Test: Linear Algebra - 2 | 19 Questions MCQ Test

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This mock test of Test: Linear Algebra - 2 for Mathematics helps you for every Mathematics entrance exam. This contains 20 Multiple Choice Questions for Mathematics Test: Linear Algebra - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Linear Algebra - 2 quiz give you a good mix of easy questions and tough questions. Mathematics students definitely take this Test: Linear Algebra - 2 exercise for a better result in the exam. You can find other Test: Linear Algebra - 2 extra questions, long questions & short questions for Mathematics on EduRev as well by searching above.

QUESTION: 1

If {v₁, v₂, v₃} is a linearly independent set of vectors in a vector space over which one of the following sets is also linearly independent?

A.
{v₁ + v₂ + v₃, 2v₁ + v₂ + 3v₃, 5v₁ + 4v₂}
B.
{v₁ - v₂, v₂ - v₃, v₃ - v₁}
C.
{v₁ + v₂ - v₃, v₂ + v₃ - v₁ ,v₃ + v₁ - v₂, v₁ + v₂ + v₃)
D.
{v₁ + v₂, v₂ + 2v₃, v₃ + 3v₁}

Solution:

QUESTION: 2

Consdierthe subspace W = {(x₁, x₂,.... ,x₁₀) x_n = x_n-1 + x_{n - 2} for 3 < n < 10) of the vector space the dimension of W is

A.
2
B.
9
C.
3
D.
10

Solution:

QUESTION: 3

Which one of the following is a subspace of the vector space

A.
{(x, y, z) : x + 2y = 0,2x + 3z = 0}
B.
{(x,y,z) : 2x + 3y + 4z - 3 = 0, z = 0}
C.
{(x, y, z) : x >,y > 0)
D.
{(x, y ,z) : x - 1 = 0,y = 0)

Solution:

QUESTION: 4

If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be

A.
3
B.
0
C.
2
D.
1

Solution:

Here A & B a re 3 × 2 real matrices such that rank (AB) = 1
So, |AB| = 0 ⇒ |A| |B| = 0 (∴ |AB| = |A| |B|)
⇒ either |A| or |B| should be zero
So, |BA| = |B||A| = 0
⇒ BA is singular
Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)

QUESTION: 5

Which of the following sets of functions from is a vector space over

A.
Only S₁
B.
Only S₂
C.
S₁ and S₃ but not S₂
D.
all the three are vector spaces

Solution:

QUESTION: 6

Let T : R³ → R³ be the linear transformation define by T(x, y, z) = (x + y, + z, z + x) for all (x, y, z) ∈ 3. Then

A.
rank (T) = 0, nullity (T) = 3
B.
rank (T) = 2, nullity (T) = 1
C.
rank (T) = 3, nullity (T) = 0
D.
rank (T) = 1, nullity (T) = 2

Solution:

Let T : R² → R³ be the L.T. defined as
T (x, y, z) = (x + y, y + z, z + x) ∀ (x, y, z) ∈ R³
N (T ) = {( x, y, z)|T ( x, y, z) = 0}
= {(x, y, z)|x + y = 0,y + z = 0, z + x = 0}
= {(0, 0, 0)| x, y, z ∈R}
⇒ dim N (T) = 0
Rank (N) + Nullity T = dim R³= 3
Rank (T) + 0 = 3 ⇒ Rank of T = 3

QUESTION: 7

If the matrixthe A

A.
Skew-Hermitian
B.
Hermitian
C.
Unitary
D.
None of these

Solution:

Here,

So, Answer is skew-Hermitian matrix.

QUESTION: 8

Let V be a 3 dimensional vector space over the field of 3 elements. The number of distinct 1 dimensional subspaces of V is

A.
13
B.
26
C.
9
D.
15

Solution:

QUESTION: 9

Let M = {(a₁,a₂,a₃): a_i ∈ (1,2,3,4); a₁ + a₂ + a₃ = 6} then the number of elements in M is

A.
8
B.
9
C.
10
D.
12

Solution:

QUESTION: 10

The dimension of the vector space of all symmetric matrices A = (a_ij) of order n x n (n > 2) with real entries, a_ii = 0 and trace zero is

Solution:

The dimension of the space of n×n symmetric matrices with diagonal equal to zero is . Now, in your case the diagonal is not zero but the sum of its elements is zero, that means that you have n−1 elements which can vary. SO you get as expected.

QUESTION: 11

Let n be a positive integer and let H_n be the space of all n x n matrices A = (a_ij) with entries in satisfying a_ij = a_rs whenever i + j = r + s (i, j, r, s = 1, 2,..., n) then the dimention of H_n, as a vector space over

A.
n²
B.
n² - n + 1
C.
2n + 1
D.
2n - 1

Solution:

QUESTION: 12

The dimension of the vector space of all symmetric matrices of order n x n (n > 2) with real entries and trace equal to zero is

Solution:

QUESTION: 13

Let {X,Y,Z) be a basis of Consider the following statements P and Q
P : {X + Y,Y + Z, X - Z) is a basis of
Q : {X + Y + Z,X + 2Y - Z, X - 3Z} is a basis of

Which of the above statements hold true?

A.
only P
B.
only Q
C.
both P and Q
D.
neither P nor Q

Solution:

QUESTION: 14

Let V denote the vector space C⁵[a, b] over R and

A.
dim V = ∞ and dim W = ∞
B.
dim V = ∞ and dim W = 4
C.
dim V = 6 and dim W = 5
D.
dim V = 5 and dim W = 4

Solution:

QUESTION: 15

Let M be the space of all 4 x 3 matrices with entries in the finite field of three elements. Then the number of matrices of Rank three In M is

A.
(3⁴ - 3) (3⁴ - 3²)(3⁴ - 3³)
B.
(3⁴ - 1)(3⁴ - 2)(3⁴ - 3)
C.
(3⁴ - 1)(3⁴ - 3) (3⁴ - 3²)
D.
3⁴(3⁴ - 1)(3⁴ - 2)

Solution:

QUESTION: 16

Consider the subspace W = {[a_ij] : a_ij = 0, if i is even) of all 10 x 10 matrices(real) then the dimension of W is

A.
25
B.
50
C.
75
D.
100

Solution:

QUESTION: 17

The dimension of the vector space over the field

A.
n²
B.
n² - 1
C.
n² - n
D.

Solution:

QUESTION: 18

Then the dim V is

A.
0
B.
1
C.
2
D.
3

Solution:

QUESTION: 19

and let V = {(x, y, z) : |A| = 0}.Then the dimension of V equals

A.
0
B.
1
C.
2
D.
3

Solution:

QUESTION: 20

A basis of V = {(x, y, z, w) : x + y - z = 0, y + z + w = 0, 2x + y - 3z = 0)

A.
{(1,1, -1,0), (0,1,1,1), (2,1, -3,1)}
B.
{(1, -1, 0, 1)}
C.
{(1, 0,1, -1)}
D.
{(2, -1,1,0) , (1, -1, 0,1)}

Solution:

The system of 3 equations in 4 unknowns is

1x + 1y +(-1)z +0w =0……..(1)

0x + 1y + 1z + 1w = 0……….(2)

2x + 1y +(-3)z +1w = 0……..(3). The row operation (3) - 2×(1) reduces (3) to

0x +(-1)y+(-1)z+(-1)w =0…….(4), which is just (-1)× (3). Hence the given system is equivalent to the pair of equations:

x = -y+z and y= -z -w. This pair is equivalent to x=(z+w)+z = 2z+w and y = -z -w, where z and w are completely arbitrary. Thus

[x,y,z,w] =[2z+w,-z-w,z,w] =

z[2,-1,1,0] + w[1,-1,0,1].

We had already observed that the rank of the echelon matrix equivalent to the given coefficient matrix is 2, and hence its nullity = 4–2 = 2, which is the same as the dimension of the solution space V. Thus a basis of V may be taken as

B={[2,-1,1,0], [1,-1,0,1]}.