PGTRB PRACTICE QUESTIONS ON UNIT 1 ALGEBRA

Algebra Concepts MCQ Quiz

Test your knowledge of groups, rings, fields, and vector spaces

1. The set \( \mathbb{R} \) of all real numbers forms a group with respect to the binary operation \( * \) defined by \( a * b = a + b + 1 \). What is the identity element of this group?

a) 0

b) 1

c) -1

d) 2

Answer: c) -1

For \( e \) to be the identity element, we need \( a * e = a \) for all \( a \in \mathbb{R} \).

\( a * e = a + e + 1 = a \) ⇒ \( e + 1 = 0 \) ⇒ \( e = -1 \).

Verification: \( a * (-1) = a + (-1) + 1 = a \), and \( (-1) * a = (-1) + a + 1 = a \).

2. Which of the following sets forms a group under multiplication?

a) \( \{1, \omega, \omega^2\} \) where \( \omega \) is a complex cube root of unity

b) \( \{1, -1, 2, -2\} \)

c) \( \mathbb{Z} \) (the set of all integers)

d) \( \mathbb{N} \) (the set of natural numbers)

Answer: a) \( \{1, \omega, \omega^2\} \) where \( \omega \) is a complex cube root of unity

• Option a: Forms a cyclic group of order 3 under multiplication. It satisfies closure (\( \omega \cdot \omega^2 = \omega^3 = 1 \)), associativity, has identity (1), and every element has an inverse (\( \omega^{-1} = \omega^2 \)).

• Option b: Not closed under multiplication (\( 2 \cdot 2 = 4 \) which is not in the set).

• Option c: Most integers don't have multiplicative inverses in \( \mathbb{Z} \).

• Option d: Natural numbers don't have multiplicative inverses or additive identity for multiplication.

3. Let \( G = \{a \in \mathbb{R} \mid a \neq 1\} \) with binary operation \( a * b = a + b - ab \). What is the inverse of an element \( a \) in this group?

a) \( \frac{1}{a} \)

b) \( \frac{a}{a-1} \)

c) \( \frac{a}{1-a} \)

d) \( 1-a \)

Answer: b) \( \frac{a}{a-1} \)

Let \( a^{-1} \) be the inverse of \( a \). Then \( a * a^{-1} = e \), where \( e \) is the identity.

First, find the identity: \( a * e = a + e - ae = a \) ⇒ \( e - ae = 0 \) ⇒ \( e(1-a) = 0 \) ⇒ \( e = 0 \) (since \( a \neq 1 \)).

Now, \( a * a^{-1} = a + a^{-1} - a \cdot a^{-1} = 0 \)

⇒ \( a + a^{-1}(1 - a) = 0 \)

⇒ \( a^{-1}(1 - a) = -a \)

⇒ \( a^{-1} = \frac{-a}{1-a} = \frac{a}{a-1} \)

4. Which of the following is a normal subgroup of \( S_3 \) (the symmetric group of degree 3)?

a) \( \{e, (1 2)\} \)

b) \( \{e, (1 2 3), (1 3 2)\} \)

c) \( \{e, (1 3)\} \)

d) \( \{e, (2 3)\} \)

Answer: b) \( \{e, (1 2 3), (1 3 2)\} \)

\( S_3 = \{e, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)\} \) has order 6.

• Option b is \( A_3 \), the alternating group of degree 3, which is the unique subgroup of order 3 in \( S_3 \).

• Subgroups of order 2 in \( S_3 \) (options a, c, d) are not normal. For example, for \( H = \{e, (1 2)\} \):

Take \( g = (1 3) \), then \( gHg^{-1} = (1 3)\{e, (1 2)\}(1 3) = \{e, (2 3)\} \neq H \).

Thus \( H \) is not normal in \( S_3 \).

5. The set of all \( 2 \times 2 \) matrices of the form \( \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \) where \( a, b \in \mathbb{R} \) and \( a^2 + b^2 \neq 0 \), forms a group under matrix multiplication. What type of group is this?

a) Finite non-abelian group

b) Finite abelian group

c) Infinite non-abelian group

d) Infinite abelian group

Answer: d) Infinite abelian group

• The set is infinite because there are infinitely many real numbers \( a \) and \( b \).

• The group is abelian because:

Let \( A = \begin{pmatrix} a & b \\ -b & a \end{pmatrix} \), \( B = \begin{pmatrix} c & d \\ -d & c \end{pmatrix} \)

\( AB = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)

\( BA = \begin{pmatrix} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{pmatrix} \)

So \( AB = BA \), hence the group is abelian.

This group is isomorphic to the multiplicative group of nonzero complex numbers.

6. Which of the following is an integral domain but not a field?

a) \( \mathbb{Z}_7 \)

b) \( \mathbb{Q} \)

c) \( \mathbb{Z} \)

d) \( \mathbb{R} \)

Answer: c) \( \mathbb{Z} \)

An integral domain is a commutative ring with unity and no zero divisors. A field is a commutative ring with unity where every nonzero element has a multiplicative inverse.

• \( \mathbb{Z}_7 \): Is a field (since 7 is prime)

• \( \mathbb{Q} \): Is a field

• \( \mathbb{Z} \): Is an integral domain but not a field (only ±1 have multiplicative inverses in \( \mathbb{Z} \))

• \( \mathbb{R} \): Is a field

7. Let \( V \) be a vector space of all \( 2 \times 2 \) matrices over \( \mathbb{R} \). What is the dimension of \( V \)?

a) 2

b) 3

c) 4

d) 8

Answer: c) 4

A basis for the vector space of all \( 2 \times 2 \) matrices over \( \mathbb{R} \) is:

\( \left\{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right\} \)

This set has 4 elements and any \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) can be written as a linear combination:

\( a\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + b\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} + c\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + d\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \)

Thus, the dimension is 4.

8. Which of the following polynomials is irreducible over \( \mathbb{Q} \)?

a) \( x^2 - 2 \)

b) \( x^2 + x + 1 \)

c) \( x^2 - 4 \)

d) Both a and b

Answer: d) Both a and b

A polynomial is irreducible over a field if it cannot be factored into polynomials of lower degree with coefficients in that field.

• \( x^2 - 2 \): Irreducible over \( \mathbb{Q} \) (no rational roots)

• \( x^2 + x + 1 \): Irreducible over \( \mathbb{Q} \) (discriminant = 1 - 4 = -3 < 0, no rational roots)

• \( x^2 - 4 \): Reducible over \( \mathbb{Q} \) since \( x^2 - 4 = (x-2)(x+2) \)

Thus, both a and b are irreducible over \( \mathbb{Q} \).

9. What is the splitting field of the polynomial \( x^4 - 5x^2 + 6 \) over \( \mathbb{Q} \)?

a) \( \mathbb{Q}(\sqrt{2}) \)

b) \( \mathbb{Q}(\sqrt{3}) \)

c) \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \)

d) \( \mathbb{Q}(\sqrt{6}) \)

Answer: c) \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \)

Factor the polynomial: \( x^4 - 5x^2 + 6 = (x^2 - 2)(x^2 - 3) \)

The roots are \( \pm\sqrt{2}, \pm\sqrt{3} \).

The smallest field containing \( \mathbb{Q} \) and all these roots is \( \mathbb{Q}(\sqrt{2}, \sqrt{3}) \).

This is the splitting field because the polynomial factors completely into linear factors over this field:

\( x^4 - 5x^2 + 6 = (x - \sqrt{2})(x + \sqrt{2})(x - \sqrt{3})(x + \sqrt{3}) \)

10. Let \( T: \mathbb{R}^3 \to \mathbb{R}^2 \) be a linear transformation. If the dimension of the kernel of \( T \) is 1, what is the dimension of the range of \( T \)?

a) 1

b) 2

c) 3

d) Cannot be determined

Answer: b) 2

By the Rank-Nullity Theorem:

\( \text{dim}(V) = \text{rank}(T) + \text{nullity}(T) \)

Here, \( \text{dim}(V) = 3 \) (since \( V = \mathbb{R}^3 \))

\( \text{nullity}(T) = \text{dim}(\text{ker}(T)) = 1 \)

So, \( 3 = \text{rank}(T) + 1 \) ⇒ \( \text{rank}(T) = 2 \)

The rank of \( T \) is the dimension of its range, so the dimension of the range is 2.