\(x\frac { dy }{ dx } =ylogy\)

order 2;degree not defined

\(\frac { 13 }{ 3 } \)

\(\frac { 1 }{ 2 } \)

\(\frac { 1 }{ 6 } \left[ log\left( \frac { 35 }{ 8 } \right) \right] \)

concave downward everywhere

(1, 0)

Given u(x, y) = x² + 3xy + y²
u(x_o, y_o) = u(2,1)
= 2² + 3(2)(1) + 1²
= 4 + 6 + 1 = 11
\(\frac { \partial u }{ \partial x } \) = 2x+ 3y
\({ \left( \frac { \partial u }{ \partial x } \right) }_{ (2,1) }\)= 2 + 3 = 5
\(\frac { \partial u }{ \partial y } \) = 3x+ 2y
\({ \left( \frac { \partial u }{ \partial y } \right) }_{ (2,1) }\) = 6 + 2 = 8
Linear approximation 
L(x,y) = U(x_o, y_o) + \({ \left( \frac { \partial u }{ \partial x } \right) }_{ ({ x }_{ 0 },{ y }_{ 0 }) }\) (x - x_o) + \({ \left( \frac { \partial u }{ \partial y} \right) }_{ ({ x }_{ 0 }{ ,y }_{ 0 }) }\)(y - y_o)
L (x,y) = 11 + 5 (x - 2) + 8 (y - 1)
= 11 + 5x - 10 + 8y - 8
L(x,y) = 5x + 8y - 7

Given * on A is commutative
Given b * a = c ⇒ a * b = c
Given c * a = a ⇒ a * c = a
Given b * c = a ⇒ c * b = a
Hence

Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

In the given differential equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 1.
Therefore, the given differential equation is of order 2.
The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

We are given that, T = 10x − x²
Hence, the rate of change at any distance from one end is given by \(\frac{dT}{dx}=10-2x \) 
The mid point of the rod is at x = 5
Substituting x = 5, we get \(\frac{dT}{dx}=0\)

y''− 5y' + 6y = 0 ......(1)
Given y = e^mx  .....(2)
Differentiating cquation (2) w.r.t 'x', we get
\(\frac{dy}{dx} = em^x . m\)
To find the value of m:
Given y" - 5y' + 6y = 0 
e^mx . m² -5e^mx+ 6e^mx = 0
e^mx [m- 5m +6] = 0
m - 5m + 6 = 0
(m - 3) (m - 2) = 0
m = 3, 2

Let z = -2
z = 2(-1) = 2(cos π + i sin π)
∴ arg(z) = π

(cos θ + i sin θ )² = cos 2θ + isin 2θ
[By De moivre's theorem]
⇒ cos 2θ + isin 2θ = x + iy
Equating the real and imaginary parts we get,
x = cos 2θ, y = sin 2θ
∴ x² + y² = cos²2θ + sin²2θ = 1
Hence proved

Since the vertices are (0, ±7), equation of the hyperbola is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
a = 7 and e = \(\frac { 4 }{ 3 } \)
b² = a²(e² - 1) = 49\(\left( \frac { 16 }{ 9 } -1 \right) =49\left( \frac { 16-9 }{ 9 } \right) \)
= \(49\left( \frac { 7 }{ 9 } \right) =\frac { 343 }{ 9 } \)
∴ Equation of the hyperbola is \(\frac { { y }^{ 2 } }{ 49 } -\frac { { x }^{ 2 } }{ \frac { 343 }{ 9 } } =1\) 
⇒ \(\frac { { y }^{ 2 } }{ 49 } -\frac { 9{ x }^{ 2 } }{ 343 } =1\)

Let P(x, y) be the movable point.
By focal property of ellipse, PA + PB = 2a
∴ 2a = 10 ⇒ a = 5
Since focus is (4, 0), ae = 4 ⇒ 5e = 4 ⇒ e = \(\frac45\)
Also b² = a²(1 - e²) = 25\(\left( 1-\frac { 16 }{ 25 } \right) =25\left( \frac { 9 }{ 25 } \right) \) = 9
Equation of ellipse is \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\)
\(\frac { { x }^{ 2 } }{ 25 } +\frac { { y }^{ 2 } }{ 9 } =1\)

Given
\({ cot }^{ -1 }\left( \frac { 1 }{ 7 } \right) =0\Rightarrow \theta =\frac { 1 }{ 7 } \)
\(\Rightarrow tan\theta =7\)
\(\Rightarrow sec\theta =\sqrt { 1+{ tan }^{ 2 }\theta } =\sqrt { 1+{ 7 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } \)
\(\Rightarrow cos\theta ={ \frac { 1 }{ 5\sqrt { 2 } } }\)

Let \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =y\) where \(\frac{-\pi}{2}<y<\frac{\pi}{2}\)
\(\Rightarrow tan\ y=\frac { -1 }{ 3 } =-tan\frac { \pi }{ 6 } =tan\left( \frac { -\pi }{ 6 } \right) \)
\(y=\frac { -\pi }{ 6 } \) \(\left[ \because \frac { -\pi }{ 6 } \epsilon \left( \frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } \right) \right] \)
The principal value of \({ tan }^{ -1 }\left( \frac { -1 }{ \sqrt { 3 } } \right) =6\)

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

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

The matrix form of the system is
\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where
A=\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
|A| =\(\left| \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right| =3\left| \begin{matrix} 2 & 12 \\ 1 & 7 \end{matrix} \right| -1\left| \begin{matrix} 3 & 12 \\ 2 & 7 \end{matrix} \right| +9\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| \)
= 3 (14 - 12) - 1 (21 -24) + 9 (3 -4)
= 3 (2) -1 (-3) + 9 (-1)
= 6 + 3 - 9 = 9 - 9 = 0
Since |A| = 0, the homogeneous system of equations have non-trivial solutions also.

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)

\(\overset { \rightarrow }{ OA } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) , \(\overset { \rightarrow }{ OB } =\overset { \wedge }{ i } -\overset { \wedge }{ j } -3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ OC } =4\overset { \wedge }{ i } -3\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
Area of △ ABC = \(\frac { 1 }{ 2 } \left| \overset { \rightarrow }{ AB } \times \overset { \rightarrow }{ AC } \right| \)
\(\overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA } =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } -3\overset { \wedge }{ k } \right) -\left( 3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =-2\overset { \wedge }{ i } -5\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ AC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OA } =\left( 4\overset { \wedge }{ i } -3\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) -\left( 3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\overset { \wedge }{ i } -2\overset { \wedge }{ j } -\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ AB } \overset { \rightarrow }{ \times AC } =\left| \begin{matrix} \overset { \wedge }{ i } \\ -2 \\ 1 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ 0 \\ -2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ -5 \\ -1 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } \)(0-10) - \(\overset { \wedge }{ j } \) (2+5) + \(\overset { \wedge }{ k } \) (4-0)
\(=10\overset { \wedge }{ i } -7\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ AB } \overset { \rightarrow }{ \times AC } \right| =\sqrt { 101+49+16 } =\sqrt { 165 } \)
∴ Area of Δ ABC \(=\frac { 1 }{ 2 } \sqrt { 165 } \) sq. units 

Sum of the roots = sin ∝ + cos ∝ = \(\frac{-b}{a}\)
Product of the roots = sin ∝ cos ∝ = \(\frac{c}{a}\)
Now 1 = cos²∝ + sin² ∝
= (sin ∝ +cos ∝)² - 2 sin ∝ cos ∝
\(1=\frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } \Rightarrow 1=\frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \)
⇒ a² = b² - 2ac ⇒ a² + 2ac = b²
Adding c² both sides, a² +2ac+c² = b²+c²
⇒ (a+c)² = b² + c²

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

We know that the volume of the parallelepiped whose coterminus edges are \(\vec { a } ,\vec { b } ,\vec { c } \) is given by |\([\vec { a } ,\vec { b } ,\vec { c } ]\)|. Here, \(\vec { a } =\hat { 2i } -\hat { 3j } +\hat { 4k } ,\vec { b } =\hat { i } +\hat { 2j } -\hat { k } ,\vec { c } =\hat { 3i } -\hat { j } +\hat { 2k } \)
Since \([\vec { a } ,\vec { b } ,\vec { c } ]\)  = \(\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| =-7\) , the volume of the given parallelepiped is \(\left| -7 \right| =7\) cubic units.

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

Reducing 9x²-16y² = 144 to the standard form,
we have, \(\frac { { x }^{ 2 } }{ 16 }- \frac { { y }^{ 2 } }{ 9 } =1\)
With the transverse axis is along x-axis vertices are (−4, 0) and (4, 0); and c² = a²+b² = 16 + 9 = 25, c = 5  
Hence the foci are (−5, 0) and (5, 0) 

\(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
Let A = \(\left[ \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} -2 & 4 \\ 1 & -3 \end{matrix} \right| \)= 6 - 4 = 2 ≠ 0
Since A is nonsingular, A^-1 exists
A^-1 = \(\frac { 1 }{ |A| } \)
Now, adj A = \(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \)
[Interchange the entries in leading diagonal and change the sign of elements in the off diagonal]
∴ A^-1 = \(\frac{1}{2}\)\(\left[ \begin{matrix} -3 & -4 \\ -1 & -2 \end{matrix} \right] \).

cos^-1(-x) = \(\pi\)-cos⁻¹(x)
Let cos^-1(-x) = \(\theta \)   ..(1)
\(\Rightarrow -x=cos\theta \)
\(\Rightarrow x=-cos\theta =cos\theta =cos\left( \pi -\theta \right) \)
\(\Rightarrow \pi -\theta ={ cos }^{ -1 }\left( x \right) \)
\(\Rightarrow \theta =\pi -{ cos }^{ -1 }x\) ...(2)
From (1) & (2) \({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \)
\({ cos }^{ -1 }\left( -x \right) =\pi -{ cos }^{ -1 }\left( x \right) \) is true.

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

Since the domain of y = sin^-1 is −[11], and 2\(\notin \)[-1, 1], sin⁻¹(2) does not exist.

Equation of the circle with end points of the diameter as (x₁, y₁₎ and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x² + y² + 3x + y − 6 = 0 which is the required equation of the circle.

Equation of the given curve is y = e^-2x
Required Volume = \(\pi \int _{ 0 }^{ 1 }{ { { (e }^{ -2x }) }^{ 2 }dx } \)
\(=\pi \int _{ 0 }^{ 1 }{ { e }^{ -4x } } dx=\pi { \left[ \frac { { e }^{ -4x } }{ -4 } \right] }_{ 0 }^{ 1 }\)
\(=\frac { -\pi }{ 4 } \left[ { e }^{ -4 }-{ e }^{ -0 } \right] =-\frac { \pi }{ 4 } \left( { e }^{ -4 }-1 \right) \)
\(V=\frac { \pi }{ 4 } (1-{ e }^{ -4 })\) cubic units

Given U(x, y, z) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ xy } +3{ z }^{ 2 }y\)
\(\frac { \partial U }{ \partial x } =\frac { xy(2x)-({ x }^{ 2 }+{ y }^{ 2 })(y) }{ { x }^{ 2 }{ y }^{ 2 } } +0\)
\(=\frac { 2{ x }^{ 2 }y-{ x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }y-{ y }^{ 3 } }{ { x }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { y({ x }^{ 2 }-{ y }^{ 2 }) }{ { x }^{ 2 }{ y }^{ 2 } } =\frac { { x }^{ 2 }-{ y }^{ 2 } }{ { x }^{ 2 }y } \)
\(\frac { \partial U }{ \partial y } =\frac { xy(2y)-({ x }^{ 2 }+{ y }^{ 2 })(x) }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
= \(\frac { { 2xy }^{ 2 }-{ x }^{ 3 }-{ xy }^{ 2 } }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { xy }^{ 2 }-{ x }^{ 3} }{ { x }^{ 2 }{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(=\frac { { y }^{ 2 }-{ x }^{ 2 } }{ x{ y }^{ 2 } } +3{ z }^{ 2 }\)
\(\frac { \partial U }{ \partial z } =0+3y(2z)=6yz\)

Let X be the random variable denotes number of| girl child among 4 children
X = {0, 1, 2, 3, 4}
X =2) (0) {BBBB}
X(1) = {GBBB, BGBB, BBGB, BBBG}
X(2) = {GGBB, BBGG, GBGB, BGBG, BGGB, GBBG}
X(3) = {BGGG, GGGB, GBGG, GGBG}
X(4) = {GGGG}

(i) Probability mass function

(ii) Cumulative distribution function
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)P(X = x_i)
P(X<0) = 0 for -\(\infty\) < x < 0
\(F(0)=\frac { 1 }{ 16 } \) 
\(F(1)=\frac { 1 }{ 16 } +\frac { 1 }{ 4 } =\frac { 5 }{ 16 } \) 
\(F(2)=\frac { 5 }{ 16 } +\frac { 3 }{ 8 } =\frac { 5 }{ 16 } +\frac { 6 }{ 16 } =\frac { 11 }{ 16 } \)
\(F(3)=\frac { 11 }{ 6 } +\frac { 1 }{ 4 } =\frac { 11 }{ 16 } +\frac { 4 }{ 16 } =\frac { 15 }{ 16 } \) 
\(F(4)=\frac { 15 }{ 16 } +\frac { 1 }{ 16 } =\frac { 16 }{ 16 } =1\)
\(F(x)=\left\{\begin{array}{lll} \frac{0}{16} & \text { for } & x<0 \\ \frac{1}{16} & \text { for } & x \leq 0 \\ \frac{5}{16} & \text { for } & x \leq 1 \\ \frac{11}{16} & \text { for } & x \leq 2 \\ \frac{15}{16} & \text { for } & x \leq 3 \\ 1 & \text { for } & x \leq 4 \end{array}\right.\)

Rearrange the terms as
(2x-1)(2x+ 3) (x + 3) (x - 2) + 20 = 0
⇒ (4x² + 6x - 2x- 3)(x² - 2x + 3x - 6) + 20 = 0
⇒ (4x² + 4x - 3) (x² + x - 6) + 20 = 0
put x²+ x = y
⇒ (4y - 3) (y - 6) + 20 = 0
⇒ 4y² - 24y - 3y + 18 + 20 = 0
⇒ 4y²-27y +38 = 0
⇒ (y - 2)( 4y - 19) = 0
\(y=2,\frac { 19 }{ 4 } \)

Case (i) 
When y = 2
x²+ x = 2
x² + x - 2 = 0
⇒ (x + 2)(x - 1) = 0
⇒ x = -2, 1
Case (ii) 
When \(y=\frac { 19 }{ 4 } ,{ x }^{ 2 }+x=\frac { 19 }{ 4 } \)
\(\Rightarrow { 4x }^{ 2 }+4x=19\)
\(\Rightarrow { 4x }^{ 2 }-4x-19=0\)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16-4(4)(-19) } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 16+304 } }{ 8 } \)
\(\Rightarrow x=\frac { -4\pm \sqrt { 320 } }{ 8 } \)
\(\Rightarrow x=\frac { 4\pm 8\sqrt { 5 } }{ 8 } \)
\(\Rightarrow x=\frac { -4(-1\pm 2\sqrt { 5 } ) }{ 8 } \)
\(\frac{-1 \pm 2 \sqrt{5}}{2}\)
Hence the roots are 1, -2, \(\frac{-1 \pm 2 \sqrt{5}}{2}\)

Let f, g : (a, b)→R be differentiable functions and h(x) = f (x)g(x).
Then h being product differentiable functions, is differentiable on (a,b) 
So by definition dh = h'(x)dx.
Now by using product rule we have h'(x) = f (x)g'(x) + f'(x)g(x).
Thus dh = h'(x)dx = ( f (x)g'(x) + f''(x)g(x))dx = f (x)g'(x)dx + f '(x)g(x) dx
= f (x)dg + g(x)df = fdg + gdf

Let X be the random variable of getting apples Given 5 mangoes and 4 apples are in an urn 
= {0, 1,2,3}
The sample space consists of ⁹C₃ = 84
X = 0, X (3 mangoes) = ⁵C₃ = 10
X = 1, X (2 mangoes and 1 apples) = ⁵C₂ x ⁴C₁ = 40
X = 2, X (1 mangoes and 2 apples) = ⁵C₁ x ⁴C₂ = 30
X = 3, X (apples) = ⁴C₃ = 4

We have,
\(f'(x)=2x>0, \forall x\in(2,7)\) and
\(f'(x)=2x>0, \forall x\in(-2,0)\)
and hence the proof is completed.

Clearly there are 2 sign changes for the given polynomial P(x) and hence number of positive roots of P(x) cannot be more than two. Further, as P(-x) = -9x⁹- 2x⁵- x⁴- 7x²+ 2, there is one sign change for P(-x) and hence the number of negative roots cannot be more than one. Clearly 0 is not a root. So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.