\(\begin{matrix} lim \\ (x,y)\rightarrow (0,0) \end{matrix}cos=\left( \frac { { e }^{ x }siny }{ y } \right) \) = \(cos\left( { e }^{ 0 }\frac { siny }{ y } \right) \)
= cos[(1)(1)] = cos (1) \(\left[ \because \begin{matrix} lim \\ y\rightarrow 0 \end{matrix}\frac { siny }{ y } =1 \right] \)

Take 2a = 2\(\pi\) and f(x) = g(cosx)
Then, f (2a−x) = f(2\(\pi\)-x) = g(cos(2\(\pi\)-x)) = g(cos x) = f(x)
\(\therefore \int _{ 0 }^{ 2a }{ f(x)dx=2 } \int _{ 0 }^{ a }{ f(x)dx } \)
\(\therefore \int _{ 0 }^{ 2\pi }{ g(cosx)dx=2\int _{ 0 }^{ \pi }{ g(cosx)dx } } \)

(i) At the maximum height, the velocity v(t) of the particle is zero.
Now, we find the velocity of the particle at time t.
\(v(t)=\frac{ds}{dt}=128-32t\)
\(v(t)=0 \Rightarrow 128-32t=0 \Rightarrow t=4.\)
After 4 seconds, the particle reaches the maximum height.
The height at t = 4 is s(4) = 128(4) - 16(4)² = 256 ft.
(ii) When the particle hits the ground then s = 0 .
s = 0 ⇒ 128t −16t² = 0
⇒ t = 0, 8 seconds.
The particle hits the ground at t = 8 seconds. The velocity when it hits the ground v(8) = –128 ft /s.

z²+ 2zw +w²
(5-2i)²+2(5-2i)(-1+3i)+(-1+3i)²
= 25 + 4i²- 20i + 2 [-5 + 15i + 2i - 6i²] + 1 + 9i²-6i [∴ i² = -1]
= 25 - 4 - 20i + 2(-5 + 17i + 6) +1-9 -6i
= 21- 20i + 2(1+17i) -8 -6i
= 21- 20i + 2 + 34i -8 -6i
= 15 + 8i

\(\overline { (2-i)^{ 12 }+(2+i)^{ 12 } } =\overline { (2-i)^{ 12 } } +\overline { (2+i)^{ 12 } } \)
\(\left( 2+i\sqrt { 3 } \right) ^{ 10 }-\left( 2-i\sqrt { 3 } \right) ^{ 10 }\)
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

Prove the following properties
Z is real if and only if z =\(\bar { z } \)
Let z = x+iy
Then \(\bar { z } \) = x+iy
z = \(\bar { z } \)
⇔ x + iy = x-iy
⇔ x + iy - x + iy = 0
⇔ 2iy = 0
⇔ y = 0
[∴ 2 and i are constants]
when y = 0, z = x which is real
∴ z is purely ⇔ z = \(\bar { z } \)

Since α and β are the roots of 17x²+ 43x −73 = 0 , we have α + β =\(\frac { -43 }{ 17 } \) and  αβ =\(\frac { -73 }{ 17 } \).
We wish to construct a quadratic equation with roots α + 2 and  β + 2. Thus, to construct such a quadratic equation, calculate
the sum of the roots =  α + β + 4 = \(\frac { -4 }{ 17 } +4=\frac { 25 }{ 17 } \) and
the product of the roots = αβ + 2(α+β)+4 = \(\frac { -73 }{ 17 } +2\left( \frac { -43 }{ 17 } \right) +4=\frac { -91 }{ 17 } \)
Hence a quadratic equation with required roots is x²-\(\frac { 25 }{ 17 } x-\frac { 91 }{ 17 } \) = 0
Multiplying this equation by 17, gives 17x²−25x−91 = 0
which is also a quadratic equation having roots α + 2 and β + 2

This is an indeterminate of the form \((\frac{\infty}{\infty})\)
To evaluate this limit, we apply l’Hôpital Rule m times
\(\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{x^{m}}=\underset{x\rightarrow \infty}{lim}\frac{e^{x}}{m!} = \infty\) 

Given m (x) = \(\sqrt { 3 } x\) = \(\sqrt { 3 } .{ x }^{ \frac { 1 }{ 2 } }\)
Differentiating with respect to 'x' we get,

when x = 3, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 3 } } =\frac { 1 }{ 2 } \) Kg/m
when x = 27, \(\frac { dm }{ dx } =\frac { \sqrt { 3 } }{ 2\sqrt { 27 } } \)
\(=\frac{\sqrt{\not 3}}{2(3) \sqrt{\not 3}}=\frac{1}{6} \mathrm{Kg} / \mathrm{m}\)

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

From figure the midpoint of line joining foci is the centre C(0, 0).
Transverse axis is y-axis 
AA′ = 2a \(\Rightarrow \) 2a = 8,
SS′ = 2c = 12, c = 6
a = 4
b² = c²−a² = 36−16 = 20
Hence the equation of the required hyperbola is \(\frac { { y }^{ 2 } }{ 16 }- \frac { { x }^{ 2 } }{ 20 } =1\)

Parabola is open left and axis of symmetry as x-axis and vertex (0, 0)  
Then the equation of the required parabola is
(y - 0)² = -4\(\sqrt { 2 } \) (x - 0)
y² = -4\(\sqrt { 2 } \) x

By definition, the domain of yx = cos^-1 x is -1. This leads to \(-1\le\frac{2+sinx}{3}\le1\) which is same as -3\(\le\)2+sinx\(\le\)3
so, -5\(\le sin\ x\le1 \) reduces to -1\(\le sin\ x\le1 \), which gives
-sin^-1(1)\(\le x\le sin^-1(1) or -\frac{\pi}{2}\le x\le \frac{\pi}{2}\)
Thus, the domain of cos-1\((\frac{2+sin\ x}{3}) is [-\frac{\pi}{2},\frac{\pi}{2}].\)

Given ∝, β and ૪ are the roots of ax³ + bx² + cx + d = 0
\(\therefore \alpha +\beta +\gamma =\frac { -b }{ a } \)
\(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\(\alpha \beta \gamma =\frac { -d }{ a } \)
Now, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)

C(2, 1) is the centre and 3x + 4y + 10 = 0 cuts a chord AB on the circle. Let M be the midpoint of AB,
then AM = BM = 3. Now BMC is a right triangle.
So, we have CM = \(\frac { \left| 3\left( 2 \right) +4\left( 1 \right) +10 \right| }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 } } } =4\)
By Pythagoras theorem BC² = BM² + MC² = 3² + 4² = 25
BC = 5 = radius
Equation of the required circle is
(x−2)²+(y−1) = 5²
x²+y²−4x−2y−20 = 0 .

With usual notations in triangle ABC, we have \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \) \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Then applying dot product, we get
\(\vec { BC } .\vec { BC } =(-\vec { CA } -\vec { AB } ).(-\vec { CA } -\vec { AB } )\)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }={ \left| \vec { CA } \right| }^{ 2 }+{ \left| \vec { AB } \right| }^{ 2 }+\vec { 2CA } .\vec { AB } \)
⇒ a² = b²+c²+2bc cos (\(\pi\) - A)
⇒ a² = b²+c²−2bc cos A.
The results (ii) and (iii) are proved in a similar way.

We know that the domain of sin⁻¹(x) is [-1, 1].
This leads to −1\(\le\)2 - 3x²\(\le\)1, Which implies -3\(\le\) -3x²\(\le\)-1
Now, -3\(\le\) -3x², gives x²\(\le\)1 and ............(1)
-3\(\le\)-3x²\(\le\)-1, gives x²\(\ge\)\(\frac{1}{2}\) .........(2)
Combining the equations (1) and (2), we get \(\frac{1}{3}\le x^2\le 1\). That is \(\frac{1}{\sqrt3}\le |x|\le1\), Which gives \(x\in[-1,-1\frac{1}{\sqrt3}]\cup[\frac{1}{\sqrt3},1]\)
since  a\(\le|x|\le b\) implies x \(\in[-b,-a]\cup[a,b]\).

Let A = \(\left[ \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right] \). Then |A| = \(\left| \begin{matrix} 2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3 \end{matrix} \right| \) = 2(7) + (-12) + 3(-1) = -1 ≠ 0.
Therefore, A⁻¹ exists. Now, we get
adj A = \({ \left[ \begin{matrix} +\left| \begin{matrix} 3 & 1 \\ 2 & 3 \end{matrix} \right| & -\left| \begin{matrix} -5 & 1 \\ -3 & 3 \end{matrix} \right| & +\left| \begin{matrix} -5 & 3 \\ -3 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} -1 & 3 \\ 2 & 3 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ -3 & 3 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ -3 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} -1 & 3 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ -5 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ -5 & 3 \end{matrix} \right| \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} 7 & 12 & -1 \\ 9 & 15 & -1 \\ -10 & -17 & 1 \end{matrix} \right] }^{ T }=\left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] \).
Hence, A^-1 = \(\frac { 1 }{ \left| A \right| } \)(adj A) = \(\frac { 1 }{ \left( -1 \right) } \left[ \begin{matrix} 7 & 9 & -10 \\ 12 & 15 & -17 \\ -1 & -1 & 1 \end{matrix} \right] =\left[ \begin{matrix} -7 & -9 & 10 \\ -12 & -15 & 17 \\ 1 & 1 & -1 \end{matrix} \right] \).

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

Given f(0) = -1, f(2) = 4
∴ f(x) is a continuous function in [0,2]
By Lagrange's mean value theorem,
f'(x) = \(\frac { f(b)-f(a) }{ b-a } \) = \(\frac{f(2)-f(0)}{2-0}\)
= \(\frac{4-(-1)}{2}\) = \(\frac{5}{2}\) = 2.5
= 2.5 ∉ [0, 2]
Since f'(x) cannot be 2.5 at any point in [0, 2], there does not exist a differentiable function f(x).

Given \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =-\hat { i } +2\hat { j } -4\hat { k } ,\vec { c } =\hat { i } +\hat { j } +\hat { k } \)
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ 2 & -4 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ -1 & -4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ -1 & 2 \end{matrix} \right| \)
= \(\hat { i } (-12+2)-\hat { j } (-8-1)+\hat { k } (4+3)\)
= \(-10\hat { i } +9\hat { j } +7\hat { k } \)
\(\vec { a } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= \(\hat { i } (3+1)-\hat { j } (2+1)+\hat { k } (2-3)\)
= \(4\hat { i } -3\hat { j } -\hat { k } \)
∴ \((\vec { a } \times \vec { b } ).(\vec { a } \times \vec { c } )=(-10\hat { i } +9\hat { j } +7\hat { k } ).(4\hat { i } -3\hat { j } -\hat { k } )\)
= -40-27-7 = -74

Equation of parabola is x² + 6x + 4y + 5 = 0.
x² + 6x + 9 − 9 + 4y + 5 = 0
(x + 3)² = -4(y-1) ... (1)
Let X = x + 3, Y = y −1
Equation (1) takes the standard form
X² = -4Y
Equation of tangent is XX₁ = −2(Y+ Y₁)
At (1, -3) X₁ = 1 + 3 = 4; y₁ = -3 - 1 = -4
Therefore, the equation of tangent at (1,-3) is
(x +3)4 = -2(y-1- 4)
2x + 6 = −y + 5.
2x + y + 1 = 0
Slope of tangent at(1, -3) is -2, so slope of normal at (1, -3) is \(\frac { 1 }{ 2 } \)
Therefore, the equation of normal at (1, -3) is given by y + 3 = \(\frac { 1 }{ 2 } \)(x-1)
2y + 6 = x -1
x-2y-7 = 0 .

Let the position vector of the vertices of ΔABC be  \(\vec { a } \), \(\vec { b } \) and  \(\vec { c } \) respectively.
Since G is the centroid of ΔABC, \(\vec { OG } =\frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } \)
Are of ΔGAB, = \(|\vec { AB } \times \vec { AG } |=|(\vec { OB } -\vec { OA } )\times (\vec { OG } -\vec { OA } )|\)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } -\vec { a } \right) \right| \)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -3\vec { a } }{ 3 } \right) \right| \)
= \(\left| (\vec { b } -\vec { a } )\times \left( \frac { \vec { b } +\vec { c } +2\vec { a } }{ 3 } \right) \right| \)
= \(\frac { 1 }{ 3 } |(\vec { b } -\vec { a } )\times (\vec { a } +\vec { 0 } -2\vec { a } )|\)
= \(\frac { 1 }{ 3 } |\vec { b } \times \vec { b } +\vec { b } \times \vec { c } -2\vec { b } \times \vec { a } -\vec { a } \times \vec { b } -\vec { a } \times \vec { c } +2\vec { a } \times \vec { a } |\)
[∵ cross product is distributive]
= \(\frac { 1 }{ 3 } |\vec { b } \times \vec { c } +2\vec { a } \times \vec { b } -\vec { a } \times \vec { b } +\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)  \( [\because \vec { b } \times \vec { b } =\vec { 0 } ,\vec { a } \times \vec { a } =\vec { 0 } ,.(1)\vec { a } \times \vec { b } =-\vec { b } \times \vec { a } ]\)
Area of ΔGAC = \(|\vec { CA } \times \vec { AG } |\)
= \(|(\vec { OA } -\vec { OC } )\times (\vec { OG } -\vec { OA } )|\)
\(\left| (\vec { a } -\vec { c } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } }{ 3 } -\vec { a } \right) \right| \)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -3\vec { a } }{ 3 } \right) \right| \)
= \(\frac { 1 }{ 3 } |(\vec { a } -\vec { c } )\times (\vec { b } +\vec { c } -2\vec { a } )|\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { a } \times \vec { c } -2\vec { a } \times \vec { a } -\vec { c } \times \vec { b } -\vec { c } \times \vec { c } +2\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } -\vec { c } \times \vec { a } +\vec { b } \times \vec { c } +2\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)
Also area of ΔGBC = \(|\vec { BC } \times \vec { BG } |\)
= \(|\vec { OC } -\vec { OB } )\times (\vec { OG } -\vec { OB } )|\)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } +\vec { c } -\vec { b } }{ 3 } \right) \right| \)
= \(\left| (\vec { c } -\vec { a } )\times \left( \frac { \vec { a } +\vec { b } -2\vec { a } }{ 3 } \right) \right| \)
\(\frac { 1 }{ 3 } |(\vec { c } -\vec { b } )\times (\vec { a } +\vec { c } -2\vec { b } )|\) 
= \(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +\vec { c } \times \vec { c } -2\vec { c } \times \vec { b } -\vec { b } \times \vec { a } -\vec { b } \times \vec { c } +2\vec { b } \times \vec { b } |\)
= \(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +2\vec { b } \times \vec { c } +\vec { a } \times \vec { b } -\vec { b } \times \vec { c } |\)
\(\frac { 1 }{ 3 } |\vec { c } \times \vec { a } +\vec { b } \times \vec { c } +\vec { a } \times \vec { b } |\)
From (1), (2) and (3),
Area of ΔGAB = Area of ΔGAC = Area of ΔGBC
= \(\frac { 1 }{ 3 } |\vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } |\)
= \(\frac { 1 }{ 3 } \) Area of ΔABC.

Given A =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
|A| = 0-1\(\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \)
= -1(0-1) + 1(1-0) = 1 + 1 = 2
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} (0-1) & -(0-1) & +(1-0) \\ -(0-1) & +(0-1) & -(0-1) \\ +(1+0) & -(0-1) & +(0-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) ................(1)
Now A² =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] \)
A²- 3I =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) .............(2)
From (1) and (2), it is proved that A^-1 = \(\frac{1}{2}\) [A² - 3I]

Given equation are \({ x }^{ 2 }px+q=0\)  ............(1)
and \({ x }^{ 2 }+p'x+q'=0\)    ..........(2)
Let ∝ be the common root for (1) and (2)
∴ ∝² + p∝ + q = 0   .........(3)
and ∝²+ p'∝ + q' = 0    .............(4)
Solving (3) and (4) by cross multiplication method we get
p  q 1 p
p' q' 1 p'
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \)
consider \(\frac { { \alpha }^{ 2 } }{ pq'-p'q } =\frac { \alpha }{ q-q' } \)
\(\Rightarrow \frac { { \alpha }^{ 2 } }{ \alpha } =\frac { pq'-p'q }{ q-q' } \)
\(\alpha =\frac { pq'-p'q }{ q-q } \)
Consider \(\frac { \alpha }{ q-q' } =\frac { 1 }{ p'-p } \Rightarrow \alpha =\frac { q-q' }{ p'-p } \)
Hence its roots are \(\frac { pq'-p'q }{ q-q' } or\quad \frac { q-q' }{ p'-p } \)