQ Test - IIT JAM Mathematics Practice Test- 6

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IIT JAM Mathematics Practice Test- 6 - Question 1

The linear operation L(x) is defined by the cross product L(x) = bx, where b = [0, 1, 0] T and X = [x1 x2, x3]T are three dimensional vectors. The 3 x 3 matrix M o f the operation satisfies  Then the eigen values of M are

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 1

from LHS and RHS , we get

Now we find the eigen values of M

Hence eigen values are 0, i , -i

IIT JAM Mathematics Practice Test- 6 - Question 2

Find the condition when the following system of linear equations have no solution.

x + 4z = 2

x + w = 0

x +y = 0

x + 2y + 3w + tz = s

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 2

Augmented matrix 

Applying elementary row transformations.

Therefore, when t = -16 s = -8, there are no solutions.

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IIT JAM Mathematics Practice Test- 6 - Question 3

The value of the determinant of nth order, being given by  is

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 3

we have

[Applying R2 -> R2 - R1, R3 -> R3 - R1] ::: Rn -> Rn - RJ]

[Expanding along R1]

= x(x - 1 )n- 1 + (x - 1 )n- 1 [1 + 1 + ...+(n - 1) times ] = (x - 1 )n-1 (x + n - 1 )

IIT JAM Mathematics Practice Test- 6 - Question 4

The system of linear equations

x + y + z = 2, 2x + y - z = 3, 3x + 2y + kz = 4 has a unique solution, if

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 4

The given system of equations has a unique solution, if

IIT JAM Mathematics Practice Test- 6 - Question 5

For the matrix  one of the eigen value is equal to -2. Which of the following is an eigen vector?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 5

eigen values are λ= 3, -2, 1 then eigen vector corresponding to eigen value λ = -2 is _______.

A x = - 2x => (A + 2I) x = 0

let x2 = k then 

 Hence eigen vector is 

IIT JAM Mathematics Practice Test- 6 - Question 6

The linear system has

x1 + 2x2 - 3x4 + x5 = 2

x1 + 2x2 + x3 - 3x4 + 2x6 + x5 = 3

x1 + 2x2 - 3x4 + 2x5 + x6 = 4

3x1 + 6x2 + x3 - 9x4 + 4x5 + 2x6 = 9

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 6

Step 1. The augmented matrix of this linear system is

Step 2. The augmented matrix is row equivalent to the matrix (verify)

Step 3 The linear system represented by (2)

x1 + 2x2 - 3x4- x6 = 0

x3 + 2x6 = 1 x5 + x6 = 2 .

Solving each equation for the unknown that corresponds to the leading entry in each row of (2), we obtain 

Letting x6 = r, x4 = s, and x2 = t, a solution to the linear system (1) is 


Where r, s, and t are any real numbers. Thus (3) is the solution to the given linear system (1). Since r, s, and t can be assigned any real number, the given linear system (1) has infinitely many solutions.

IIT JAM Mathematics Practice Test- 6 - Question 7

The System of equations,

x + y + z = 8 

x - y + 2z = 6

3x + 5y+ 7z= 14 has,

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 7

The augmented matrix C = [A : B]

  

Rank of A = 3, rank of C = 3.
So, rank of A = rank of C = 3 = number of unknowns. Hence the equations are consistent with unique solution.

IIT JAM Mathematics Practice Test- 6 - Question 8

Suppose

Which of these subsets of the vector space R4 is/are subspace (s) ?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 8

we have

Let

 and α, β, ∈ F with b1 - c1 = 4 and b2 - c2 = 4

if 

X is not a subspace of R4 , now Y and Z are subspaces . W e can check them with same condition.

IIT JAM Mathematics Practice Test- 6 - Question 9

Which of the following sets is not linearly independent?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 9

(A) Let a, b, c be scalars such that 

bers.

(B) Let be any finite subset of S having n vectors .

Here, m1,m2 .....mn  are some non-negative integers. Let a1,a2 .....an, be scalars such that

By definition of equality of two polynomials,

∴ S is linearly independent.

(C) Let a, b, c be scalars such that a (1 , 1 ,0 , 0) + b(0, 1, - 1 , 0) + c(0, 0, 0, 3) = ( 0 , 0 , 0 , 0 )
⇒ (a, a + b, - b , 3c) = (0, 0, 0, 0)

⇒ a = 0, b = 0, c = 0

Hence, given set of vectors in V4 (R) is linearly independent.

(D) Let a, b, c be scalars such that a(1, 2 , 1 ) + b ( 3 , 1 , 5 ) + c (3, - 4 , 7) = (0, 0, 0)

⇒ (a + 3b + 3c, 2a + b - 4c, a + 5b + 7c) = (0, 0, 0)

⇒ a + 3b + 3c = 0, 2a + b - 4c = 0, a + 5b + 7c = 0

⇒ |A| = 0

∴ Rank (A) < 3 =* given set of vectors is linearly dependent.

IIT JAM Mathematics Practice Test- 6 - Question 10

if  is a basis of C3(C), then which of the following set is also a basis of C3(C) ?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 10

Any subset of C3 having three linearly independent vectors will form a basis of C3.

(Since, α, β, and γ are independent)

⇒ a = 0, b = 0, c = 0

 are linearly independent

 is the basis of C3

IIT JAM Mathematics Practice Test- 6 - Question 11

Let y(x) be the solution of differential equation,

which satisfy the condition y(1) = 0. Then which of the following’s is true ?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 11

we have

Given y(1) = 0

⇒(-1) (2 ) = c  ⇒ c = -2

So we have

(y-1) (x+1) = - 2x

 ⇒ y is decreasing on R\ {-1}.

IIT JAM Mathematics Practice Test- 6 - Question 12

if  Then,

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 12

we  have

and

therefore fx(0,0) = fy(0,0) = 0 Hence option (a) is correct.

IIT JAM Mathematics Practice Test- 6 - Question 13

If  then  is

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 13

Hence option (b) is correct.

IIT JAM Mathematics Practice Test- 6 - Question 14

For what value of k , the function  is continuous ?

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 14

Hence option (b) is correct.

IIT JAM Mathematics Practice Test- 6 - Question 15

If Then, the directional derivative at c = (0,0) along the direction u (a,b),a ≠ 0 ≠ b is

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 15

we have

Hence option (a) is correct

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 6 - Question 16

Let T be linear operator on R3- the matrix of which in the standard ordered basis is Then

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 16

Det A = 1 (4 -3 ) -2 (1 )+ 1(1)= 1 - 2 + 1 = 0

∴  A is not invertible and so T is not invettible.

Let 

be standard ordered basis of R3.

⇒ every e le m e n t in Ker T is multiple of ( - 1 , 1 , - 1 ) 

⇒ Ker T is spanned by ( - 1 , 1 , - 1 )

Since ( - 1 . 1 . - 1 ) ≠ o. { ( - 1 , 1, - 1 ) } is a basis of Ker T.

∴ dim Ket T = 1  => dim Range T = 2

Since T ∈1 = (1, 0, - 1 )

T∈2 = (2 ,1 ,3 )

belong to Range

we find

a(1, 0 , - 1 ) + b(2, 1, 3) = 0 

⇒ b = 0, a = 0

⇒    is a linearly in dependent set in Range T .As dim Range T = 2, {(1,0, -1), (2, 1, 3)} is basis of Range T.

IIT JAM Mathematics Practice Test- 6 - Question 17

If the Linear Transformation is defined as and T (1, 0) is equal to,

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 17

and

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 6 - Question 18

The "Cyclic" transformation T is defined by T(v1,v2,v3) = (v2,v3,v1), then T100 (v) is not equal to

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 18

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 6 - Question 19

Let L : P2 -> P2 be the linear transformation defined as, L (at2 + bt + c) = (a+ 2b)t+(b+ c), then,

(I) - 4t2 + 2t - 2 is in th e ker (L)

(II) Basis for ker (L) is 2t2 - 1+ 1

which of the following options is / are not true.

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 19

Since L(-4t2 + 2t - 2) = (-4 + 2.2)t + (-2 + 2) = 0, 

We conclude that - 4t2 + 2t - 2 is in ker L.

The vector at2 + bt + c is in ker L if 

L(at2 + bt + c) = 0, that is, if (a + 2b)t + (b + c) = 0.

Then a + 2b = 0 b + c = 0.

Transforming the augmented matrix of this linear system to reduced row echelon form, we find (verify) that a basis for the solution space is

So a basis for ker L is {2t2 - t + 1}.

*Multiple options can be correct
IIT JAM Mathematics Practice Test- 6 - Question 20

Let V be the vector space of all polynomial functions of degree ≤ 2 from the field of real numbers R into itself. Let {f1, f2, f3} be an ordered basis of V(R), where f1(x) = 1, f2(x) = x + 1, f3(x) = (x + 1)2, then the co-ordinates of 1 + x + x2 in this basis is

Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 20

∵ 1 + X + X2 = 1.1 + (-1 ).(x + 1) + 1. (x + 1 )2 = 1 .f1 + (-1 )f2 + 1 .f3 

∴ Required co-ordinates are(1, -1, 1).

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 21

If  and  

be subspaces of R5, then dim  is equal to _______.


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 21

Here, It is given that

∴ Which is subspace of R5 and clearly

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 22

Let  M2x2 (R) be the vector  space of all 2 * 2 matrices over R

Let

dim  is equal to ________ .


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 22

Here

then

which is a sub space of M2x2(R)

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 23

Find the dimension of the subspace 


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 23

The subspace W can be written as

Since the set of vectors   is a linearly independent set, thus it forms a basis of W. Thus, W is a subspace of Cwith dimension 4.

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 24

Find the dimension of the subspace   of M2x2  (R)


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 24

we have

a + b = c
b = c
b + c = d
c + d = a

When we solve above system then we get a = b = c = d = 0

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 25

Given the linear transformation

Find rank of T.


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 25

To find the range of T, apply T to the elements of a spanning set for c3. W e will use the standard basis vectors

Each of these vectors is a scalar multiple of  the others, so we can toss two of them in reducing the spanning set to a linearly independent set. The result is the basis of the range,

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 26

  find 


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 26

Given

 then

and

again differentiating both side

then  except at (a,b) 

[Note → at(a,b) FΔPΔ are not defined]

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 27

If   then the value of the expression  at the point ( 1 ,2 ) i s -----------


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 27

Here u is a hemogenous function with degree 1 then by Eulers the.

at (1,2) 

= √3 sin-1(1/2)

= √3 * π/6

= π/2√3

= 0.924

 

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 28

find the value of  where ω = x2 + y2


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 28

Given

ω = x2 + y2

then

At t =1, x = 0,  y = 1/2

then 

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 29

Find the value of the  where w = xy + yz + zx, x = t2,y = tet z = te-t at t = 0


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 29

ω= x y + y z = z x

then 

= (y + z ) 2t + (x + z ) ( tet + et ) + (y+x) (e-t - te-t) 

At t = 0

x = t2 => x = 0

y = tet => y = 0

z = te-t => z = 0

then

*Answer can only contain numeric values
IIT JAM Mathematics Practice Test- 6 - Question 30

 is equal to _______________


Detailed Solution for IIT JAM Mathematics Practice Test- 6 - Question 30

Take

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