Let z = x+iy be the given complex number
∴ \(\bar { z } \) = x-iy
z\(\bar { z } \) = (x+iy) (x-iy) = x²+y²
∴ z\(\bar { z } \) -(2+3i)z -(2-3i)\(\bar { z } \)+9
⇒ x²+y²-(2+3i)(x+iy)-(2-3i)(x-iy)+9 = 0
⇒ x²+y²-[2x+2iy+3ix+i²y] - [2x-2iy-3ix+3i²y]+9 = 0
\(\Rightarrow x^{2}+y^{2}-2 x -\not 2 i y-\not 3i x +3 y-2 x+\not 2 i y+\not 3 i x+3 y+9=0 \)
⇒ x²+y²-4x+6y+9 = 0
Here 2u = -4 ⇒ u = -2
2v = 6 ⇒ v = 3 and d = 9
∴ Centre of the circle is (-u, -v) = (2, -3)
Radius =\(\sqrt { { u }^{ 2 }+{ v }^{ 2 }-d } =\sqrt { 4+9-9 } \)
=\(\sqrt { 4 } \) = 2 units
Hence, the centre of the circle is (2, -3) and radius is 2 units.

Let the required numbers be x, y and z
By the given data,
x + y + z = 20  ....(1)
2z + x = 23 ⇒ x + 2z = 23...(2)
y + z + 3x = 46 ⇒  3x + y + z = 46..(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix} \right| =1\left| \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -2 + 5 + 1 =4
Δ₁ = \(\left| \begin{matrix} 20 & 1 & 1 \\ 23 & 0 & 2 \\ 46 & 1 & 1 \end{matrix} \right| =20\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| -1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| +1\left| \begin{matrix} 23 & 0 \\ 46 & 1 \end{matrix} \right| \)
= -40 + 69 + 23 = 52
Δ₂ = \(\left| \begin{matrix} 1 & 20 & 1 \\ 1 & 23 & 2 \\ 3 & 46 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| -20\left| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| \)
= -69 + 100 - 23 = 8
Δ₃ = \(\left| \begin{matrix} 1 & 1 & 20 \\ 1 & 2 & 23 \\ 3 & 1 & 46 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 0 & 23 \\ 1 & 46 \end{matrix} \right| -1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| +20\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -23 + 23 + 20 = 20
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 52 }{ 4 } \) = 13
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 8 }{ 4 } \) = 2 and z =\(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 20 }{ 4 } \) = 5
Hence the required numbers are 13, 2 and 5.

Let ∝, β be the roots x²+ bx + ca = 0
∝ +β = -b, ∝β = ca
and a, ૪ be the roots of x² + cx + ab = 0
∝+૪ = -c, ∝૪ = ab
Then, a² + b∝ + ca = 0 and
∝²+ c∝ + ab = 0
⇒ (b-c)∝+a(c-b) = 0 ⇒ ∝ = a
Also ∝β = ca and ∝૪ = ab
∴ β = c and ૪ = c
since ∝ = a is a root of x² + bx + ca = 0,
we get a² + ba + ca = 0 ⇒ a + b + c = 0
Thus, β+૪ = b + c = -a and β૪ = bc
Hence β, ૪ are the roots of the equation
x² + ax + bc = 0

\(\frac { 3 }{ x } -\frac { 4 }{ y } -\frac { 2 }{ z } \) - 0, \(\frac { 1 }{ x } +\frac { 2 }{ y } +\frac { 1 }{ z } \) - 2 = 0, \(\frac { 2 }{ x } -\frac { 5 }{ y } -\frac { 4 }{ z } \) + 1 = 0
Put \(\frac { 1 }{ x } =u,\frac { 1 }{ y } =v,\frac { 1 }{ z } =w\)
We get 3u - 4v - 2w = 1, u + 2v + w = 2, 2u - 5v - 4w = -1
∴ \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| =3\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(- 8 + 5) + 4'(- 4 - 2) - 2(- 5 - 4)
= 3(- 3) + 4(- 6) - 2(- 9)
= - 9 - 24 + 18 = -15
Δ₁ = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ -1 & -5 \end{matrix} \right| \)
= 1(- 8 + 5) + 4(- 8 + 11 -2(-10 + 2)
= 1(- 3) + 4(-7) - 2(- 8)
= - 3 - 28 + 16= -15
Δ₂ = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| \)
= 3(- 8 + 1) - 1(- 4 - 2) - 2(- 1 - 4)
= 3(-7) - 1(- 6) - 2(- 5)
= -21 + 6 + 10 = -5
Δ₃ = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 2 \\ -5 & -1 \end{matrix} \right| +4\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 - 4)+ 1(-5 - 4)
= 3(8) + 4(- 5) + 1(- 9)
= 24 - 20 - 9 = - 5
∴ \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -15 }{ -15 } =1\Rightarrow \frac { 1 }{ x } =1\Rightarrow \)x = 1
v = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \Rightarrow \)y = 3
w = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 3 } \Rightarrow \)z = 3
∴ Solution set is {1, 3, 3}

Now, \(\frac { z-1 }{ z+1 } =\frac { x+iy-1 }{ x+iy+1 } =\frac { \left( x-1 \right) +iy }{ \left( x+1 \right) +iy } =\frac { \left[ \left( x-1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] }{ \left[ \left( x+1 \right) +iy \right] \left[ \left( x+1 \right) -iy \right] } \)
\(\Rightarrow \frac { z-1 }{ z+1 } =\frac { \left( { x }^{ 2 }+{ y }^{ 2 }-1 \right) +i\left( 2y \right) }{ \left( x+1 \right) ^{ 2 }+{ y }^{ 2 } } \)
Since, arg \(\left( \frac { z-1 }{ z+2 } \right) =\frac { \pi }{ 2 } \Rightarrow { tan }^{ -1 }\left( \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } \right) \)= \(\frac { \pi }{ 2 } \)
\(\Rightarrow \frac { 2y }{ { x }^{ 2 }+{ y }^{ 2 }-1 } =tan\frac { \pi }{ 2 } \) ⇒ x²+ y² − 1 = 0
\(\Rightarrow { x }^{ 2 }+{ y }^{ 2 }=1\)

Here the number of equations is equal to the number of unknowns.
Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -2 & -5 \\ 0 & -4 & -9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow R_{ 2 }\div \left( -1 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div 7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) }{ { \longrightarrow } } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ([A | O]) = 3 = Number of unknowns.
Hence, the system has a unique solution. Since x = 0, y = 0, z = 0 is always a solution of the homogeneous system, the only solution is the trivial solution x = 0, y = 0, z = 0.
Note
In the above example, we find that
|A| = \(\left| \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix} \right| \) = 1(48 - 40) - 2(36 - 28) + 3(30 - 28) = 8 - 16 + 6 = -2 ≠ 0.

x = 1 
y = x² -5x + 6
y = 1 -5 + 6 = 2
x = 2
y = 4 -10 + 6 = 0
x = 0 
y = 6
x = 3
y = 9 -15 + 6 = 12
x = -1 
y = 1 + 5 + 6 = 12
x = 4
y = 16 - 20 + 6 = 2

(1, 2), (0, 6), (-1, 12)
P(x) = (x²-5x + 6) (x²-5x+16)
= x⁴- 5x³+16x²-5x+25x²- 80x + 6x²- 30x + 96 = 0
x⁴-10x³+ 47x² -110x + 96 = 0
It has two sign changes
\(\therefore\) it has two positive real roots
P(-x) = x⁴-10x³+ 47x² -110x + 96
It has no sign changes, no negative real roots
y = x²- 5x + 16

x	0	1	-1	2	4
y	16	12	23	10	12

 

First we evaluate the determinants
Δ = \(\left| \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right| \) = 6 ≠ 0, Δ₁ = \(\left| \begin{matrix} 3 & -1 & 0 \\ 17 & 3 & 4 \\ 7 & 1 & 2 \end{matrix} \right| \) = 12, Δ₂ = \(\left| \begin{matrix} 1 & 3 & 0 \\ 2 & 17 & 4 \\ 0 & 7 & 2 \end{matrix} \right| \) = -6, Δ₃ = \(\left| \begin{matrix} 1 & -1 & 3 \\ 2 & 3 & 17 \\ 0 & 1 & 7 \end{matrix} \right| \) = 24
By Cramer’s rule, we get  x₁ = \(\frac { { \Delta }_{ 1 } }{ \Delta } =\frac { 12 }{ 6 } \) = 2, x₂ = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { -6 }{ 6 } \) = -1, x₃ = \(\frac { 24 }{ 6 } \) = 4.
So, the solution is (x₁ = 2, x₂ = -1, x₃ = 4).

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₂] =\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 5 & -2 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -\frac { 5 }{ 2 } & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 2 }{ \longrightarrow } \left[ \begin{matrix} 1 & -\frac { 1 }{ 2 } \\ 0 & 1 \end{matrix}|\begin{matrix} \frac { 1 }{ 2 } & 0 \\ -5 & 2 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)
∴ We get A^-1=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Given that \(\left| { z }_{ 1 } \right| =\left| { z }_{ 2 } \right| =\left| { z }_{ 3 } \right| =r\Rightarrow { z }_{ 1 }\bar { { z }_{ 1 } } ={ z }_{ 2 }\bar { { z }_{ 2 } } ={ r }^{ 2 }\)
\(\Rightarrow { z }_{ 1 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 1 } } } ,{ z }_{ 2 }=\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } ,{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 3 } } \)
Therefore \({ z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 }=\frac { { r }^{ 2 } }{ { \bar { z } }_{ 1 } } +\frac { { r }^{ 2 } }{ \bar { { z }_{ 2 } } } +\frac { { r }^{ 2} }{ \bar { { z }_{ 3 } } } \)
= \({ r }^{ 2 }\left( \frac { \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } \bar { { z }_{ 3 } } +\bar { { z }_{ 1 } } { \overline { z } }_{ 2 } }{ \overline { { z }_{ 1 } } \bar { { z }_{ 2 } } \bar { { z }_{ 3 } } } \right) \) 
\(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| =\left| { r }^{ 2 } \right| \left| \frac { \overline { { z }_{ 2}{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } } }{ \overline { { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } } } \right| \) \(\left(\because \bar{z}_{1}+\bar{z}_{2}=\overline{z_{1}+z_{2}}\right)\)
= \({ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| } \)  \(\left( \because |z|=|\bar { z } |and\ \left| { z }_{ 1 }{ z }_{ 2 }{ z }_{ 3 } \right| =\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| \left| { z }_{ 3 } \right| \right) \)
= \(\left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| ={ r }^{ 2 }\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ { r }^{ 3 } } =\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ r } \)
\(\frac { \left| { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } \right| }{ \left| { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } \right| } \) = r   (given that \(z_{1}+z_{2}+z_{3} \neq 0\))
Thus,\(\left| \frac { { z }_{ 2 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 3 }+{ z }_{ 1 }{ z }_{ 2 } }{ { z }_{ 1 }+{ z }_{ 2 }+{ z }_{ 3 } } \right| \) = r

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).

The discriminant Δ =((4p)² - 4(4)(p+2) = 16(p²-p-2) = 16(p+1)(p-2). So we get
Δ < 0 if -1 < p < 2
Δ = 0 if p = -1 or p = 2
Δ >0 if \(\infty\). Thus the given polynomial has
imaginary roots if -1 < p < 2
equal real roots if p = −1 or p = 2;
distinct real roots if -\(\infty\) < p < -1 or 2 < p < \(\infty\)

Let  α, β and γ  be the roots of x³+ ax²+ bx + c = 0
Then, we get
Σ₁ = α + β + γ = -a ....(1)
Σ₂ = αβ + βγ + γα = b ...(2)
Σ₃ = αβγ = -c ...(3)
We have to form the equation whose roots are  α², β² and γ².
Using (1), (2) and (3), we find the following
Σ₁ = α² + β² + γ² = (α + β + γ )² - 2( αβ + βγ + γα) = (-a)² -2(b) = a²-2b,
Σ₂ = α²β² +  β²γ² + γ²α² = (αβ + βγ + γα)² - 2((αβ)( βγ)(γα) + (γα)(αβ))
= (αβ + βγ + γα)² - 2αβγ (β + γ + α) = (b)² - 2(-c)(-a) = b²-2ca
Σ₃ = α²β²γ² = (αβγ)² = (-c)² = c²
Hence, the required equation is
x³-(α² + β² + γ²)x² + (α²β² + β²γ² + γ²α²)x -  α²β²γ² = 0
That is, x³-(a²-2b)x² + (b²-2ca)x-c² = 0

Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\theta \Rightarrow cos\theta =\frac { 1 }{ 2 } \)
\(\Rightarrow sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 4 } } =\sqrt { \frac { 3 }{ 4 } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) =\frac { \sqrt { 3 } }{ 2 } \)

We have \(x=\sqrt { 2+x } \)
\(\Rightarrow { x }^{ 2 }=2+x \Rightarrow { x }^{ 2 }-x-2=0\)
\(\Rightarrow x=\frac { 1\pm \sqrt { 1+8 } }{ 2 } \Rightarrow x=\frac { 1\pm 3 }{ 2 } \)
\(\Rightarrow x=\frac { 1+3 }{ 2 } ,\frac { 1-3 }{ 2 } \Rightarrow x=2,-1\)
Also x>0, we get x = 2

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

(i)^-1924+ (i)²⁰¹⁸ = (i)^{-1924 + 0} + (i)^{2016 + 2} = (i)⁰ + (i)² = 1 - 1 = 0

\({ sin }^{ -1 }(-1)+{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }(2)\)
Let \({ sin }^{ -1 }\left( -1 \right) =x\)
\(\Rightarrow -1=sin\ x\)
\(\Rightarrow sin\ x=-1=-sin\frac { \pi }{ 2 } =sin\left( \frac { -\pi }{ 2 } \right) \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow x=\frac { -\pi }{ 2 } \)
\(\Rightarrow \frac { 1 }{ 2 } =cos\ y\Rightarrow cos\frac { \pi }{ 3 } \)
\(\Rightarrow y=\frac { \pi }{ 3 } \)
\(\therefore { sin }^{ -1 }\left( -1 \right) +{ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) +{ cot }^{ -1 }\left( 2 \right) \)
\(\frac { -\pi }{ 2 } +\frac { \pi }{ 3 } +{ cot }^{ -1 }\left( 2 \right) \)
\({ cot }^{ -1 }(2)+\frac { -3\pi +2\pi }{ 0 } ={ cot }^{ -1 }\left( 2 \right) -\frac { \pi }{ 6 } \)

\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 2
∴ \(\rho \)(A) ≤ min (3, 2) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} -1 & 3 \\ 4 & -7 \end{matrix} \right| \)= 7-12 = 5 ≠ 0
∴ \(\rho \)(A) = 2

Using the given values forr z₁, z₂ and z₃ we get |z₁| = |3+4i| =\(\sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } =5\)
|z₂| = |5-12i| = \(\sqrt { { 5 }^{ 2 }+(-12)^{ 2 } } =13\)
|z₃| = |6+8i| = \(\sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } =10\)
|z₁+z₂| = |(3+4i)+(5-12i)| = |8-8i| = \(\sqrt { 128 } =8\sqrt { 2 } \)
|z₂-z₃| = |(5-12i)-(6+8i)| = |1-20i| = \(\sqrt { 401 } \)
|z₁+z₃| = |(3+4i)+(6+8i)| = |9+12i| = \(\sqrt { 225 } =15\)
Note that the triangle inequality is satisfied in all the cases
|z₁+z₃| = |z₁|+|z₃| = 15 

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
Now \(\overline { z } \) = \(\overline { (2+\sqrt { 3 } )^{ 10 }-(2-i\sqrt { 3 } )^{ 10 } } \)
\(\overline { z } \) = \(\overline { (2+i\sqrt { 3 } )^{ 10 } } -(2-i\sqrt { 3 } )^{ 10 }\)
[∵ \(\overline { { z }_{ 1 }-{ z }_{ 2 } } =\overline { { z }_{ 1 } } -\overline { { z }_{ 2 } } \)]
= \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)
= -\(\left[ (2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 } \right] \)
∴ \(\overline { z } \) = -\(\ { z } \) ⇒ z is purely imaginary
Hence \((2-i\sqrt { 3 } )^{ 10 }-(2+i\sqrt { 3 } )^{ 10 }\)is purely imaginary

It is known that cos^-1 x : [-1, 1]\(\rightarrow\)[0, \(\pi\)] is given by
cos⁻¹x = y if and only if x = cos y for -1\(\le x\le1 and 0\le y \le\pi\) 
Thus, we have
cos^-1 \((-\frac{1}{\sqrt2})\) = \(\frac{3\pi}{4}\), since \(\frac{3\pi}{4}\)\(\in[0,\pi]\)cos\(\frac{3\pi}{4}\) = cos\((\pi=\frac{\pi}{4})=-cos \frac{\pi}{4}=-\frac{1}{\sqrt2}\)

Let cos⁻¹\(\left( \frac { \sqrt { 3 } }{ 2 } \right) \) = y. Then, cos y = \( \frac { \sqrt { 3 } }{ 2 } \)
The range of the principal values of y = cos⁻¹x is [0, \(\pi\)].
So, let us find y in [0, \(\pi\)] such that cos y =\( \frac { \sqrt { 3 } }{ 2 } \)
But, cos\(\frac{\pi}{6}=\frac{\sqrt3}{2} and \frac{\pi}{6}\in[0,\pi]\). Therefore, y = \(\frac{\pi}{6}\)
Thus, the principal value of cos^-1 \(\left( \frac { \sqrt { 3 } }{ 2 } \right) is\frac { \pi }{ 6 } \)

x	0	\(\frac {3 \pi }{ 2 } \)	\(3\pi \)	\(\frac { 9\pi }{ 2 } \)	\(6\pi \)
y	0	1	0	-1	0

The amplitude of sin x is 1 [Max of sin x curve is 1]
\(\Rightarrow \) amplitude of sin 7x is also 1
If p is the period of the function,
then f(x+p) = f(x)
Since the period of sine function is \(2\pi \)
The period of sin is \(\frac { 2\pi }{ 7 } \) 
amplitude = 1

Let the given complex number be
P (x, y) A(2, 1), and B(1, -2)
 Given PA = PB ⇒ PA² = PB²
⇒ (x - 2)² + (y-1)² = (x-1)² + (y + 2)²
\(\Rightarrow \not x^{2}-4 x+4+\not y^{2}-2 y+1= \not x^{2}-2 x+1+\not y^{2}+4 y+4 \)
⇒ -4x-2y+5 = -2x+4y+5
⇒ -2x-6y = 0
⇒ x+3y = 0

|A| =\(\left[ \begin{matrix} -2 & -3 \\ 5 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & 3 \\ -5 & 2 \end{matrix} \right] \)
(A^-1)^T = \(\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & 2 \end{matrix} \right] \)...(1)
A^T =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \)
|A^T| =\(\left[ \begin{matrix} -2 & 5 \\ -3 & -6 \end{matrix} \right] \) = 12+15 = 27
∴ (A^T)^-1 = \(\frac { 1 }{ |A^{ T }| } adj(A^{ T })=\frac { 1 }{ 27 } \left[ \begin{matrix} -6 & -5 \\ 3 & -2 \end{matrix} \right] \)...(2)
From (1) and (2), (A^-1)^T = (A^T)^-1

Put y = \(\frac { x-1+x-5 }{ 2 } =-3\)
⇒ x = y + 3
∴ (x-1)⁴+(x-5)⁴ = 82
⇒ (y+3-1)⁴+(y+3-5)⁴ = 82
⇒ (y+2)⁴+(y-2)⁴ = 82
⇒ 2(y⁴+24y²+16) = 82
⇒ y⁴+24y²+16 = 41
⇒ y⁴+24y²-25 = 0
⇒ (y²+25)(y²-1) = 0
⇒ y = 土5i, y = 士1
∴ x = 3土5i, 4, 2.

tan^-1\((tan(\frac{3\pi}{4})\)
Observe that \(\frac{3\pi}{4}\) is not in the interval \(\left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \), the principal range of tan^-1 x.
So, we write \(\frac{3\pi}{4}=\pi-\frac{\pi}{4}\)
Now, \(tan\left( \frac { 3\pi }{ 4 } \right) =tan\left( \pi -\frac { \pi }{ 4 } \right) =-tan\frac { \pi }{ 4 } =tan\left( -\frac { \pi }{ 4 } \right) and-\frac { \pi }{ 4 } \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)
Hence, \({ tan }^{ -1 }\left( tan\left( \frac { 3\pi }{ 4 } \right) \right) ={ tan }^{ -1 }\left( tan\left( -\frac { \pi }{ 4 } \right) \right) =-\frac { \pi }{ 4 } ,since-\frac { \pi }{ 4 } \in \left( -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right) \)

\(3-i\sqrt { 3 } \)
Let x + iy = \(3-i\sqrt { 3 } \)
= r(cos θ + i sin θ)
r = \(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } =\sqrt { 3^{ 2 }+(\sqrt { 3 } )^{ 2 } } =\sqrt { 9+3 } \)
= \(\sqrt { 12 } =2\sqrt { 3 } \)
α = \(tan^{ -1 }\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { -\sqrt { 3 } }{ 3 } \right| =tan^{ -1 }\left| \frac { 1 }{ \sqrt { 3 } } \right| =\frac { \pi }{ 6 } \)
Since the complex number \(3-i\sqrt { 3 } \) lies in the IV quadrant, [∵ x ⟶ +ve y ⟶ -ve]
Its principal value θ = -α
⇒ θ = \(\frac { \pi }{ 6 } \)
∴ Its polar form is
\(3-i\sqrt { 3 } \) = 2\(\sqrt { 3 } \)\(\left[ cos\left( 2k\pi -\frac { \pi }{ 6 } \right) +isin\left( 2k\pi -\frac { \pi }{ 6 } \right) \right] ,k\in Z\)

\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 4
∴ \(\rho \)(A) ≤ min(3, 4) = 3 
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{matrix} \right| \) = 0+0-8(4-4) = 0
[Expanded along C₁]
Also, \(\left| \begin{matrix} 0 & 2 & 1 \\ 0 & 4 & 3 \\ 8 & 0 & 2 \end{matrix} \right| =0+0-8\left| \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right| \)
[Expanded along C₁]
= -8(6-4) = -8(2) = -16 ≠ 0
∴ \(\rho \)(A) = 3

(1+i)¹⁸
Let 1+ i = \(r(cos\theta +isin\theta )\). Then , we get
\(r=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } ;\alpha ={ tan }^{ -1 }\left( \frac { 1 }{ 1 } \right) =\frac { \pi }{ 4 } \)
\(\theta =\alpha =\frac { \pi }{ 4 } \) (\(\because\) 1+i lies in the first Quadrant)
Therefore 1+ i = \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
Raising the power 18 on both sides
\(\left( 1+i \right) ^{ 18 }=\left[ \sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \right] ^{ 18 }=\sqrt { 12 } ^{ 18 }\left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
By de Moivre’s theorem
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \({ 2 }^{ 9 }\left( cos\left( 4\pi +\frac { \pi }{ 2 } \right) +isin\left( 4\pi +\frac { \pi }{ 2 } \right) \right) ={ 2 }^{ 9 }\left( cos\frac { \pi }{ 2 } +isin\frac { \pi }{ 2 } \right) \)
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }(i)=512i\)

If P(x) denotes the polynomial in the equation, then P(2) = 0.
Hence 2 is a root of the polynomial.
To find other roots, we divide the given polynomial x³−5x²−4x + 20 by x − 2 and get Q(x) = x² −3x−10 as the quotient.
Solving Q(x) = 0 we get −2 and 5 as roots.
Thus 2, −2, 5 are the solutions of the given equation.

Since 2-\(\sqrt{3}\)i is a root of the required polynomial equation with real coefficients, 2 +\(\sqrt{3}\)i is also a root. Hence the sum of the roots is 4 and the product of the roots is 7. Thus x²-4x + 7= 0 is the required monic polynomial equation.

For the given A, We get |A| = (2)(7) - (9)(1) = 14 - 9 = 5. So, A^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -9 \\ -1 & 2 \end{matrix} \right] =\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 9 }{ 5 } \\ -\frac { 1 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] \).
Then, (A^-1)^T = \(\left[ \begin{matrix} \frac { 7 }{ 5 } & -\frac { 1 }{ 5 } \\ -\frac { 9 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).           ....(1)
For the given A, We get A^T = \(\left[ \begin{matrix} 2 & 1 \\ 9 & 7 \end{matrix} \right] \). So |A^T| = (2)(7) - (1)(9) = 5
Then, (A^T)^-1 = \(\frac { 1 }{ 5 } \left[ \begin{matrix} 7 & -1 \\ -9 & 2 \end{matrix} \right] \).             ...(2)
From (1) and (2), we get (A^-1)^T = (A^T)^-1. Thus, we have verified the given property.