\(\overline { \left( 5+9i \right) +\left( 2-4i \right) } \)
= \(\overline { (5+2)+(9i-4i) } =\overline { 7+5i } \)
= 7 - 5i [∵ Conjugate of 7 + 5i is 7 - 5i]

i¹⁹⁴⁷+ i¹⁹⁵⁰
i¹⁹⁴⁷+i¹⁹⁵⁰ = i¹⁹⁴⁴.i³+i¹⁹⁴⁸.i²
[∴ 1944 is a multiple of 4, or 1948 is also a multiple of 4]
= (i⁴)486.i².i¹+(i⁴)487.i² [i⁴ = 1]
= (1⁴⁸⁶)(-1) + (1)⁴⁸⁷(-1) [i²= -1]
= -i-1
= -(1- i)

Since α, β, and γ are the roots of the equation x³+ px²+ qx + r = 0, we have
Σ₁ α + β + γ = -p and Σ₃ αβγ = -r
\(\Sigma \frac { 1 }{ \beta \gamma } =\frac { 1 }{ \beta \gamma } +\frac { 1 }{ \gamma \alpha } +\frac { 1 }{ \alpha \beta } =\frac { \alpha +\beta +\gamma }{ \alpha \beta \gamma } =\frac { -p }{ -r } =\frac { p }{ r } \).

The roots of x³+2x²+3x+4 = 0 are ∝, β, ૪
∴ ∝ + β + ૪ = -co-efficient of x² = -2   ...(1)
∝β + β૪ + ૪∝ = co-effficient of x = 3   ...(2)
-∝β૪ = +4 ⇒ ∝β૪ = -4      ...(3)
Form the equation whose roots are ∝-β-૪
∴ -∝-β-૪ = -(∝+β+૪)
= -(-2) = 2
∝β + β૪ + ૪∝ = 3
(-∝)(-β)(-૪) = -(∝β૪) = -(-4) = 4
∴ The required cubic equation is 
x³-(-∝-β-૪)x²+(∝β+β૪+૪∝)
x-[(-∝)(-β)(-૪)] = 0
⇒ x³-(2)x²+3x-4 = 0
⇒ x³-2x²+3x-4 = 0

We have |2z-3-i| = 3
|2(x + iy)-3 - i| = 3
Squaring on both sides, we get
|(2x - 3) + (2y - 1)i|² = 9
\(\Rightarrow\) (2x - 3)² + (2y - 1)² = 9
\(\Rightarrow\) 4x² + 4y² -12x - 4y + 1 = 0, the locus of z in Cartesian form

\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) +isin\left( \frac { 9\pi }{ 4 } -\left( \frac { -3\pi }{ 2 } \right) \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) +isin\left( \frac { 9\pi }{ 4 } +\frac { 3\pi }{ 2 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { 15\pi }{ 4 } \right) +isin\left( \frac { 15\pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( cos\left( 4\pi -\frac { \pi }{ 4 } \right) +isin\left( 4\pi -\frac { \pi }{ 4 } \right) \right) \)
= \(\frac { 1 }{ 2 } \left( cos\left( \frac { \pi }{ 4 } \right) -isin\left( \frac { \pi }{ 4 } \right) \right) =\frac { 1 }{ 2 } \left( \frac { 1 }{ \sqrt { 2 } } +i\frac { 1 }{ \sqrt { 2 } } \right) \)
\(\frac { 2\left( cos\frac { 9\pi }{ 4 } +isin\frac { 9\pi }{ 4 } \right) }{ 4\left( cos\left( \frac { -3\pi }{ 2 } + \right) sin\left( \frac { -3\pi }{ 2 } \right) \right) } \) = \(\frac { 1 }{ 2\sqrt { 2 } } -i\frac { 1 }{ 2\sqrt { 2 } } =\frac { \sqrt { 2 } }{ 4 } +i\frac { \sqrt { 2 } }{ 4 } \) Which is in rectangular form.

5x − 2y + 16 = 0, x + 3y − 7 = 0
Given Δ = \(\left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| \) = 15+2 = 17
Δ₁ = \(\left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| \) = -48+14 = -34
Δ₂ = \(\left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| \) = 35+16 = 51
∴ x = \(\frac { \triangle _{ 1 } }{ \triangle } =\frac { -34 }{ 7 } \) = -2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 51 }{ 17 } \)
∴ Solution set is {-2, 3}

The matrix form of the system is AX = B , where A = \(\left[ \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] \)
We find |A| = \(\left| \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right| \) = 10 - 6 = 4 ≠ 0. So, A⁻¹ exists and A⁻¹ = \(\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \)
Then, applying the formula X = A⁻¹B, we get
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} -4 \\ 16 \end{matrix} \right] =\left[ \begin{matrix} \frac { -4 }{ 4 } \\ \frac { 16 }{ 4 } \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 4 \end{matrix} \right] \).
So the solution is (x = −1, y = 4).

|z²-3| ≤ |z²|+|-3| [Triangle law of inequality]
≤ |z|²+3≤1+3  [∴ |z| = 1]
|z²-3| ≤ 4 ..............(1)
Also, |z²- 3| ≥ ||z²|-|-3||
≥ ||z|²-3|   [∵ |-3| = 3]
≥ |1²-3|  [∵ |z| = 1]
≥ |-2| .
|z²-3| ≥ 2.............(2)
From (1) and (2) we get 2 ≤ |z²-3| ≤ 4
Hence proved.

2cos²x - 9cosx + 4 = 0 ............ (1)
The left hand side of this equation is not a polynomial in x. But it looks like a polynomial. In fact, we can say that this is a polynomial in cos x. However, we can solve the equation (1) by using our knowledge on polynomial equations. If we replace cos x by y, then we get the polynomial equation 2y²- 9y + 4 = 0 for which 4 and \(\frac{1}{2}\) are solutions.
From this we conclude that x must satisfy cos x = 4 or cos x = \(\frac{1}{2}\).
But cos x = 4 is never possible, if we take cos x = \(\frac{1}{2}\), then we get infinitely many real numbers x satisfying cos x = \(\frac{1}{2}\); in fact, for all n\(\in \)Z, x = 2nπ ±\(\frac { \pi }{ 3 } \) are solutions for the given equation (1).
If we repeat the steps by taking the equation cos²x - 9 cosx + 20 = 0, we observe that this equation has no solution.

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
A =\(\left[ \begin{matrix} 1 \\ 2 \\ 5 \end{matrix}\begin{matrix} 1 \\ -1 \\ -1 \end{matrix}\begin{matrix} 1 \\ 3 \\ 7 \end{matrix}\begin{matrix} 3 \\ 4 \\ 11 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { { R } }_{ 2 }-2{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 5 \end{matrix}\begin{matrix} 1 \\ -3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 7 \end{matrix}\begin{matrix} 3 \\ -2 \\ 11 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-5{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -6 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\begin{matrix} 3 \\ -2 \\ -4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { { R } }_{ 3 }-2{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 3 \\ -2 \\ 0 \end{matrix} \right] \)
The last equivalent matrix is in row echelon form it ha two non-zero row \(\rho \)(A) = 2

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

\(\left| z+3+4i \right| \le \left| z \right| +\left| 3+4i \right| =2+5=7\)
\(\left| z+3+4i \right| \le 7\) .............. (1)
\(\left| z+3+4i \right| \ge \left| \left| z \right| -\left| 3+4i \right| \right| =\left| 2-5 \right| =3\)
\(\left| z+3+4i \right| \ge 3\) ............ (2)
From (1) and (2) we get, \(3\le \left| z+3+4i \right| \le 7\) 

The distance between origin to z = i, −2 + i, and 3 are
|z| = |i| = 1
|z| = |−2+i| = \(\sqrt { \left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 5 } \)
|z| = |3| = 3
Since \(1<\sqrt { 5 } <3\), the farthest point from the origin is 3 .

Let A be the matrix. Performing elementary row operations, we get
A = \(\left[ \begin{matrix} 2 \\ \begin{matrix} -3 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -1 \\ 7 \end{matrix} \end{matrix} \right] \)\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} -6 \\ 6 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 8 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} -4 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} -2 \\ 7 \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -13 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -2 \end{matrix} \end{matrix} \right] \).
\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ -45 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ -30 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -15 \right) }{ \longrightarrow } \left[ \begin{matrix} 2 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 4 \\ \begin{matrix} 8 \\ 3 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 7 \\ 2 \end{matrix} \end{matrix} \right] \).
The last equivalent matrix is in row-echelon form. It has three non-zero rows. So, ρ(A) = 3.

Given A\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
Let B =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \) and 
C = \(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
∴ AB = C
Post multiply by B^-1 we get
A(BB^-1) = CB^-1
⇒ A = CB^-1     [∵ BB^-1 = 1]
|B| = \(\left| \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right| \)
= -10 + 3 = -7 ≠ 0
∴ B^-1 exists
B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { -1 }{ 7 } \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
A = CB^-1
=\(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(7\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} -4+1 & -6+5 \\ -2+1 & -3+5 \end{matrix} \right] =-\left[ \begin{matrix} -3 & -1 \\ -1 & 2 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 3 & 1 \\ 1 & -2 \end{matrix} \right] \).

The roots are rational if Δ = b²−4ac = (−2p)²−4(p²−q²+2qr−r²).
But this expression reduces to 4(q²−2qr+r²) or 4(q−r)² which is a perfect square.
Hence the roots are rational.

Since \(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a root, x-\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) is a factor. To remove the outermost square root, we take x +\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) as another factor and find their product.
\(\left( x+\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) \left( x-\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \right) ={ x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \)
Still we didn’t achieve our goal. So we include another factor x²+\(\sqrt { \frac { \sqrt { 2 } }{ \sqrt { 3 } } } \) and get the product.
\(\left( { x }^{ 2 }-\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) \left( { x }^{ 2 }+\frac { \sqrt { 2 } }{ \sqrt { 3 } } \right) ={ x }^{ 4 }-\frac { 2 }{ 3 } \)
So, 3x⁴- 2 = 0 is a required polynomial equation with the integer coefficients.
Now we identify the nature of roots of the given equation without solving the equation. The idea comes from the negativity, equality to 0, positivity of Δ = b²- 4ac.

Let z₁ = (2+i)x + (1−i)y + 2i−3 = (2x+y−3) + i(x−y+ 2)and
z₂ = x+(−1+2i)y+1+i = (x−y+1) + i(2y+1)
Given that z₁ = z₂ 
Therefore (2x+y−3) + i(x−y+2) = (x−y+1) + i(2y+1).
Equating real and imaginary parts separately, gives
2x+y−3 = x−y+1 \(\Rightarrow\) x+2y = 4
x−y+2 = 2y +1 \(\Rightarrow\) x−3y = −1
Solving the above equations, gives
x = 2 and y = 1.

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0

We have to find \(1^{\frac{1}{4}}\). Let z⁴ = \(1^{\frac{1}{4}}\). Then z⁴ = 1.
In polar form, the equation z = 1 can be written as
\(z^4=cos\left( 0+2k\pi \right) +isin\left( 0+2k\pi \right) ={ e }^{ i2k\pi }\), k = 0, 1, 2,...
Therefore,\({ \left( z \right) }^{ \frac { 1 }{ 4 } }=cos\left( \frac { 2k\pi }{ 4 } \right) +isin\left( \frac { 2k\pi }{ 4 } \right) ={ e }^{ i\frac { 2k\pi }{ 4 } }\), k=0,1,2,3.
Taking k = 0, 1, 2, 3, we get
k = 0, z = cos 0 + isin 0 = 1
k = 1, \(z=cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) =i\)
k = 2, \(z=cos\pi +isin\pi =-1\)
k = 3, \(z=cos\frac { 3\pi }{ 2 } +isin\frac { 3\pi }{ 2 } =-cos\frac { \pi }{ 2 } -isin\frac { \pi }{ 2 } =-i\)
Fourth roots of unity are 1, i, −1, −i \(\Rightarrow\) 1, \(\omega \), \({ \omega }^{ 2 }\) and \({ \omega }^{ 3 }\), where \(\omega ={ e }^{ i\frac { 2\pi }{ 4 } }=i\)

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 unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 25 \\ 13 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 4 & 3 & 6 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 13 \\ 25 \\ 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & -1 & -13 \end{matrix}|\begin{matrix} 13 \\ -27 \\ -25 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 1 & 13 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 25 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 17{ R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 0 & 199 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 398 \end{matrix} \right] \).
The equivalent system is written by using the echelon form:
x + 5y + 7 = 13, … (1)
17y + 22z = 27, … (2)
199z = 398..... (3)
From (3), we get z = \(\frac { 398 }{ 199 } \) = 2.
Substituting z = 2 in (2), we get y = \(\frac { 27-22\times 2 }{ 17 } =\frac { -17 }{ 17 } \) = -1
Substituting z = 2, y = -1, in (1), we get x = 13 - 5 x (-1) - 7 \(\times\) 2 = 4.
So, the solution is (x = 4, y = -1, z = 2).

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

Applying Gauss-Jordan method, we get
[A | I₃] =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \frac { 1 }{ 2 } { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 3 & 2 & 1 \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & \left( 1/2 \right) & -\left( 1/2 \right) \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -\left( 3/2 \right) & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\longrightarrow 2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & \left( 1/2 \right) & \left( 1/2 \right) \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} \left( 1/2 \right) & 0 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 2 & -1 & 0 \\ -3 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 1 }\longrightarrow { R }_{ 1 }-{ R }_{ 3 } \\ { R }_{ 2 }\longrightarrow { R }_{ 2 }+{ R }_{ 3 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).
So, A^-1 = \(\left[ \begin{matrix} 3 & -1 & -1 \\ -4 & 2 & 1 \\ -1 & 0 & 1 \end{matrix} \right] \).

We can solve this fourth degree equation by rewriting it suitably and adopting a technique of substitution. Rewriting the equation as
(x−2)(x−3)(x−7)(x+2)+19 = 0
the given equation becomes
(x²−5x+6)(x²−5x−14)+19 = 0 .
If we take x² − 5x as y, then the equation becomes (y+6)(y−14)+19 = 0;
that is, y²-8y-65 = 0
Solving this we get solutions y = 13 and y = −5. Substituting this we get two quadratic equations
x²-5x-13 = 0 and x²-5x+5 = 0
which can be solved by usual techniques. The solutions obtained for these two equations together give solutions as \(\frac { 5\pm \sqrt { 77 } }{ 2 } ,\frac { 5\pm \sqrt { 5 } }{ 2 } \).

Let the roots be in A.P. Then, we can assume them in the form α-d, α, α+d
Applying the Vieta’s formula (α-d)+α+(α+d) = \(\frac{p}{1}\) = p ⇒ 3α = -p ⇒ α = -\(\frac{p}{3}\).
But, we note that α is a root of the given equation. Therefore, we get
\(\left( -\frac { p }{ 3 } \right) ^{ 3 }+p\left(- \frac { p }{ 3 } \right) ^{ 3 }+q\left(- \frac { p }{ 3 } \right) ^{ 3 }+r\) = 0 ⇒ 9 pq = 2p³+27r.

The given equation is
x⁴- 9x² + 20 = 0
This is a fourth degree equation. If we replace x² by y then we get the quadratic equation
y²- 9y + 20 = 0
It is easy to see that 4 and 5 as solutions for y²- 9y + 20 = 0. Now taking x² = 4 and x² = 5, we get 2, -2, \(\sqrt{5}\), -\(\sqrt{5}\) as solutions of the given equation.
We note that the technique adopted above can be applied to polynomial equations like x⁶-17x³+30 = 0, ax^2k+ bx^k + c = 0 and in general polynomial equations of the form a_nx^kn + a_n-1x^k(n-1) + .... + a₁x^k + a₀ = 0 where k is any positive integer.

Given equation is 2x² + kx + k = 0
Here a = 2, b = k, c = k
\(\therefore\) Discriminant 
∆ = b²-4ac = k²-4 (2)(k)
= k²-8k = 0
k (k - 8) = 0
Since k is real, the possible values of k are
k < 0, k = 0 or 8
case (i) when 0< k< 8, ∆ < 0
⇒  The roots are imaginary
case (ii) when k = 0 or 8, ∆ = 0,
∴ The roots are real and equal
Case (iii) When k > 8, ∆ = k (k - 8) > 0
⇒ The roots are real and distinct

By choosing the coordinate axes suitably, we take the equation of the circle as x²+ y² = r² and the equation of the straight line as y = mx + c. We know that the points of intersections of the circle and the straight line are the points which satisfy the simultaneous equation
x² + y² = r²
y = mx + c .......... (2)
If we substitute mx + c for y in (1), we get
x²+(mx + c)²- r² = 0
which is same as the quadratic equation
(1+m²)x²+2mcx+(c²- r²) = 0 ............(3)
This equation cannot have more than two solutions, and hence a line and a circle cannot intersect at more than two points.
It is interesting to note that a substitution makes the problem of solving a system of two equations in two variables into a problem of solving a quadratic equation.
Further we note that as the coefficients of the reduced quadratic polynomial are real, either both roots are real or both imaginary. If both roots are imaginary numbers, we conclude that the circle and the straight line do not intersect. In the case of real roots, either they are distinct or multiple roots of the polynomial. If they are distinct, substituting in (2), we get two values for y and hence two points of intersection. If we have equal roots, we say the straight line touches the circle as a tangent. As the polynomial (3) cannot have only one simple real root, a line cannot cut a circle at only one point.

Given equation is 2x⁴- 8x + 6x²- 3 = 0
Here a = 2, b = -8, c = 6, d = 0, e = -3
Let ∝, β, ૪ and \(\delta \) be the roots of equation (1)
Then by Vieta's formula,
\(\sum { _{ 1 }= } \alpha +\beta +\gamma +\delta =\frac { -b }{ a } =\frac { -(-8) }{ 2 } =4\)
\(\sum { _{ 2 } } =\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\frac { c }{ a } =\frac { 6 }{ 2 } =3\)
\(\sum { _{ 3 } } =\alpha \beta \gamma +\alpha \beta \delta +\alpha \gamma \delta +\beta \gamma \delta =\frac { -d }{ a } =\frac { 0 }{ a } \)
\(\sum { _{ 4 } } =\alpha \beta \gamma \delta =\frac { e }{ a } =\frac { -3 }{ 2 } \)
Now, (a+b+c+d)² = a²+b²+c²+d²+2(ab+ac+ad+bc+cd)
⇒ n∝²+β²+૪²+\(\delta\)² = (∝ + β + ૪ + \(\delta\))²-2(\(\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta \))
∝² + β² + ૪² = 4²-2(3) = 16 - 6 = 10

Comparing the two given lines with
\(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have, \(\vec { a } =-\hat { -1 } -3\hat { j } -5\hat { k } ,\vec { b } =3\hat { i } +5\hat { j } +7\hat { k } ,\vec { c } =2\hat { i } +4\hat { j } +6\hat { k } \) and \(\vec { d } =\hat { i } +4\hat { j } +7\hat { k } \)
We know that the two given lines are coplar, if \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
Here, \(\vec { b } \times \vec { d } \left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 5 & 7 \\ 1 & 4 & 7 \end{matrix} \right| =7\hat { i } -14\hat { j } +7\hat { k } \) and \(\vec { c } -\vec { a } =3\hat { i } +7\hat { j } +11\hat { k } \)
Then, \((\vec { c } -\vec { a } ).(\vec { b } \times \vec { d } )=(3\hat { i } +7\hat { j } +11\hat { k } )(7\hat { i } -14\hat { j } +7\hat { k } )\)
Therefore the two given lines are coplanar.Then we find the non parametric form of vector equation of the plane containing the two given coplanar lines. We know that the plane containing the two given coplanar lines is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { d } )\)=0
which implies that \((\vec { r } -(-\hat { i } -3\hat { j } -5\hat { k } )).(7\hat { i } -14\hat { j } +7\hat { k } )\)=0. Thus, the required non-parametric vector equation of the plane containing the two given coplanar lines is \(\vec { r } .(\hat { i } -2\hat { j } +\hat { k } )\)=0.