\(\Rightarrow 2 *(x * 5)=10\)
\( \Rightarrow 2 *\left(\frac{5 x}{5}\right)=10 \)
\( \Rightarrow 2 * x=10 \)
\( \Rightarrow \frac{2 x}{5}=10 \)
\( \Rightarrow x=\frac{50}{2} \)
\(\mathrm{x}=25 \)

Let the relation \(\bot \) on the set \(\angle \) be denoted by R
For symmetry : \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\bot l_{ 1 }\Rightarrow l_{ 2 }\bot l_{ 1 }Rl_{ 1 }\)
The relation R is symmetric
For reflective \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\bot l\)
No line can be perpendicular to it self
For transitive : \(l_{ 1 }Rl_{ 2 }\Rightarrow l_{ 1 }\) for \(I\ \in \ \angle \)
It is not true
Relation is not transitive

Given A = Set of all human beings in a town at a particular time and R = {(x, y); x is a wife of y; x, y \(\in \) A} 
(i) Since x is a wife of x, is not true (x, x) \(\notin \)  R
So, R is not reflexive. 
(ii) x is a wife of y, but y not wife of (y, x) \(\notin \) : R. So, R is not symmetric. 
(iii) x is a wife of y, and y is wife of z But this situation does not exist. So, R is not transitive.

(i) Let
A = (1,2,3)
A x A = (1, 1) (2, 2) (3, 3) (1, 2) (2, 1) (3, 2) (2, 3) (1, 3), (3, 1)
\(\therefore\) R = {(1, 1) (1, 2) (2, 1) (2, 2) (2, 3) (1,3) (3, 3) (3,2) (3, 1)}.
(ii) R = {(1, 1) (1,2) (2,1) (2,2) (2,3), (3, 3) (3,2)}

\(\cos \frac{\pi}{3}=\frac{1}{2} \text {, also } \cos \frac{5 \pi}{3}=\frac{1}{2} \text {, not one-one }\)

Let * be a binary on NB defined as a * b = a/b, for a, b ∈ N
We notice, if b is not  a factor a then a * b ∉ N.
Hence, division is not a binary operation on N

XRY ⇒ X∩Y = X ⇒ Y∩X = X ⇒ YRX
Hence symmetric

Given (R= (a, b)in N x N : a < b
Not reflexive as for( a, a) in R, a < a, not true

f is not onto, as for some y ∈ R from co-domain, there is no x ∈ R from domain such that y = f{x), e.g. for - 2 ∈ R (co-domain) there is no x ∈ R (domain) such that f{x) = -2, i.e. |x|= -2. Hence, not onto.

\( a * b=2 a+b-a b \\ 3 * 4=2 \times 3+4-3 \times 4 \\ 3 * 4=-2 \)

22 * 4 = HCF (22,4) = 2

\( a * b=a+3 b^2 \\ 2 * 4=2+3(4)^2 \\ =2+3 \times 16=50 \)

Given * on Q, defined by a * b = ab + 1
Let, a \(\in Q\), b E Q then
ab \(\in Q\),
and (ab + 1)  \(\in Q\),
\(\Rightarrow \) a * b = ab + 1is defined on Q
" is a binary operation on Q.
Commutative:
a*b = ab + 1
= ab + 1 
= a*b = b*a
So * is commutative on Q.
Associative:
(a "b) * c = (ab + 1) * c = (ab + l)c + 1
= abc + c + 1
a " (b " c) = a " (bc + 1)
= a(bc + 1) + 1
= abc + a + 1
So * is associative on Q. 1
Identity Element: Let e \(\in Q\) be the identity element,
then for every a 
a * e = a and e * a = a
ac + 1 = a and ca + 1 = a
\(\Rightarrow e=\frac { a-1 }{ a } ,\frac { a-1 }{ a } \)
e is not unique as it depend on 'a', hence identity element does not exist for *.
Inverse: since there is not identity element, hence there is no inverse.

Let a, b \(\in A\), a " b = a + b + ab
Commutatively : for all a, b \(\in A\)
.a * b = a + b + ab
= b + a + ba
= b * a
Associatively: Let a, b, c \(\in A\)
(a * b) " c = (a + b + ab) " c
= (a + b + ab) + c + (a + b + ab)c
(a * b) " c = a + b + c + ab + bc + ac + abc
a * (b * c) = a * (b + c + bc)
= a + b + c + ab + bc + ac + abc
Clearly
(a * b) * c = a * (b "c)  \(\forall \) a, b, c  \(\in A\)}
 * is associative.
Identity: Let e \(\in A\)}such that
a * c = a
c + a + ea = a
c = 0
Identity element of A is e = O.
Inverse: Let b \(\in A\)such that
a*b = b*a = e
\(\Rightarrow \) a + b + ab = 0 and b + a + ba = 0
\(\Rightarrow \)  a = -b-ab
\(\Rightarrow \)  a = -b(l + a)
\(\Rightarrow \) b = \(\frac { -a }{ 1+a } \) \([\because \quad a\in A\therefore a\neq -1]\)
Invertible element of A is \(\frac { -a }{ 1+a } \) for all \(a\in A\)

Let \(A=\left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] \)
A = AI
\(\therefore \left[ \begin{matrix} -1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 1 & 3 \\ 1 & 3 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\1 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 1 }\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ 2C }_{ 1 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 2 & 3 & -1 \\ 1 & 4 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ 1 & 1 & -2 \\ 0 & 0 & 1 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ 2C }_{ 3 }\)
\({ C }_{ 2}\rightarrow { C }_{ 2 }+{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ -1 & 2 & -1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 0 \\ -3 & -3 & -2 \\ 2 & 2 & 1 \end{matrix} \right] \)
\({ C }_{ 3}\rightarrow { C }_{ 3 }+{ C }_{ 2 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -1 & 2 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 0 & 1 & 1 \\ -3 & -3 & -5 \\ 2 & 2 & 3 \end{matrix} \right] \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 }\)
\({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ 2C }_{ 3 }\)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =A\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)
\(\Rightarrow { A }^{ -1 }=\left[ \begin{matrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ 5 & -4 & 3 \end{matrix} \right] \)

We know that A = I₃A
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow 2{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 2 & 4 & 4 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow 4{ R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 8 & 4 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 0 & 1 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 4 & 0 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 3 & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 0 & -1 & 2 \end{matrix} \right] A\)
Applying \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2}\)
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }+{ 3R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 4 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ -1 & 3 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ 4R }_{ 3 }\)
\(\left[ \begin{matrix} 8 & 4 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 4 & -12 & 0 \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] A\)
Applying \({ R }_{ 1 }\rightarrow { R }_{ 1 }-\frac { 4 }{ 3 } { R }_{ 2 }\)
\(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 16 }{ 3 } & -\frac { 8 }{ 3 } \\ 3 & -13 & 2 \\ 1 & -4 & 0 \end{matrix} \right] \)
Applying \({ R }_{ 1 }\rightarrow \frac { 1 }{ 8 } { R }_{ 1 },{ R }_{ 2 }\rightarrow \frac { 1 }{ 3 } { R }_{ 2 }\) and R₃ = - R₃
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] A\)
\({ A }^{ -1 }=\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 3 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \)
Matrix representation of given linear equation is
\(\left[ \begin{matrix} 8 & 4 & 3 \\ 2 & 1 & 1 \\ 1 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \)  AX = B
\(\Rightarrow \)  X = A^-1B
\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 & \frac { 2 }{ 3 } & -\frac { 1 }{ 3 } \\ 1 & -\frac { 13 }{ 3 } & \frac { 2 }{ 3 } \\ -1 & 4 & 0 \end{matrix} \right] \left[ \begin{matrix} 19 \\ 5 \\ 7 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0+\frac { 10 }{ 3 } -\frac { 7 }{ 3 } \\ 19-\frac { 65 }{ 3 } +\frac { 14 }{ 3 } \\ -19+20+0 \end{matrix} \right] \)
\(\therefore \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \)
\(\therefore \)  x = 1, y = 2, z = 1

For getting A² = \(\left( \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right) \)
For getting A³ = \(\left( \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right) \)
Simplifying A³ - 6A² + 7A + kI³ as
\(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
Equating  \(\left( \begin{matrix} k-2 & 0 & 0 \\ 0 & k-2 & 0 \\ 0 & 0 & k-2 \end{matrix} \right) \)
\(=\left( \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)
\(\Rightarrow\)  k - 2 = 0
\(\Rightarrow\)  k = 2
Alternative Method :
\(A=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\({ A }^{ 2 }=\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \)
A³ = A² . A = \(\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \)
\(=\left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] \)
Now A³ - 6A² + 7A + kP₃ = 0
\(\Rightarrow \left[ \begin{matrix} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{matrix} \right] -\left[ \begin{matrix} 30 & 0 & 48 \\ 12 & 24 & 30 \\ 48 & 0 & 78 \end{matrix} \right] +\left[ \begin{matrix} 7 & 0 & 14 \\ 0 & 14 & 7 \\ 14 & 0 & 21 \end{matrix} \right] +\left[ \begin{matrix} k & 0 & 0 \\ 0 & k & 0 \\ 0 & 0 & k \end{matrix} \right] \)
\(=\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right] \)
\(\Rightarrow\)   21 - 30 + 7 + k = 0
\(\Rightarrow\)  k = 2

Given a * b = min {a, b]
Operation table for * is 

\(f(x_{ 1 })=f(x_{ 2 })\)
\(\Rightarrow \) \(9x_{ 1 }^{ 2 }+6x_{ 1 }-5=9x_{ 2 }^{ 2 }+6x_{ 2 }-5\)
\(\Rightarrow \) \(9(x_{ 1 }^{ 2 }-x_{ 2 }^{ 2 })+6(x_{ 1 }-x_{ 2 })=0\)
\(\Rightarrow \) \((x_{ 1 }-x_{ 2 })\quad (9(x_{ 1 }+x_{ 2 })+6=0\)
\(\Rightarrow \) \(x_{ 1 }-x_{ 2 }=0\Rightarrow x_{ 1 }=x_{ 2 }\)
\(\Rightarrow \) \(f\) is one-one \(\Rightarrow \) \(f\) is invertible
Let \(y=f(x)=9x^{ 2 }+6x-5\)
\(\Rightarrow \) \(​​​​9x^{ 2 }+6x-(5+y)=0\)
\(\Rightarrow \) \(x=\frac { -6\pm \sqrt { 36+36(5+y) } }{ 18 } \)
\(=\frac { -6+6\sqrt { 6+y } }{ 18 } =\frac { -1\pm \sqrt { y+6 } }{ 3 } \)
Since \(x\ge 0,\)
\(\therefore \) \(x=\frac { -1-\sqrt { y+6 } }{ 3 } \) is not possible.
\(\therefore \) \(x=\frac { -1+\sqrt { y+6 } }{ 3 } \).
Hence \(f^{ -1 }(y)=\frac { \sqrt { y+6 } -1 }{ 3 } \)

(a) It is easy to see that f is one-one and onto, so that f is invertible with the inverse f ^–1 of f given by f ^–1 = {(1, 1), (2, 2), (3, 3)} = f.
(b) Since f (2) = f (3) = 1, f is not one-one, so that f is not invertible.
(c) It is easy to see that f is one-one and onto, so that f is invertible with f ^–1 = {(3, 1), (2, 3), (1, 2)}.

Let \(z\in C\) be the given arbitrary element.
Then there exists pre-image y of z  under g such that
\(g(y)=z.\)  \(\left[ \because \quad g\quad is\quad onto \right] \)
Again for \(y\in B\) , there exists pre-image x of y under  f such that \(f(x)=y\).
\(\left[ \because \quad f\quad is\quad onto \right] \)
\(\because \)   \(gof(x)=g(f(x)=g(y)=z\)
Hence, gof is onto.  

suppose  \((gof)(x_{ 1 })=(gof)(x_{ 2 })\)
\(\Rightarrow \)  \(g(f(x_{ 1 }))=g(f(x_{ 2 }))\)
\(\Rightarrow \)  \(f(x_{ 1 })=f(x_{ 2 })\)   [\(\because \) g is one-one]
​​​\(\Rightarrow \)  \(x_{ 1 }=x_{ 2 }\)   [​​​​​​\(\because \)f is one-one]
Hence, gof is one-one.