IIT JAM Mathematics Practice Test- 6 - Mathematics MCQ

from LHS and RHS , we get

Now we find the eigen values of M

Hence eigen values are 0, i , -i

Augmented matrix 

Applying elementary row transformations.

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

we have

[Applying R₂ -> R₂ - R₁, R₃ -> R₃ - R₁] ::: R_n -> R_n - R_J]

[Expanding along R₁]

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

The given system of equations has a unique solution, if

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 

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.

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.

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.

(A) Let a, b, c be scalars such that 

bers.

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

Here, m₁,m₂ .....m_n  are some non-negative integers. Let a₁,a₂ .....a_n, 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.

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

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}.

we  have

and

therefore f_x(0,0) = f_y(0,0) = 0 Hence option (a) is correct.

Hence option (b) is correct.

Hence option (b) is correct.

we have

Hence option (a) is correct

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.

and

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}.

∵ 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).

Here, It is given that

∴ Which is subspace of R⁵ and clearly

Here

then

which is a sub space of M_2x2(R)

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 C⁴ with dimension 4.

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

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,

Given

 then

and

again differentiating both side

then  except at (a,b) 

[Note → at(a,b) FΔPΔ are not defined]

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

Given

ω = x² + y²

then

At t =1, x = 0,  y = 1/2

then 

ω= x y + y z = z x

then 

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

At t = 0

x = t² => x = 0

y = te^t => y = 0

z = te^-t => z = 0

then

Take