UPSC Optional (Maths) Paper-02 Algebra Solution

2013

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UPSC Algebra

1(a) Show that the set of matrices $S=\left\{\left(\begin{array}{cc}a & -b \\ b & a\end{array}\right) \quad a, b \in \mathbb{R}\right\}$ is a field under the usual binary operations of matrix addition and matrix multiplication. What are the additive and multiplicative identities and what is the inverse of $\left(\begin{array}{lr}1 & -1 \\ 1 & 1\end{array}\right)$? Consider the map $f: \mathbb{C} \rightarrow S$ defined by $f(a+i b)=\left(\begin{array}{cc}a & -b \\ b & a\end{array}\right)$. Show that $f$ is an isomorphism.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To show that the set

S

is a field under the usual binary operations of matrix addition and matrix multiplication, we need to establish the following properties:

Closure under Addition and Multiplication
Associativity of Addition and Multiplication
Commutativity of Addition and Multiplication
Existence of Additive and Multiplicative Identity
Existence of Additive and Multiplicative Inverse
Distributive Law
Closure under Addition and Multiplication

Addition
Let

A = (\begin{array}{cc} a_{1} & - b_{1} \\ b_{1} & a_{1} \end{array})

and

B = (\begin{array}{cc} a_{2} & - b_{2} \\ b_{2} & a_{2} \end{array})

be two matrices in

S

Let’s substitute the values into the formula for matrix addition:

A + B = (\begin{array}{cc} a_{1} & - b_{1} \\ b_{1} & a_{1} \end{array}) + (\begin{array}{cc} a_{2} & - b_{2} \\ b_{2} & a_{2} \end{array})

After simplifying, we get:

A + B = (\begin{array}{cc} a_{1} + a_{2} & - (b_{1} + b_{2}) \\ b_{1} + b_{2} & a_{1} + a_{2} \end{array})

Since

a_{1}, a_{2}, b_{1}, b_{2}

are real numbers,

a_{1} + a_{2}

and

b_{1} + b_{2}

are also real numbers. Therefore,

A + B

is also in

S

Multiplication
Similarly, for multiplication, we have:

A \times B = (\begin{array}{cc} a_{1} & - b_{1} \\ b_{1} & a_{1} \end{array}) \times (\begin{array}{cc} a_{2} & - b_{2} \\ b_{2} & a_{2} \end{array})

After simplifying, we get:

A \times B = (\begin{array}{cc} a_{1} a_{2} - b_{1} b_{2} & - (a_{1} b_{2} + a_{2} b_{1}) \\ a_{1} b_{2} + a_{2} b_{1} & a_{1} a_{2} - b_{1} b_{2} \end{array})

Since

a_{1}, a_{2}, b_{1}, b_{2}

are real numbers,

a_{1} a_{2} - b_{1} b_{2}

and

a_{1} b_{2} + a_{2} b_{1}

are also real numbers. Therefore,

A \times B

is also in

S

Associativity of Addition and Multiplication
Associativity of addition and multiplication follows directly from the associativity of addition and multiplication of matrices.
Commutativity of Addition and Multiplication
Commutativity of addition and multiplication also follows directly from the commutativity of addition and multiplication of matrices.
Existence of Additive and Multiplicative Identity
The additive identity is $(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array})$ and the multiplicative identity is $(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array})$ .
Existence of Additive and Multiplicative Inverse

Additive Inverse
For each

A = (\begin{array}{cc} a & - b \\ b & a \end{array})

, the additive inverse is

- A = (\begin{array}{cc} - a & b \\ - b & - a \end{array})

Multiplicative Inverse
For each

A = (\begin{array}{cc} a & - b \\ b & a \end{array})

, the multiplicative inverse

A^{- 1}

is given by:

A^{- 1} = \frac{1}{a^{2} + b^{2}} (\begin{array}{cc} a & b \\ - b & a \end{array})

Let’s find the multiplicative inverse of

(\begin{array}{cc} 1 & - 1 \\ 1 & 1 \end{array})

First, we calculate

a^{2} + b^{2}

for

a = 1

and

b = 1

a^{2} + b^{2} = 1^{2} + 1^{2}

After calculating, we get

a^{2} + b^{2} = 2

Now, we can find

A^{- 1}

A^{- 1} = \frac{1}{2} (\begin{array}{cc} 1 & 1 \\ - 1 & 1 \end{array})

Distributive Law
The distributive law $A \times (B + C) = (A \times B) + (A \times C)$ also holds, as it does for matrices.

Isomorphism

f : C \to S

The map

f (a + i b) = (\begin{array}{cc} a & - b \\ b & a \end{array})

is an isomorphism if it preserves addition and multiplication, and is bijective.

Preservation of Addition

f ((a_{1} + i b_{1}) + (a_{2} + i b_{2})) = f (a_{1} + a_{2} + i (b_{1} + b_{2})) = (\begin{array}{cc} a_{1} + a_{2} & - (b_{1} + b_{2}) \\ b_{1} + b_{2} & a_{1} + a_{2} \end{array})

f (a_{1} + i b_{1}) + f (a_{2} + i b_{2}) = (\begin{array}{cc} a_{1} & - b_{1} \\ b_{1} & a_{1} \end{array}) + (\begin{array}{cc} a_{2} & - b_{2} \\ b_{2} & a_{2} \end{array}) = (\begin{array}{cc} a_{1} + a_{2} & - (b_{1} + b_{2}) \\ b_{1} + b_{2} & a_{1} + a_{2} \end{array})

Preservation of Multiplication

f ((a_{1} + i b_{1}) \times (a_{2} + i b_{2})) = f (a_{1} a_{2} - b_{1} b_{2} + i (a_{1} b_{2} + a_{2} b_{1})) = (\begin{array}{cc} a_{1} a_{2} - b_{1} b_{2} & - (a_{1} b_{2} + a_{2} b_{1}) \\ a_{1} b_{2} + a_{2} b_{1} & a_{1} a_{2} - b_{1} b_{2} \end{array})

f (a_{1} + i b_{1}) \times f (a_{2} + i b_{2}) = (\begin{array}{cc} a_{1} & - b_{1} \\ b_{1} & a_{1} \end{array}) \times (\begin{array}{cc} a_{2} & - b_{2} \\ b_{2} & a_{2} \end{array}) = (\begin{array}{cc} a_{1} a_{2} - b_{1} b_{2} & - (a_{1} b_{2} + a_{2} b_{1}) \\ a_{1} b_{2} + a_{2} b_{1} & a_{1} a_{2} - b_{1} b_{2} \end{array})

Bijectiveness
The map

f

is clearly bijective, as each complex number maps to a unique matrix in

S

, and each matrix in

S

maps to a unique complex number.

Therefore,

f

is an isomorphism.

In summary,

S

is a field under the usual binary operations of matrix addition and matrix multiplication, and the map

f : C \to S

is an isomorphism.

Verified Answer

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2(a) What are the orders of the following permutations in $\mathrm{S}_{10}$? $\left(\begin{array}{cccccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9\end{array}\right)$ and $(12345)(67)$.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

To find the order of a permutation, we need to find the smallest positive integer

n

such that applying the permutation

n

times returns us to the original arrangement of elements.

For the permutation $(\begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9 \end{array})$

First, let’s write this permutation in disjoint cycle notation. The cycles are:

$(2 8)$
$(3 7 4)$
$(5 10 9 6)$

The lengths of these cycles are

2, 3,

and

4

respectively.

The order of a permutation in disjoint cycle notation is the least common multiple (LCM) of the lengths of its disjoint cycles.

Let’s substitute the values into the formula for LCM:

LCM (2, 3, 4)

After calculating, we get

LCM (2, 3, 4) = 12

So, the order of the first permutation is

12

For the permutation $(12345) (67)$

This permutation is already in disjoint cycle notation. The lengths of these cycles are

5

and

2

Let’s substitute the values into the formula for LCM:

LCM (5, 2)

After calculating, we get

LCM (5, 2) = 10

So, the order of the second permutation is

10

In summary, the order of the first permutation is

12

and the order of the second permutation is

10

Verified Answer

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3(a) Let $J=\{a+b i \mid a, b \in \mathbb{Z}\}$ be the ring of Gaussian integers (subring of $\mathbb{C}$). Which of the following is $J$: Euclidean domain, principal ideal domain, unique factorization domain? Justify your answer.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

The ring

J = {a + b i ∣ a, b \in Z}

of Gaussian integers is a subring of

C

. We will examine whether

J

is a Euclidean Domain, a Principal Ideal Domain (PID), or a Unique Factorization Domain (UFD).

Euclidean Domain

A Euclidean Domain is an integral domain equipped with a Euclidean function

f

that satisfies the division algorithm property: for any two elements

a, b

in the domain with

b \neq 0

, there exist

q, r

in the domain such that

a = b q + r

and either

r = 0

f (r) < f (b)

For Gaussian integers, we can define the Euclidean function

f (a + b i) = a^{2} + b^{2}

. Using this function, we can show that

J

satisfies the division algorithm property. Therefore,

J

is a Euclidean Domain.

Principal Ideal Domain (PID)

A Principal Ideal Domain is an integral domain in which every ideal is generated by a single element. Since

J

is a Euclidean Domain, it is also a PID. This is because in a Euclidean Domain, any ideal

I

can be generated by its element of smallest Euclidean value, making it a principal ideal.

Unique Factorization Domain (UFD)

A Unique Factorization Domain is an integral domain in which every non-zero non-unit element can be uniquely factored into irreducible elements, up to ordering and units.

Since

J

is a PID, it is also a UFD. This is a general property: every PID is a UFD.

In summary, the ring

J

of Gaussian integers is a Euclidean Domain, a Principal Ideal Domain, and a Unique Factorization Domain.

Verified Answer

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3(b) Let $\mathrm{R}^{\mathrm{C}}=$ ring of all real-valued continuous functions on $[0,1]$, under the operations \[ \begin{aligned} &(f+g) x=f(x)+g(x) \\ &(f g) x=f(x) g(x) \end{aligned} \] Is $M=\left\{f \in R^C \mid f\left(\frac{1}{2}\right)=0\right\}$ a maximal ideal of $R$? Justify your answer.

Expert Answer

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e

An ideal

M

in a ring

R

is said to be a maximal ideal if

M \neq R

and there is no ideal

N

such that

M ⊊ N ⊊ R

Step 1: Verify

M

is an Ideal

First, let’s check if

M

is an ideal of

R^{C}

M

is the set of all functions

f

such that

f (\frac{1}{2}) = 0

Closed under addition: If $f, g \in M$ , then $f (\frac{1}{2}) = 0$ and $g (\frac{1}{2}) = 0$ . Therefore, $(f + g) (\frac{1}{2}) = f (\frac{1}{2}) + g (\frac{1}{2}) = 0 + 0 = 0$ . So, $f + g \in M$ .
Closed under multiplication by an element in $R^{C}$ : If $f \in M$ and $g \in R^{C}$ , then $f (\frac{1}{2}) = 0$ . Therefore, $(f \cdot g) (\frac{1}{2}) = f (\frac{1}{2}) \cdot g (\frac{1}{2}) = 0$ . So, $f \cdot g \in M$ .

Since

M

is closed under addition and multiplication by an element in

R^{C}

M

is an ideal of

R^{C}

Step 2: Check for Maximal Ideal

To show that

M

is a maximal ideal, we need to show that there is no ideal

N

such that

M ⊊ N ⊊ R^{C}

Suppose there exists an ideal

N

such that

M ⊊ N ⊊ R^{C}

. Then there must be a function

g \in N

but

g \notin M

. This means

g (\frac{1}{2}) \neq 0

Consider the function

h (x) = 1 - g (x)

for

x \in [0, 1]

. Since

N

is an ideal and

g \in N

h \in N

as well. Now,

h (\frac{1}{2}) = 1 - g (\frac{1}{2}) \neq 0

and

h (x) \neq 0

for all

x

[0, 1]

Now consider

f (x) = \frac{1}{h (x)}

. This function is well-defined and continuous on

[0, 1]

because

h (x) \neq 0

for all

x

. Moreover,

f (x) \cdot h (x) = 1

, a unit in

R^{C}

Since

f (x)

and

h (x)

are both in

N

, and their product is a unit in

R^{C}

, it follows that

N = R^{C}

. This contradicts the assumption that

N ⊊ R^{C}

Therefore, no such ideal

N

can exist, and

M

must be a maximal ideal of

R^{C}

Verified Answer

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2013

Expert Answer

2(a) What are the orders of the following permutations in \(\mathrm{S}_{10}\)? \(\left(\begin{array}{cccccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9\end{array}\right)\) and \((12345)(67)\).

Expert Answer

3(a) Let \(J=\{a+b i \mid a, b \in \mathbb{Z}\}\) be the ring of Gaussian integers (subring of \(\mathbb{C}\)). Which of the following is \(J\): Euclidean domain, principal ideal domain, unique factorization domain? Justify your answer.

Expert Answer

Expert Answer

Get Complete Solution Mathematics Optional Paper | UPSC

UPSC Previous Years Maths Optional Papers with Solution | Paper-02

UPSC Previous Years Maths Optional Papers with Solution | Paper-01

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