# 2013

Estimated reading: 48 minutes 86 views

UPSC Algebra

$$cos\left(\theta +\phi \right)=cos\:\theta \:cos\:\phi -sin\:\theta \:sin\:\phi$$

### 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.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To show that the set $S$$S$SS$S$ is a field under the usual binary operations of matrix addition and matrix multiplication, we need to establish the following properties:
1. Closure under Addition and Multiplication
2. Associativity of Addition and Multiplication
3. Commutativity of Addition and Multiplication
4. Existence of Additive and Multiplicative Identity
5. Existence of Additive and Multiplicative Inverse
6. Distributive Law
7. Closure under Addition and Multiplication
Let $A=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)$$A=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)$A=([a_(1),-b_(1)],[b_(1),a_(1)])A = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right)$A=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)$ and $B=\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$$B=\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$B=([a_(2),-b_(2)],[b_(2),a_(2)])B = \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)$B=\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$ be two matrices in $S$$S$SS$S$.
Let’s substitute the values into the formula for matrix addition:
$A+B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$$A+B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$A+B=([a_(1),-b_(1)],[b_(1),a_(1)])+([a_(2),-b_(2)],[b_(2),a_(2)])A + B = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) + \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)$A+B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$
After simplifying, we get:
$A+B=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$$A+B=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$A+B=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])A + B = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)$A+B=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$
Since ${a}_{1},{a}_{2},{b}_{1},{b}_{2}$${a}_{1},{a}_{2},{b}_{1},{b}_{2}$a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2${a}_{1},{a}_{2},{b}_{1},{b}_{2}$ are real numbers, ${a}_{1}+{a}_{2}$${a}_{1}+{a}_{2}$a_(1)+a_(2)a_1 + a_2${a}_{1}+{a}_{2}$ and ${b}_{1}+{b}_{2}$${b}_{1}+{b}_{2}$b_(1)+b_(2)b_1 + b_2${b}_{1}+{b}_{2}$ are also real numbers. Therefore, $A+B$$A+B$A+BA + B$A+B$ is also in $S$$S$SS$S$.
Multiplication
Similarly, for multiplication, we have:
$A×B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$$A×B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$A xx B=([a_(1),-b_(1)],[b_(1),a_(1)])xx([a_(2),-b_(2)],[b_(2),a_(2)])A \times B = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) \times \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right)$A×B=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)$
After simplifying, we get:
$A×B=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$$A×B=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$A xx B=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])A \times B = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)$A×B=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$
Since ${a}_{1},{a}_{2},{b}_{1},{b}_{2}$${a}_{1},{a}_{2},{b}_{1},{b}_{2}$a_(1),a_(2),b_(1),b_(2)a_1, a_2, b_1, b_2${a}_{1},{a}_{2},{b}_{1},{b}_{2}$ are real numbers, ${a}_{1}{a}_{2}-{b}_{1}{b}_{2}$${a}_{1}{a}_{2}-{b}_{1}{b}_{2}$a_(1)a_(2)-b_(1)b_(2)a_1a_2 – b_1b_2${a}_{1}{a}_{2}-{b}_{1}{b}_{2}$ and ${a}_{1}{b}_{2}+{a}_{2}{b}_{1}$${a}_{1}{b}_{2}+{a}_{2}{b}_{1}$a_(1)b_(2)+a_(2)b_(1)a_1b_2 + a_2b_1${a}_{1}{b}_{2}+{a}_{2}{b}_{1}$ are also real numbers. Therefore, $A×B$$A×B$A xx BA \times B$A×B$ is also in $S$$S$SS$S$.
1. Associativity of Addition and Multiplication
Associativity of addition and multiplication follows directly from the associativity of addition and multiplication of matrices.
2. Commutativity of Addition and Multiplication
Commutativity of addition and multiplication also follows directly from the commutativity of addition and multiplication of matrices.
3. Existence of Additive and Multiplicative Identity
The additive identity is $\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$([0,0],[0,0])\left(\begin{array}{cc} 0 & 0 \\ 0 & 0 \end{array}\right)$\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$ and the multiplicative identity is $\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$([1,0],[0,1])\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right)$\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$.
4. Existence of Additive and Multiplicative Inverse
For each $A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$$A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$A=([a,-b],[b,a])A = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, the additive inverse is $-A=\left(\begin{array}{cc}-a& b\\ -b& -a\end{array}\right)$$-A=\left(\begin{array}{cc}-a& b\\ -b& -a\end{array}\right)$-A=([-a,b],[-b,-a])-A = \left(\begin{array}{cc} -a & b \\ -b & -a \end{array}\right)$-A=\left(\begin{array}{cc}-a& b\\ -b& -a\end{array}\right)$.
Multiplicative Inverse
For each $A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$$A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$A=([a,-b],[b,a])A = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$A=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$, the multiplicative inverse ${A}^{-1}$${A}^{-1}$A^(-1)A^{-1}${A}^{-1}$ is given by:
${A}^{-1}=\frac{1}{{a}^{2}+{b}^{2}}\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$${A}^{-1}=\frac{1}{{a}^{2}+{b}^{2}}\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$A^(-1)=(1)/(a^(2)+b^(2))([a,b],[-b,a])A^{-1} = \frac{1}{a^2 + b^2} \left(\begin{array}{cc} a & b \\ -b & a \end{array}\right)${A}^{-1}=\frac{1}{{a}^{2}+{b}^{2}}\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$
Let’s find the multiplicative inverse of $\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$$\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$([1,-1],[1,1])\left(\begin{array}{cc} 1 & -1 \\ 1 & 1 \end{array}\right)$\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$.
First, we calculate ${a}^{2}+{b}^{2}$${a}^{2}+{b}^{2}$a^(2)+b^(2)a^2 + b^2${a}^{2}+{b}^{2}$ for $a=1$$a=1$a=1a = 1$a=1$ and $b=1$$b=1$b=1b = 1$b=1$:
${a}^{2}+{b}^{2}={1}^{2}+{1}^{2}$${a}^{2}+{b}^{2}={1}^{2}+{1}^{2}$a^(2)+b^(2)=1^(2)+1^(2)a^2 + b^2 = 1^2 + 1^2${a}^{2}+{b}^{2}={1}^{2}+{1}^{2}$
After calculating, we get ${a}^{2}+{b}^{2}=2$${a}^{2}+{b}^{2}=2$a^(2)+b^(2)=2a^2 + b^2 = 2${a}^{2}+{b}^{2}=2$.
Now, we can find ${A}^{-1}$${A}^{-1}$A^(-1)A^{-1}${A}^{-1}$:
${A}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$${A}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$A^(-1)=(1)/(2)([1,1],[-1,1])A^{-1} = \frac{1}{2} \left(\begin{array}{cc} 1 & 1 \\ -1 & 1 \end{array}\right)${A}^{-1}=\frac{1}{2}\left(\begin{array}{cc}1& 1\\ -1& 1\end{array}\right)$
1. Distributive Law
The distributive law $A×\left(B+C\right)=\left(A×B\right)+\left(A×C\right)$$A×\left(B+C\right)=\left(A×B\right)+\left(A×C\right)$A xx(B+C)=(A xx B)+(A xx C)A \times (B + C) = (A \times B) + (A \times C)$A×\left(B+C\right)=\left(A×B\right)+\left(A×C\right)$ also holds, as it does for matrices.
Isomorphism $f:\mathbb{C}\to S$$f:\mathbb{C}\to S$f:Crarr Sf: \mathbb{C} \rightarrow S$f:\mathbb{C}\to S$
The map $f\left(a+ib\right)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$$f\left(a+ib\right)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$f(a+ib)=([a,-b],[b,a])f(a + ib) = \left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$f\left(a+ib\right)=\left(\begin{array}{cc}a& -b\\ b& a\end{array}\right)$ is an isomorphism if it preserves addition and multiplication, and is bijective.
$f\left(\left({a}_{1}+i{b}_{1}\right)+\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}+{a}_{2}+i\left({b}_{1}+{b}_{2}\right)\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$$f\left(\left({a}_{1}+i{b}_{1}\right)+\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}+{a}_{2}+i\left({b}_{1}+{b}_{2}\right)\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$f((a_(1)+ib_(1))+(a_(2)+ib_(2)))=f(a_(1)+a_(2)+i(b_(1)+b_(2)))=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])f((a_1 + ib_1) + (a_2 + ib_2)) = f(a_1 + a_2 + i(b_1 + b_2)) = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)$f\left(\left({a}_{1}+i{b}_{1}\right)+\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}+{a}_{2}+i\left({b}_{1}+{b}_{2}\right)\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$
$f\left({a}_{1}+i{b}_{1}\right)+f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$$f\left({a}_{1}+i{b}_{1}\right)+f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$f(a_(1)+ib_(1))+f(a_(2)+ib_(2))=([a_(1),-b_(1)],[b_(1),a_(1)])+([a_(2),-b_(2)],[b_(2),a_(2)])=([a_(1)+a_(2),-(b_(1)+b_(2))],[b_(1)+b_(2),a_(1)+a_(2)])f(a_1 + ib_1) + f(a_2 + ib_2) = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) + \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right) = \left(\begin{array}{cc} a_1 + a_2 & -(b_1 + b_2) \\ b_1 + b_2 & a_1 + a_2 \end{array}\right)$f\left({a}_{1}+i{b}_{1}\right)+f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)+\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}+{a}_{2}& -\left({b}_{1}+{b}_{2}\right)\\ {b}_{1}+{b}_{2}& {a}_{1}+{a}_{2}\end{array}\right)$
Preservation of Multiplication
$f\left(\left({a}_{1}+i{b}_{1}\right)×\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}{a}_{2}-{b}_{1}{b}_{2}+i\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$$f\left(\left({a}_{1}+i{b}_{1}\right)×\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}{a}_{2}-{b}_{1}{b}_{2}+i\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$f((a_(1)+ib_(1))xx(a_(2)+ib_(2)))=f(a_(1)a_(2)-b_(1)b_(2)+i(a_(1)b_(2)+a_(2)b_(1)))=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])f((a_1 + ib_1) \times (a_2 + ib_2)) = f(a_1a_2 – b_1b_2 + i(a_1b_2 + a_2b_1)) = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)$f\left(\left({a}_{1}+i{b}_{1}\right)×\left({a}_{2}+i{b}_{2}\right)\right)=f\left({a}_{1}{a}_{2}-{b}_{1}{b}_{2}+i\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$
$f\left({a}_{1}+i{b}_{1}\right)×f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$$f\left({a}_{1}+i{b}_{1}\right)×f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$f(a_(1)+ib_(1))xx f(a_(2)+ib_(2))=([a_(1),-b_(1)],[b_(1),a_(1)])xx([a_(2),-b_(2)],[b_(2),a_(2)])=([a_(1)a_(2)-b_(1)b_(2),-(a_(1)b_(2)+a_(2)b_(1))],[a_(1)b_(2)+a_(2)b_(1),a_(1)a_(2)-b_(1)b_(2)])f(a_1 + ib_1) \times f(a_2 + ib_2) = \left(\begin{array}{cc} a_1 & -b_1 \\ b_1 & a_1 \end{array}\right) \times \left(\begin{array}{cc} a_2 & -b_2 \\ b_2 & a_2 \end{array}\right) = \left(\begin{array}{cc} a_1a_2 – b_1b_2 & -(a_1b_2 + a_2b_1) \\ a_1b_2 + a_2b_1 & a_1a_2 – b_1b_2 \end{array}\right)$f\left({a}_{1}+i{b}_{1}\right)×f\left({a}_{2}+i{b}_{2}\right)=\left(\begin{array}{cc}{a}_{1}& -{b}_{1}\\ {b}_{1}& {a}_{1}\end{array}\right)×\left(\begin{array}{cc}{a}_{2}& -{b}_{2}\\ {b}_{2}& {a}_{2}\end{array}\right)=\left(\begin{array}{cc}{a}_{1}{a}_{2}-{b}_{1}{b}_{2}& -\left({a}_{1}{b}_{2}+{a}_{2}{b}_{1}\right)\\ {a}_{1}{b}_{2}+{a}_{2}{b}_{1}& {a}_{1}{a}_{2}-{b}_{1}{b}_{2}\end{array}\right)$
Bijectiveness
The map $f$$f$ff$f$ is clearly bijective, as each complex number maps to a unique matrix in $S$$S$SS$S$, and each matrix in $S$$S$SS$S$ maps to a unique complex number.
Therefore, $f$$f$ff$f$ is an isomorphism.
In summary, $S$$S$SS$S$ is a field under the usual binary operations of matrix addition and matrix multiplication, and the map $f:\mathbb{C}\to S$$f:\mathbb{C}\to S$f:Crarr Sf: \mathbb{C} \rightarrow S$f:\mathbb{C}\to S$ is an isomorphism.
5/5
$$cos^2\left(\frac{\theta }{2}\right)=\frac{1+cos\:\theta }{2}$$
$$2\:sin\:\theta \:sin\:\phi =-cos\:\left(\theta +\phi \right)+cos\:\left(\theta -\phi \right)$$

### 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 $$(12345)(67)$$.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
To find the order of a permutation, we need to find the smallest positive integer $n$$n$nn$n$ such that applying the permutation $n$$n$nn$n$ times returns us to the original arrangement of elements.
1. For the permutation $\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)$$\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 8& 7& 3& 10& 5& 4& 2& 6& 9\end{array}\right)$([1,2,3,4,5,6,7,8,9,10],[1,8,7,3,10,5,4,2,6,9])\left(\begin{array}{cccccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 1 & 8 & 7 & 3 & 10 & 5 & 4 & 2 & 6 & 9\end{array}\right)$\left(\begin{array}{cccccccccc}1& 2& 3& 4& 5& 6& 7& 8& 9& 10\\ 1& 8& 7& 3& 10& 5& 4& 2& 6& 9\end{array}\right)$
First, let’s write this permutation in disjoint cycle notation. The cycles are:
• $\left(2\phantom{\rule{thinmathspace}{0ex}}8\right)$$\left(2\phantom{\rule{thinmathspace}{0ex}}8\right)$(28)(2 \, 8)$\left(2\phantom{\rule{thinmathspace}{0ex}}8\right)$
• $\left(3\phantom{\rule{thinmathspace}{0ex}}7\phantom{\rule{thinmathspace}{0ex}}4\right)$$\left(3\phantom{\rule{thinmathspace}{0ex}}7\phantom{\rule{thinmathspace}{0ex}}4\right)$(374)(3 \, 7 \, 4)$\left(3\phantom{\rule{thinmathspace}{0ex}}7\phantom{\rule{thinmathspace}{0ex}}4\right)$
• $\left(5\phantom{\rule{thinmathspace}{0ex}}10\phantom{\rule{thinmathspace}{0ex}}9\phantom{\rule{thinmathspace}{0ex}}6\right)$$\left(5\phantom{\rule{thinmathspace}{0ex}}10\phantom{\rule{thinmathspace}{0ex}}9\phantom{\rule{thinmathspace}{0ex}}6\right)$(51096)(5 \, 10 \, 9 \, 6)$\left(5\phantom{\rule{thinmathspace}{0ex}}10\phantom{\rule{thinmathspace}{0ex}}9\phantom{\rule{thinmathspace}{0ex}}6\right)$
The lengths of these cycles are $2,3,$$2,3,$2,3,2, 3,$2,3,$ and $4$$4$44$4$ 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:
$\text{LCM}\left(2,3,4\right)$$\text{LCM}\left(2,3,4\right)$“LCM”(2,3,4)\text{LCM}(2, 3, 4)$\text{LCM}\left(2,3,4\right)$
After calculating, we get $\text{LCM}\left(2,3,4\right)=12$$\text{LCM}\left(2,3,4\right)=12$“LCM”(2,3,4)=12\text{LCM}(2, 3, 4) = 12$\text{LCM}\left(2,3,4\right)=12$.
So, the order of the first permutation is $12$$12$1212$12$.
1. For the permutation $\left(12345\right)\left(67\right)$$\left(12345\right)\left(67\right)$(12345)(67)(12345)(67)$\left(12345\right)\left(67\right)$
This permutation is already in disjoint cycle notation. The lengths of these cycles are $5$$5$55$5$ and $2$$2$22$2$.
Let’s substitute the values into the formula for LCM:
$\text{LCM}\left(5,2\right)$$\text{LCM}\left(5,2\right)$“LCM”(5,2)\text{LCM}(5, 2)$\text{LCM}\left(5,2\right)$
After calculating, we get $\text{LCM}\left(5,2\right)=10$$\text{LCM}\left(5,2\right)=10$“LCM”(5,2)=10\text{LCM}(5, 2) = 10$\text{LCM}\left(5,2\right)=10$.
So, the order of the second permutation is $10$$10$1010$10$.
In summary, the order of the first permutation is $12$$12$1212$12$ and the order of the second permutation is $10$$10$1010$10$.
5/5
$$a=b\:cos\:C+c\:cos\:B$$
$$b=c\:cos\:A+a\:cos\:C$$

### 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.

upsc-algebra-2bd23ae3-3e62-420b-a4ae-2e7475b0ac9e
The ring $J=\left\{a+bi\mid a,b\in \mathbb{Z}\right\}$$J=\left\{a+bi\mid a,b\in \mathbb{Z}\right\}$J={a+bi∣a,b inZ}J = \{ a + bi \mid a, b \in \mathbb{Z} \}$J=\left\{a+bi\mid a,b\in \mathbb{Z}\right\}$ of Gaussian integers is a subring of $\mathbb{C}$$\mathbb{C}$C\mathbb{C}$\mathbb{C}$. We will examine whether $J$$J$JJ$J$ is a Euclidean Domain, a Principal Ideal Domain (PID), or a Unique Factorization Domain (UFD).
1. Euclidean Domain
A Euclidean Domain is an integral domain equipped with a Euclidean function $f$$f$ff$f$ that satisfies the division algorithm property: for any two elements $a,b$$a,b$a,ba, b$a,b$ in the domain with $b\ne 0$$b\ne 0$b!=0b \neq 0$b\ne 0$, there exist $q,r$$q,r$q,rq, r$q,r$ in the domain such that $a=bq+r$$a=bq+r$a=bq+ra = bq + r$a=bq+r$ and either $r=0$$r=0$r=0r = 0$r=0$ or $f\left(r\right)$f\left(r\right)f(r) < f(b)f(r) < f(b)$f\left(r\right).
For Gaussian integers, we can define the Euclidean function $f\left(a+bi\right)={a}^{2}+{b}^{2}$$f\left(a+bi\right)={a}^{2}+{b}^{2}$f(a+bi)=a^(2)+b^(2)f(a + bi) = a^2 + b^2$f\left(a+bi\right)={a}^{2}+{b}^{2}$. Using this function, we can show that $J$$J$JJ$J$ satisfies the division algorithm property. Therefore, $J$$J$JJ$J$ is a Euclidean Domain.
1. Principal Ideal Domain (PID)
A Principal Ideal Domain is an integral domain in which every ideal is generated by a single element. Since $J$$J$JJ$J$ is a Euclidean Domain, it is also a PID. This is because in a Euclidean Domain, any ideal $I$$I$II$I$ can be generated by its element of smallest Euclidean value, making it a principal ideal.
1. 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$$J$JJ$J$ is a PID, it is also a UFD. This is a general property: every PID is a UFD.
In summary, the ring $J$$J$JJ$J$ of Gaussian integers is a Euclidean Domain, a Principal Ideal Domain, and a Unique Factorization Domain.
5/5
$$2\:sin\:\theta \:cos\:\phi =sin\:\left(\theta +\phi \right)+sin\:\left(\theta -\phi \right)$$
$$2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)$$

### 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.

An ideal $M$$M$MM$M$ in a ring $R$$R$RR$R$ is said to be a maximal ideal if $M\ne R$$M\ne R$M!=RM \neq R$M\ne R$ and there is no ideal $N$$N$NN$N$ such that $M⊊N⊊R$$M⊊N⊊R$M⊊N⊊RM \subsetneq N \subsetneq R$M⊊N⊊R$.
Step 1: Verify $M$$M$MM$M$ is an Ideal
First, let’s check if $M$$M$MM$M$ is an ideal of ${R}^{C}$${R}^{C}$R^(C)R^C${R}^{C}$. $M$$M$MM$M$ is the set of all functions $f$$f$ff$f$ such that $f\left(\frac{1}{2}\right)=0$$f\left(\frac{1}{2}\right)=0$f((1)/(2))=0f\left(\frac{1}{2}\right) = 0$f\left(\frac{1}{2}\right)=0$.
1. Closed under addition: If $f,g\in M$$f,g\in M$f,g in Mf, g \in M$f,g\in M$, then $f\left(\frac{1}{2}\right)=0$$f\left(\frac{1}{2}\right)=0$f((1)/(2))=0f\left(\frac{1}{2}\right) = 0$f\left(\frac{1}{2}\right)=0$ and $g\left(\frac{1}{2}\right)=0$$g\left(\frac{1}{2}\right)=0$g((1)/(2))=0g\left(\frac{1}{2}\right) = 0$g\left(\frac{1}{2}\right)=0$. Therefore, $\left(f+g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)+g\left(\frac{1}{2}\right)=0+0=0$$\left(f+g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)+g\left(\frac{1}{2}\right)=0+0=0$(f+g)((1)/(2))=f((1)/(2))+g((1)/(2))=0+0=0(f + g)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) + g\left(\frac{1}{2}\right) = 0 + 0 = 0$\left(f+g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)+g\left(\frac{1}{2}\right)=0+0=0$. So, $f+g\in M$$f+g\in M$f+g in Mf + g \in M$f+g\in M$.
2. Closed under multiplication by an element in ${R}^{C}$${R}^{C}$R^(C)R^C${R}^{C}$: If $f\in M$$f\in M$f in Mf \in M$f\in M$ and $g\in {R}^{C}$$g\in {R}^{C}$g inR^(C)g \in R^C$g\in {R}^{C}$, then $f\left(\frac{1}{2}\right)=0$$f\left(\frac{1}{2}\right)=0$f((1)/(2))=0f\left(\frac{1}{2}\right) = 0$f\left(\frac{1}{2}\right)=0$. Therefore, $\left(f\cdot g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)\cdot g\left(\frac{1}{2}\right)=0$$\left(f\cdot g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)\cdot g\left(\frac{1}{2}\right)=0$(f*g)((1)/(2))=f((1)/(2))*g((1)/(2))=0(f \cdot g)\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) \cdot g\left(\frac{1}{2}\right) = 0$\left(f\cdot g\right)\left(\frac{1}{2}\right)=f\left(\frac{1}{2}\right)\cdot g\left(\frac{1}{2}\right)=0$. So, $f\cdot g\in M$$f\cdot g\in M$f*g in Mf \cdot g \in M$f\cdot g\in M$.
Since $M$$M$MM$M$ is closed under addition and multiplication by an element in ${R}^{C}$${R}^{C}$R^(C)R^C${R}^{C}$, $M$$M$MM$M$ is an ideal of ${R}^{C}$${R}^{C}$R^(C)R^C${R}^{C}$.
To show that $M$$M$MM$M$ is a maximal ideal, we need to show that there is no ideal $N$$N$NN$N$ such that $M⊊N⊊{R}^{C}$$M⊊N⊊{R}^{C}$M⊊N⊊R^(C)M \subsetneq N \subsetneq R^C$M⊊N⊊{R}^{C}$.
Suppose there exists an ideal $N$$N$NN$N$ such that $M⊊N⊊{R}^{C}$$M⊊N⊊{R}^{C}$M⊊N⊊R^(C)M \subsetneq N \subsetneq R^C$M⊊N⊊{R}^{C}$. Then there must be a function $g\in N$$g\in N$g in Ng \in N$g\in N$ but $g\notin M$$g\notin M$g!in Mg \notin M$g\notin M$. This means $g\left(\frac{1}{2}\right)\ne 0$$g\left(\frac{1}{2}\right)\ne 0$g((1)/(2))!=0g\left(\frac{1}{2}\right) \neq 0$g\left(\frac{1}{2}\right)\ne 0$.
Consider the function $h\left(x\right)=1-g\left(x\right)$$h\left(x\right)=1-g\left(x\right)$h(x)=1-g(x)h(x) = 1 – g(x)$h\left(x\right)=1-g\left(x\right)$ for $x\in \left[0,1\right]$$x\in \left[0,1\right]$x in[0,1]x \in [0, 1]$x\in \left[0,1\right]$. Since $N$$N$NN$N$ is an ideal and $g\in N$$g\in N$g in Ng \in N$g\in N$, $h\in N$$h\in N$h in Nh \in N$h\in N$ as well. Now, $h\left(\frac{1}{2}\right)=1-g\left(\frac{1}{2}\right)\ne 0$$h\left(\frac{1}{2}\right)=1-g\left(\frac{1}{2}\right)\ne 0$h((1)/(2))=1-g((1)/(2))!=0h\left(\frac{1}{2}\right) = 1 – g\left(\frac{1}{2}\right) \neq 0$h\left(\frac{1}{2}\right)=1-g\left(\frac{1}{2}\right)\ne 0$ and $h\left(x\right)\ne 0$$h\left(x\right)\ne 0$h(x)!=0h(x) \neq 0