Let A = \(\left[ \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3.
The only third order minor is |A| = \(\left| \begin{matrix} -2 & 2 & -1 \\ 0 & 5 & 1 \\ 0 & 0 & 0 \end{matrix} \right| \) = (-2)(5)(0) = 0. So ρ(A) ≤ 2.
There are several second order minors. We find that there is a second order minor, for example, \(\left| \begin{matrix} -2 & 2 \\ 0 & 5 \end{matrix} \right| \) = (-2)(5) = -10 ≠ 0. So, ρ(A) = 2.
Note that there are two non-zero rows. The third row is a zero row.

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

Here A = 3, C = 2 and F = -14
A ≠ C and A and C are of the same sign.
Hence, the given equation represents an ellipse.

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 

Equation of the circle is x² + (y + 2)² = 0
Compare with(x-h)²+(y-k)² = r²
h = 0, k = -2, r² = 0
Centre (h, k) = (0, -2)
radius is 0.

Since α and β are the roots of the quadratic equation, we have α + β =\(\frac { 7 }{ 2 } \) and αβ = \(\frac { 13 }{ 2 } \).
Thus, to construct a new quadratic equation,
Sum of the roots = α²+β² = (α+β)²-2αβ =\(\frac { -3 }{ 4 } \)
Product of the roots = α²β² = (αβ)² = \(\frac { 169 }{ 4 }\)
Thus a required quadratic equation is  x²+\(\frac { 3 }{ 4 } x+\frac { 169 }{ 4 } \)= 0. 
From this we see that 4x²+3x+169 = 0 is a quadratic equation with roots α² and β².

(i)⁷= (i)⁴⁺³ = (i)³ = -i

The length a of the semi major axis of the elliptical ceiling is17 m. The height b of the semi minor axis is 8 m. Thus c² = a² -b² = 17² - 8²
then c =\(\sqrt { 289-64 } =\sqrt { 225 } =15\)
For the elliptical ceiling the foci are located on either side about 15 m from the centre, along its major axis.

AS = 94.5 × 10⁶ km,
SA' = 152 × 10⁶ km
a+c = 152 × 10⁶
a-c = 94.5 × 10⁶
Subtracting 2c = 57.5 × 10⁶ = 575 × 10⁵ km
Distance of the Sun from the other focus is SS' = 575 × 105 km.

Length of latus rectum = 4,
distance between foci = 4\(\sqrt { 2 } \) major axis is y-axis
Given \(\frac { { 2b }^{ 2 } }{ a } =4\) and distance between foci = 2ae = 4\(\sqrt { 2 } \)
⇒ ae = \(2\sqrt { 2 } \)
⇒a²e² = 8 ....(1)
\(\frac { { 2b }^{ 2 } }{ a } =4\Rightarrow { b }^{ 2 }=2a\) ...(2)
We know b² = a²(1 - e²)
b² = a² - a²e²
2a = a2 - 8
[using (1) and (2)]
a² - 2a - 8 = 0
On factorising we get
(a - 4)(a + 2) = 0
a = 4 or-2
a = 4
⇒ [∴ a = -2 is not possible]
⇒ ae = 16
∴ From (2), b² = 2(4) = 8
Hence, the equation of the ellipse is
\(\frac { { x }^{ 2 } }{ 8 } +\frac { { y }^{ 2 } }{ 16 } =1\) [∵ Major axis is y -axis]

2 +i2\(\sqrt { 3 } \)
Let 2+i2\(\sqrt { 3 } \)  = x + iy = r (cosθ + i sinθ)
r = modulus =\(\\ \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \)
=\(\\ \sqrt { { 2 }^{ 2 }+(2\sqrt { 3 } )^{ 2 } } \)
= \(\sqrt { 4+12 } =\sqrt { 16 } \) = 4
α = tan-1\(\left| \frac { y }{ x } \right| =tan^{ -1 }\left| \frac { 2\sqrt { 3 } }{ 2 } \right| \)
= \(tan^{ -1 }(\sqrt { 3 } )=\frac { \pi }{ 3 } \)
Since the complex number 2+i2 \(\sqrt { 3 } \) lies in the I quadrant, [x, y both +ve] its principal value θ = α = \(\frac { \pi }{ 3 } \)
∴ Its polar form is 2+i2\(\sqrt { 3 } \)
= 4\(\left[ cos\left( 2k\pi +\frac { \pi }{ 3 } \right) +isin\left( 2k\pi +\frac { \pi }{ 3 } \right) \right] ,k\in Z\).

|z-4| = 16
Given z = x + iy
|z - 4| = 16
⇒ |x + iy - 4| = 16
⇒ |(x- 4) + iy| = 16
⇒ \(\\ \sqrt { (x-4)^{ 2 }+{ y }^{ 2 } } \) = 16
⇒ (x - 4)² + y² = 16²
[Squaring both sides]
⇒ x²-8x + 16 + y² = 256
⇒ x²-8x + y²+ 16-256 = 0
⇒ x²-8x + y²-240 = 0 Which is the required Cartesian equation.
The locus of the point is a circle.

Since \(2+i\sqrt { 3 } \) is a root of the polynomial equation, its conjugate 2-i\(\sqrt3\) is also a root of the equation:
∴ Sum of the roots \(=2+i\sqrt { 3 } +2-i\sqrt { 3 } =4\)
Product of the roots \(=(2+i\sqrt { 3 } )(2-i\sqrt { 3 } )\)
\(={ 2 }^{ 2 }+{ (\sqrt { 3 } })^{ 2 }\)
\([\because (a+ib)(a-ib)={ a }^{ 2 }+{ b }^{ 2 }]\)
= 4 + 3 = 7
Hence, the polynomial equation of minimum degree with rational co-efficients is
x² - x (sum of the roots) + product of the roots = 0
\(\Rightarrow { x }^{ 2 }-x(4)+7=0\)
\(\Rightarrow { x }^{ 2 }-4x+7=0\)

Given A = \(\left[ \begin{matrix} 1 & \tan { x } \\ -\tan { x } & 1 \end{matrix} \right] \)
|A| = 1 + tan² x
∴ A^-1 = \(\frac { 1 }{ |A| } \)adjA
= \(\frac { 1 }{ 1+tan^{ 2 } } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of elements in the off diagonal]
A^T= \(\\ \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
∴ A^TA^-1
=\(\left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1-tan^{ 2 }x & -tanx-tanx \\ tanx+tanx & -tan^{ 2 }x+1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } & \frac { -2tanx }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tanx }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+tan^{ 2 }x } \end{matrix} \right] \)
A^TA^-1 =\(\left[ \begin{matrix} cos2x & -sin2x \\ sin2x & cos2x \end{matrix} \right] \)
Hence proved.

Since 2-\(\sqrt{3}\)i is a root and the coefficients are rational numbers, 2+\(\sqrt{3}\)i  is also a root. A required polynomial equation is given by
x² −(Sum of the roots) x + Product of the roots = 0
and hence
x²- 4x +1 = 0 is a required equation.

 y² - 4y - 8x + 12 = 0
y²-4y = 8x-12
Adding 4 both sides, we get,
y - 4y + 4 = 8x - 12 + 4 = 8x - 8
⇒ (y - 2)² = 8(x - 1)
This is a right open parabola and latus
rectum is 4a = 8 ⇒ a = 2.
(a) Vertex is (1, 2) ⇒ h = 1, k = 2
(b) focus is (h + a, 0 + k)
⇒ (1 + 2, 0 + 2)
⇒ (3, 2)
(c) Equation of directrix is x = h - a
⇒ x = 1-2
⇒ x = -1
(d) Length of latus rectum is 4a = 8 units.

 \(\left( -\sqrt { 3 } +3i \right) ^{ 31 }\)
Let \(-\sqrt { 3 } +3i=r\left( cos\theta +isin\theta \right) \). Then, we get
\(r=\sqrt { \left( -\sqrt { 3 } \right) ^{ 2 }+{ 3 }^{ 2 } } =\sqrt { 12 } =2\sqrt { 3 } \)
\(\alpha ={ tan }^{ -1 }\left| \frac { 3 }{ -\sqrt { 3 } } \right| ={ tan }^{ -1 }\sqrt { 3 } =\frac { \pi }{ 3 } \)
\(\theta =\pi -\alpha =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \) (\(\because\)  \(\sqrt { 3 } +3i\) lies in II Quadrant)
Therefore,\(-\sqrt { 3 } +3i=2\sqrt { 3 } \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
Raising power 31 on both sides,
\(\left( -\sqrt { 3 } +3i \right) ^{ 31 }=\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) ^{ 31 }\)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( 20\pi +\frac { 2\pi }{ 3 } \right) +isin\left( 20\pi +\frac { 2\pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( cos\left( \pi -\frac { \pi }{ 3 } \right) +isin\left( \pi -\frac { \pi }{ 3 } \right) \right) \)
= \(\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -cos\frac { \pi }{ 3 } +isin\frac { \pi }{ 3 } \right) =\left( 2\sqrt { 3 } \right) ^{ 31 }\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \).

|z| = 2 represents the circle with centre (0, 0) and radius 2.
Let A, B, and C be the vertices of the given triangle. Since the vertice z₁, z_2,and z₃  form an equilateral triangle inscribed in the circle|z| = 2, the sides of this triangle AB, BC, and CA subtend \(\frac { 2\pi }{ 3 } \) radians (120 degree) at the origin (circumcenter of the triangle).
(The complex number ze¹⁶ is a rotation of z by \(\theta\) radians in the counter clockwise direction about the origin.)
Therefore, we can obtain z₂ and z₃ and by the rotation of z₁ by \(\frac { 2\pi }{ 3 } and\ \frac { 4\pi }{ 3 } \) respectively.
Given that \(\vec { OA } ={ z }_{ 1 }=1+i\sqrt { 3 } \)
\(\vec { OB } ={ z }_{ 1 }e^{ i\frac { 2\pi }{ 3 } }=\left( 1+i\sqrt { 3 } \right) e^{ i\frac { 2\pi }{ 3 } }\)
= \(\left( 1+i\sqrt { 3 } \right) \left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(\left( 1+i\sqrt { 3 } \right) \left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) =-2;\)
\(\vec { OC } ={ z }_{ 1 }{ e }^{ i\frac { 4\pi }{ 3 } }={ z }_{ 2 }{ e }^{ i\frac { 2\pi }{ 3 } }=-2e3^{ i\frac { 2\pi }{ 3 } }\)
= \(-2\left( cos\frac { 2\pi }{ 3 } +isin\frac { 2\pi }{ 3 } \right) \)
= \(-2\left( -\frac { 1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) =1-i\sqrt { 3 } \)
Therefore, z₂ = -2 and z₃ = 1-i\(\sqrt { 3 } \)

By taking the vertex; at the origin, the parabola is open downward.

Its equation is x² = -4ay
It passes through (6, -4)
∴ 36 = -4a(-4) ⇒ 4a = - \(\frac { 36 }{ 4 } \) = 9
∴ (1) becomes, x² = -9y
To find the slope at (-6, -4)
Differentiating (1) with respect to 'x' we get,
2x = -9\(\frac { dy }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { -2x }{ 9 } \)
At (-6, -4), \(\frac { dy }{ dx } =-2\frac { (-6) }{ 9 } =\frac { 12 }{ 9 } =\frac { 4 }{ 3 } \)
∴ \(tan\theta =\frac { 4 }{ 3 } \Rightarrow \theta ={ tan }^{ -1 }\left( \frac { 4 }{ 3 } \right) \)
∴ The angle of projection is tan^-1 \(\left( \frac { 4 }{ 3 } \right) \)

Let V₁ be the vertex of the parabola and
V₂ be the vertex of the hyperbola.
\(\overset { \_ \_ \_ \_ \_ \_ }{ { F }_{ 1 }{ F }_{ 2 } } \) = 14−2 = 12m, 2c = 12, c = 6
The distance of centre to the vertex of the hyperbola is a = 6−1 = 5
b² = c² - a²
= 36−25 = 11.
Therefore the equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 25 } -\frac { { x }^{ 2 } }{ 11 } =1\)

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

Let the price of bond invested in 6%, 8% and 9% rates be let Rs. x, Rs. y and Rs. z respectively
∴ By the given data, x + y + z = 65000 ..........(1)
\(\frac { 6\times x\times 1 }{ 100 } +\frac { 8\times y\times 1 }{ 100 } +\frac { 9\times z\times 1 }{ 100 } \) = 4800
[∵ Intrest = \(\frac { PNR }{ 100 } \)]
⇒ \(\frac { 6x }{ 100 } +\frac { 8y }{ 100 } +\frac { 9z }{ 100 } \)= 4800
⇒ 6x+8y+9z = 480000 ............(2)
Also, \(\frac { 9z }{ 100 } =600+\frac { 8y }{ 100 } \)
⇒ \(\frac { -8y }{ 100 } +\frac { 9y }{ 100 } \) = 600
⇒ -8y+9z = 60000 ............(3)
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row operation, we get
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 8 & 9 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 480000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 21 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 420000 \end{matrix} \right] \)
Writing the equivalent from the row echelon matrix we get,
x+y+z = 65000 ...........(1)
2y+z = 90000 ...........(2)
21z = 42000
⇒ z = \(\frac { 420000 }{ 21 } \) = 20000
Substituting z = 20,000 in (2),
2y + 3(20,000) = -90000
⇒ 2y+60,000 = 90,000
⇒ 2y = 90,000 - 60,000
= 30,000
⇒ y = \(\frac { 30,000 }{ 2 } \) = 15,000
Substitutingy = 15,000 and z = 20,000 in (1) we get,
x + 15,000 + 20,000 = 65000
⇒ x + 35,000 = 65000
⇒ x = 65,000 - 35,000
⇒ 30,000
Thus the price of 6% bond is f 30,000 the price of 8% bond is f 15,000 and the price of 9% bond is f 20,000 is Rs. 20,000.

Since v(3) = 64, v(6) = 133,and v(9) = 208 , we get the following system of linear equations
9a + 3b + c = 64 ,
36a + 6b + c = 133 ,
81a + 9b + c = 208 .
We solve the above system of linear equations by Gaussian elimination method.
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row
operations, we get
[A | B] = \(\left[ \begin{matrix} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{matrix}|\begin{matrix} 64 \\ 133 \\ 208 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & -6 & -3 \\ 0 & -18 & -8 \end{matrix}|\begin{matrix} 64 \\ -123 \\ -368 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -3 \right) ,{ R }_{ 3 }\div \left( -2 \right) }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 9 & 4 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 184 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow 2{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 18 & 8 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 368 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ -1 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 1 \end{matrix} \right] \).
Writing the equivalent equations from the row-echelon matrix, we get
9a + 3b + c = 64, 2b + c = 41, c = 1.
By back substitution, we get c = 1, b = \(\frac { \left( 41-c \right) }{ 2 } =\frac { \left( 41-1 \right) }{ 2 } \) = 20, a = \(\frac { 64-3b-c }{ 9 } =\frac { 64-60-1 }{ 9 } =\frac { 1 }{ 3 } \).
So, we get v(t) = \(\frac { 1 }{ 3 }\)t² + 20t + 1. Hence, v(15) = \(\frac { 1 }{ 3 }\) (225) + 20(15) + 1 = 75 + 300 + 1 = 376.

\(\left[ \begin{matrix} 2 & -1 \\ 5 & -2 \end{matrix} \right] \)
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=\(\left[ \begin{matrix} -2 & 1 \\ -5 & 2 \end{matrix} \right] \)

Rearranging terms in the equation of hyperbola to bring it to standard form,
we have, 11(x²-4x)-25(y²-2y)-256 = 0
11(x− 2)²−25(y−1)² = 256−44+25
11(x−2 )²− 25 (y−1)² = 275
\(\frac { { \left( x-2 \right) }^{ 2 } }{ 25 } -\frac { { \left( y-1 \right) }^{ 2 } }{ 11 } =1\)
Centre (2, 1) a² = 25, b² = 11
c² = a² +b²
= 25 +11 = 36
Therefore, c = ±6
and e = \(\frac { c }{ a } =\frac { 6 }{ 5 } \)and the coordinates of foci are(8, 1) and(-4, 1) from figure.

Given cubic equation is 2x³-6x²+3x+k = 0
Here, a = 2, b = -6, c = 3, d = k
Let ∝, β, ૪ be the roots
Given ∝ = 2(β+૪) ⇒ \(\frac{\alpha}{2}\) = β+૪ ...(1)
Now, \(\alpha +\beta +\gamma =\frac { -b }{ a } =-\frac { (-6) }{ 2 } =3\)
\(\frac { \alpha }{ 2 } +\alpha =3\Rightarrow \frac { \alpha +2\alpha }{ 2 } =3\Rightarrow \frac { 3\alpha }{ 2 } =3\)
\(\Rightarrow \alpha =2\)
\(\alpha \beta \gamma =\frac { -d }{ a } =\frac { -k }{ 2 } \Rightarrow 2.\beta \gamma =\frac { -k }{ 2 } \)
\(\beta \gamma =\frac { -k }{ 4 } ...(2)\)
Also, \(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \) 
\(2\beta +\beta \gamma +2\gamma =\frac { 3 }{ 2 } \)
\(2(\beta +\gamma )+\beta \gamma =\frac { 3 }{ 2 } \)
\(\alpha \frac { -k }{ 4 } =\frac { 3 }{ 2 } \quad [from(1)\& (2)]\)
Also, \(2-\frac { k }{ 4 } =\frac { 3 }{ 2 } [\because \alpha =2]\)
\(2-\frac { 3 }{ 2 } =\frac { k }{ 4 } \Rightarrow \frac { 1 }{ 2 } =\frac { k }{ 4 } \)
\(\\ k=\frac { 4 }{ 2 } \Rightarrow k=2\)
From(2), \(\beta \gamma =\frac { -k }{ 4 } =\frac { -2 }{ 4 } =\frac { -1 }{ 2 } \)\(\Rightarrow \gamma =\frac { -1 }{ 2\beta } \)
From \((1),\beta +\gamma =\frac { \alpha }{ 2 } =\frac { 2 }{ 2 } =1\)
Substituting \(\gamma =\frac { -1 }{ 2\beta } \) We get
\(\beta -\frac { 1 }{ 2\beta } =1\Rightarrow 2{ \beta }^{ 2 }-1=2\beta \Rightarrow 2\beta -2\beta -1=0\)
\(\beta =\frac { 2\pm \sqrt { 4-4(2)(-1) } }{ 4 } =\frac { 2\pm \sqrt { 4+8 } }{ 4 } \)
\(=\frac { 2\pm \sqrt { 12 } }{ 4 } =\frac { 2\pm 2\sqrt { 3 } }{ 4 } \)
\(\beta =\frac { 1\pm \sqrt { 3 } }{ 2 } \)
Hence the roots are \(2,\frac { 1+\sqrt { 3 } }{ 2 } ,\frac { 1-\sqrt { 3 } }{ 2 } \) and k = 2

The given equation is same as
(2x-3)(3x-2)(6x-1)(x-2)-5 = 0
After a computation, the above equation becomes
(6x²-13x+6)(6x²-13x+12)-5 = 0
By taking y = 6x²-13x, the above equation becomes
(y+6)(y+12)-5 = 0
which is same as
y²+18y+7 = 0
Solving this equation, we get y = −1 and y = −7.
Substituting the values of y in y = −6x²-13x, we get
6x²-13x+1 = 0
6x²-13x+7 = 0
Solving these two equations, we get
x = 1, x = \(\frac { 7 }{ 6 } \), x = \(\frac { 13 + \sqrt { 145 } }{ 12 } \) and x = \(\frac { 13-\sqrt { 145 } }{ 12 } \)
as the roots of the given equation.

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.