n = 500
P = proportion of bad apples
\(=\frac{45}{500}=0.09\)
q = 1 - p = 1 - 0.09 = 0.91
Confidence limits for the population p of bad apples are given by
\(=p \pm Z_{c} \sqrt{\frac{p q}{n}}\)
\(=0.09 \pm(2.58) \sqrt{\frac{(0.09)(0.91)}{500}}\)
\(=0.09 \pm 0.033\)
Required interval is (0.057, 0.123) Thus the bad apples in the consignment lie between 5.7% and 12.3%

Let X be the random variable denoting the number of student who passes.
\( \mathrm{p}=\frac{60}{100}=\frac{3}{5}, \mathrm{q}=\frac{40}{100}=\frac{2}{5} \)
\(\mathrm{n}=6, \mathrm{P}(\mathrm{X}=\mathrm{x})=n C_{x} p^{x} q^{n-x} \)
\(\mathrm{P}^{\prime}(\mathrm{X} \geq 4)=\mathrm{P}(\mathrm{X}=4)+\mathrm{P}(\mathrm{X}=5)+\mathrm{P}(\mathrm{X}=6)\)
\( =6 C_{4}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)^{2}+6 C_{1}\left(\frac{3}{5}\right)^{4}\left(\frac{2}{5}\right)+\left(\frac{3}{5}\right)^{6} \)
\(=15 \frac{(81 \times 4)}{5^{6}}+\frac{(6)(243)(2)}{5^{6}}+\frac{729}{5^{6}} \)
\( =\frac{8505}{15625}\)

\(\because\) if is a p.d.f
\( \int_{-\infty}^{x} f(x) d x =1 \)
\(k \int_{0}^{1} x(1-x) d x =1 \)
\(k \int_{0}^{1}\left(x-x^{2}\right) d x =1 \)
\(k\left[\frac{x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{1} =1 \)
\(k\left(\frac{1}{2}-\frac{1}{3}\right) =1 \)
\(k\left(\frac{1}{6}\right)=1 \Rightarrow k=6\)
Cumulative distribution function
\( \mathrm{F}(x) =0 \text { for } x \geq 0 \)
\(\mathrm{F}(x) =\mathrm{P}(\mathrm{X} \leq x)=\int_{-1}^{1} f(x) d x \)
\( =\int_{0}^{x} 6 x(1-x) d x \)
\(\int_{0}^{x}\left(6 x-6 x^{2}\right) d x =\left[\frac{6 x^{2}}{2}-\frac{6 x^{3}}{3}\right]_{0}^{x}=3 x^{2}-2 x^{3} \)
\(\therefore \mathrm{F}(x) =0 \text { for } x<0 \)
\( =3 x^{2}-2 x^{3} \text { for } 0<x<1 \)
\( =1 \text { for } x \geq 1 \)

\( \left(x^{2}-\mathrm{y} x^{2}\right) \mathrm{dy} =-\left(\mathrm{y}^{2}+x^{2} \mathrm{y}^{2}\right) \mathrm{d} x \)
\(x^{2}(1-\mathrm{y}) \mathrm{dy} =-\mathrm{y}^{2}\left(1+x^{2}\right) \mathrm{d} x \)
\(\int \frac{1-y}{y^{2}} d y =-\int \frac{\left(1+x^{2}\right)}{x^{2}} d x \)
\(\int \frac{d y}{y^{2}}-\int \frac{y d y}{y^{2}} =-\int x^{-2} d x-\int d x \)
\(\int y^{-2} d y-\int \frac{d y}{y} =-\int x^{-2} d x-\int d x \)
\(\frac{y^{-1}}{-1}-\log |y| =\frac{x^{-1}}{1}-x+c \)
\(\log |y|+\frac{1}{y}+\frac{1}{x}-x =\mathrm{c} \)

\( \mathrm{p} =x^{2}+4 x+5 \)
\(\mathrm{p} =10 \)
\(x^{2}+4 x+5 =10 \)
\(x^{2}+4 x-5 =0 \)
\((x+5)(x-1) =0 \Rightarrow x=1 \)
\( \mathrm{PS} =p_{0} x_{0}-\int_{0}^{x_{1}} g(x) d x \)
\( =36-\int_{0}^{3}\left(3+x^{2}\right) d x=36-\left[3 x+\frac{x^{3}}{3}\right]_{0}^{3} \)
\(=36-[9+9]=36-18=18 \text { units }\)

\( \mathrm{t} =x^{4}, \mathrm{dt}=4 x^{3} \mathrm{~d} x \)
\(\frac{d t}{4} =x^{3} \mathrm{~d} x \)
\(\frac{1}{4} \int \sin t d t =\frac{-1}{4} \cos t+c \)
\( =-\frac{1}{4} \cos \left(\mathrm{x}^{4}\right)+c \)

Commodity	Base Year		Current Year
	p0	q0	p1	q1
A	10	12	12	15
B	7	15	5	20
C	5	24	9	20
D	16	5	14	5

 

p1q0	p0q0	p1q1	p0q1
144	120	180	150
75	105	100	140
216	120	180	100
70	80	70	80
505	425	530	470

Fisher's ideal index = \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{\sum p_{0}q_{0}\times \sum p_{0}q_{1}} \times 100\)
=   \(\sqrt{\frac{505}{425} \times \frac {430}{470}} \times 100\) 
\(P^{F}_{01}\)   = 115.75   

Since the random variable X is the probability distribution function, Σp_i = 1
∴ a + 2a + 3a + 4a + 5a + 6a = 1
21a = 1 ⇒ a = \(\frac{1}{21}\)
Now, F(4) = P(X ≤ 4)
= P(X = 0) + P(X = 1) + P(X = 2)P(X = 3) + P(X = 4)
= a + 2a + 3a + 4a + 5a = 15a
= 15\((\frac{1}{21})=\frac{5}{7}\)
∴ F(4) = \(\frac{5}{7}\)

By data, we have
x_o = 40, x₁ = 50, x₂ = 60, x₃ = 70
y_o = 31, y₁ = 73, y₂ = 124, y₃ = 159.
Using Lagrange's formula, we get
\(y={ y }_{ 0 }\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 }) } +{ y }_{ 1 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 }) } +{ y }_{ 3 }\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 }) } \)
∴ y(42) = 31\(\frac { (-8)(-18)(-28) }{ (-10)(-20)(-30) } +73\frac { (2)(-18)(-28) }{ (10)(-10)(-20) } +124\frac { (2)(-8)(-28) }{ (20)(10)(-10) } +59\frac { (2)(-8)(-28) }{ (30)(20)(10) } \)
= 20. 832 + 36. 792 - 27. 776 + 7.632
y = 37. 48

\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
\(\Rightarrow \left( \begin{matrix} x0+0 \\ 0+0+z \\ 0+y+0 \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow x=2\quad z=-1\quad y=3\)
\(\therefore\) Solution set is {2,3, -1}

Given \(\frac { a+{ x }^{ 2 } }{ 1+y } =xy\frac { dy }{ dx } \)
Separating the variables we get,
\(\frac { (1+{ x }^{ 2 })dx }{ x } \) = y(1+y)dy
⇒ \(\left( \frac { 1 }{ x } +\frac { { x }^{ 2 } }{ x } \right) \)dx = (y+y²)dy
⇒ \(\left( \frac { 1 }{ x } +x \right) \) = (y+y²)dy
Integrating both sides we get,
\(\int { \left( \frac { 1 }{ x } +x \right) dx } =\int { (y+y^{ 2 })dy } \)
⇒ logx + \(\frac { { x }^{ 2 } }{ 2 } =\frac { { y }^{ 2 } }{ 2 } +\frac { y^{ 3 } }{ 3 } \) + C.

Scale:
In x-axis 1 cm = 100 units
In y-axis 1 cm = 1000 units
Plot the points (200, 2500), (300, 2800) (400, 3100) (640; 3820), (540,3220) and (580, 3640).

At x = 350, draw a vertical line and from the intersecting point on the curve, draw a horizontal line.
From the graph, we find that when x = 350, y = 2900.
Hence, the expense at a level of 350 units is Rs. 2900

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

Task	Men	cost
P	1	9
Q	3	6
R	2	20
s	d	0
Total cost		35

The optimal assignment (minimum) cost = Rs. 35

Given transportation table is

Total Availability = Total Requirement 
Therefore the given problem is balanced transportation problem.
Hence there exists a feasible solution to the given problem.
First allocation :

Second allocation :

Third Allocation :

Fourth Allocation :

Fifth allocation :

Final allocation :

Transportation schedule : O₁⟶D₁, O₁⟶D₂, O₂⟶D₂, O₂⟶D₃, O₃⟶D₃,O₃⟶D₃.
The transportation cost
= (6\(\times\)6)+(8\(\times\)4)+(2\(\times\)9)+(14\(\times\)2)+(1\(\times\)6)+(4\(\times\)2)
= Rs.128

Commodity	Price		P = \(\frac {p_{1}}{p_{0}}\) \(\times 100\)	V	[PV
	2012 (p0)	2015 (p0)
Rice	250	280	112	10	1120
Wheat	70	85	121.42	5	607.1
Corn	150	170	113.33	6	679.98
Oil	25	35	140	4	560
Dhal	85	90	105.88	3	317.64
				28	3284.72

Using family budget method,
  C.L.I =  \(\frac {\sum PV}{\sum V}\) = \(\frac {3284.72}{28}\) = 117.31

The different methods of measurements of  trends are
(i) Free hand or graphic method
(ii) Method of semi averages
(iii) Method of moving averages
(iv) Method of least squares

Given equation can be written as \(\frac { ydx-xdy }{ { y }^{ 2 } } -{ 3x }^{ 2 }{ e }^{ { x }^{ 2 } }dx=0\)
Integrating, ഽ\(\frac { ydx-xdy }{ { y }^{ 2 } } \) - ഽ3x²e^x3 dx = c
ഽ\(d\left( \frac { x }{ y } \right) \) - ഽe^tdt = c  
(where t = x³ and dt = 3x²dx )
\(\frac { x }{ y } \) - e^t = c
\(\frac { x }{ y } \) - \({ e }^{ { x }^{ 2 } }\) = c

Hypothesis testing is also referred as "Statistical Decision Making".
There are two types of statistical hypothesis
(i) Null hypothesis: which is tested for possible rejection under the assumption that it is true
(ii) Alternative hypothesis which is complementary to the null hypothesis
For example: H_o:μ=μ₀
i) H₁:μ≠μ₀ is know as two tailed alternative test
ii) H₁:μ>μ₀ and H₁:μ<μ₀ are known as one tailed alternative
iii) H₁:μ>μ₀ is said to be right tailed test where the critical region lies entirely on the right tail of the normal curve.
iv) H₁:μ<μ0 is said to the left tailed test where the critical region lies entirely on the left tail of the normal curve.

Errors, which arise in the normal course of investigation or enumeration on account of chance, are called sampling errors. Sampling Errors arise due to the following reasons:
a) Faulty selection of the sample instead of correct sample by defective sampling technique.
b) The investigator substitutes a convenient sample if the original sample is not available while investigation.
c) In area surveys, while dealing with border lines it depends upon the investigator whether to include them in the sample or not. This is known as Faulty demarcation of sampling units.

Let X be probability of mortality rate for a certain disease
∴ p = \(\frac { 7 }{ 1000 } \) and n = 400

Hence X follows poisson distribution with p(x, λ)
= \(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \)
∴ P (just 2 deaths) = P(X = 2)
= \(\frac { e^{ -2.8 }(2.8)^{ 2 } }{ 2! } \)    [∵ p(x,λ) =\(\frac { e^{ -\lambda }.{ \lambda }^{ x } }{ x! } \), λ = 2.8, x = 2]
= \(\frac { (0.06)(2.8)^{ 2 } }{ 2 } \) = (0.03)(2.8)² = 0.2352
∴ Probability for just 2 deaths on account of this disease = 0.2352.

Let X be a random variable follows poisson distribution with number of erythrocytes.
Hence, Mean λ = 8 erythrocytes/single square
P(exactly 5 erythrocytes are in one square) = P(X = 5)=\(\\ \frac { { e }^{ -\lambda }{ \lambda }^{ x } }{ x! } =\frac { { e }^{ -8 }{ 8 }^{ 5 } }{ 5! } \)
\(=\frac { 0.000335\times 32768 }{ 120 } \)
= 0.0916
The probability that exactly 5 erythrocytes are found in one square is 0.0916. i.e there are 9.16% chances that exactly 5 erythrocytes are found in one square.

Given MR = 6 − 3x² − x³
⇒ ഽMR =ഽ(6 − 3x² − x³)dx
\(\Rightarrow \mathrm{R}=6 x-\frac{\not{3} x^{3}}{\not3}-\frac{x^{4}}{4}+k\)
\(\Rightarrow 6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } +k\)
When x = 0, R = 0 ⇒ k = 0
\(\Rightarrow R=6x-{ x }^{ 3 }-\frac { { x }^{ 4 } }{ 4 } \)
Demand function \(P=\frac { R }{ x } =6-{ x }^{ 2 }-\frac { { x }^{ 3 } }{ 4 } \)

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \),since f(x) is a density function
\(\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k\int _{ 0 }^{ \infty }{ { ke }^{ -2x } } dx=1\)
\(k{ \left[ \frac { { e }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }=1\)
⇒ k = 2
\(E(X)=\int _{ -\infty }^{ \infty }{ xf(x)dx } \)
\(=\int _{ 0 }^{ \infty }{ x{ e }^{ -2x } } dx\)
\(=2\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx } \)
\(=2\left\{ { \left[ \frac { { xe }^{ -2x } }{ -2 } \right] }_{ 0 }^{ \infty }\int _{ 0 }^{ \infty }{ \frac { { e }^{ -2x } }{ -2 } dx } \right\} \)
\((\because \int { udv=uv-\int { udv } } )\)
\(=\int _{ 0 }^{ \infty }{ { e }^{ -2x }dx } \) \(=\frac { 1 }{ 2 } \)

Given cost of overhaul of an engine is Rs. 10,000
Operating cost per hour = 2x - 240.
Total cost for the engine to run for 300 hours
after overhaul =10,000+\(\int _{ 0 }^{ 300 }{ (2x-240) } dx\)
\(=10,000+{ \left[ \frac { { 2x }^{ 2 } }{ 2 } -240x \right] }_{ 0 }^{ 300 }\)
\(=10,000+{ [{ x }^{ 2 }-240x] }_{ 0 }^{ 300 }\)
= 10,000 + 90,000 - 72,000
= 1,00,000 - 72,000
= Rs. 28,000

Sample space \((s)=\left\{ \begin{matrix} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ \begin{matrix} (3,1) \\ (4,1) \\ \begin{matrix} (5,1) \\ (6,1) \end{matrix} \end{matrix} & \begin{matrix} (3,2) \\ (4,2) \\ \begin{matrix} (5,2) \\ (6,2) \end{matrix} \end{matrix} & \begin{matrix} (3,3) \\ (4,3) \\ \begin{matrix} (5,3) \\ (6,3) \end{matrix} \end{matrix} \end{matrix}\begin{matrix} (1,4) & (1,5) & (1,6) \\ (2,4) & (2,5) & (2,6) \\ \begin{matrix} (3,4) \\ (4,4) \\ \begin{matrix} (5,4) \\ (6,4) \end{matrix} \end{matrix} & \begin{matrix} (3,5) \\ (4,5) \\ \begin{matrix} (5,5) \\ (6,5) \end{matrix} \end{matrix} & \begin{matrix} (3,6) \\ (4,6) \\ \begin{matrix} (5,6) \\ (6,6) \end{matrix} \end{matrix} \end{matrix} \right\} \)

Total outcomes : n(S) = 36

\(\int { \left( { x }^{ 2 }-2x+5 \right) } { e }^{ -x }dx =\int { udv } \)
= uv − u'v₁ + u''v₂ − u'''v₃ + ...
= (x² − 2x + 5)(−e^−x )− (2x − 2)e^−x + 2(−e^−x )+ c
= e^−x (−x² −5)+ c

Successive derivatives	Repeated integrals
Take u = x2 − 2x + 5 u' = 2x − 2 u'' = 2	and dv = e−xdx v = − e−x v1 = e-x v2 = e-x

\(\int { \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } dx } =\int { \left[ \frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \right] dx } \)
\(=20\int { \frac { dx }{ x-3 } -13\int { \frac { dx }{ x-2 } } } \)
\(20log\left| x-3 \right| -13 \log\left| x-2 \right| +c\)
[ By partial fractions,
\(\frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { A }{ x-3 } +\frac { B }{ x-2 } \Rightarrow \frac { 7x-1 }{ { x }^{ 2 }-5x+6 } =\frac { 20 }{ x-3 } -\frac { 13 }{ x-2 } \)]

Let the hourly charge for horse riding be Rs. x and the hourly charge for quad bike be Rs. y from the given data,
3x + 4y = 780
2x+ 3y = 560
\(\Delta =\left| \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right| =3(3)-2(4)=9-8=1\neq 0\)
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 780 & 4 \\ 560 & 3 \end{matrix} \right| =780(30-4(560)\)
= 2340 - 2240 = 100
\(\Delta y=\left| \begin{matrix} 3 & 780 \\ 2 & 560 \end{matrix} \right| =39560)-2(780)\)
= \(1680 - 1560\) = 120
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { 100 }{ 1 } =100\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 120 }{ 1 } =120\)
\(\therefore\)  Hourly charges for the two rides are Rs. 100 and Rs. 120 respectively.

Given 
A = \(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B = \(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
\(AB=\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-2+5 & -2+4-1 & 3-6+1 \\ 2+6+20 & -4-12+45 & 6+18-4 \\ 3+4+15 & -6-8+3 & 9+12-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) =\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)

Matrix (AB)	Elementary Transformation
\(AB=\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \)	\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+12{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & -44 & -4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }+11{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & 0 & 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (AB)=2\)
Now  \(BA=\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-4+9 & 1+6-6 & -1-8+9 \\ -2+8-18 & -2-12+12 & 2+16-18 \\ 5+2-3 & 5-3+2 & -5+4-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)

Matrix (BA)	Elementary Transformation
\(BA=\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ -2 & -12 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)	\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 0 & -20 & -4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-4R_{ 1 }\)

The number of non-zero rows is 2.
\(\therefore \rho (BA)=2\)