Scale:
In x axis 1 cm = 10 units
In y axis 1 cm = 2 units
Plot the points (40, 6.2), (50, 7.2), (60, 9.1) and (70, 12). At x = 48, draw a vertical line to the graph and from the intersecting point, draw a horizontal line to meet the y-axis
From the graph, we find that when x = 48, the value of Y is equal to 6.8.

	Estimated Conditional Profit 0
Rainfall	High	Medium	Low	Minimum payoff	Maximum payoff
Crop A	8000	4500	2000	2000	8000
Crop B	3500	4500	5000	3500	5000
Crop C	5000	5000	4000	4000	5000

(i) Max (2000,3500,4000) = 4000
∴ Crop C is the best according to maximin criteria
(ii) Min (8000,5000,5000) = 5000
∴ Crop B and C are best according to minimax criteria

Let k be the sixth term of the series in the difference table.
First we find the forward differences

x	y	∆	∆2
1	8
		4
2	12		3
		7
3	19		3
		10
4	29		3
		13
5	42		k-55
		k-42
6	k

Given that the second differences are constant
∴ k – 55 = 3
k = 58
∴ the sixth term of the series is 58

First allocation :

Second allocation :

Third allocation :

Fourth allocation :

Fifth allocation :

Sixth allocation :

\( =(3 \times 2)+(1 \times 11)+(2 \times 4)+(4 \times 7)+ (2 \times 2)+(3 \times 8)+(6 \times 12) \)
= 6 + 11 + 8 + 28 + 4 + 24 + 72 = Rs. 153 
Total Cost = 153

Alternative	Pay – offs (Conditional events)				Minimum pay off
	A1	A2	A3	A4
E1	7	12	20	27	27
E2	10	9	10	25	25
E3	23	20	14	23	23
E4	32	24	21	17	32

min( 27, 25, 23, 32) = 23. Since the minimum cost is 23, the best alternative is E₃ according to minimax principle.

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

The auxiliary equation is m² - 2m -15 = 0
⇒ (m - 5) (m + 3) = 0
⇒ m = 5,-3
∴ Complementary function CF is Ae^-3x + Be^5x
∴ The general solution is y = Ae^-3x + Be^5x...(1)
Given \(\frac { dy }{ dx } \) = 0 when x = 0
Differentiating (1) w.r.t 'x' we get,
\(\frac { dy }{ dx } \)=-3Ae^-3x+5Be^5x
0= -3Ae⁰ + 5Be^o ⇒ 0 = -3A + 5B...(2)
[∵ e⁰= 1]
Differentiating again w.r.t 'x' we get,
\(\frac { { d }^{ 2 }y }{ dx^{ 2 } } \) = 9Ae^-3x + 25Be^5x
Also, it is given that \(\frac { { d }^{ 2 }y }{ dx^{ 2 } } \) = 2 when x = 0
∴ 2 = 9A + 25B     .....(3)
(2) x 3 ⇒ 0=-9A +15B
(3) ⇒ \(\frac { 2=9A+25B }{ 2=40B } \)
⇒ B = \(\frac { 2 }{ 40 } =\frac { 1 }{ 20 } \)
From (2) ⇒ 0 = -A+\(\frac { 5 }{ 20 } \)
⇒ 3A=\(\frac { 1 }{ 4 } \Rightarrow A=\frac { 1 }{ 12 } \)
Substituting A = \(\frac { 1 }{ 12 } \), B = \(\frac { 1 }{ 20 } \) in (1) we get
y=\(\frac { 1 }{ 12 } e^{ -3x }+\frac { 1 }{ 20 } \)e^5x.

Year	Profit	3-year moving total	3-year moving trend
2001	142	-	-
2002	148	444	148
2003	154	448	149.33
2004	146	457	152.33
2005	157	505	168.33
2006	202	600	200.00
2007	241	706	235.33
2008	263	784	261.33
2009	280	845	281.67
2010	302	908	302.67
2011	326	981	327
2012	353	-	-

y = e^x (acos x + bsin x)   (1)
Differentiating (1) w.r.t x, we get
\(\frac { dy }{ dx } \) = e^x (acos x + bsin x)+ e^x (−a sin x + b cos x)
= y + ex (−asin x + bcos x) (from (1))
⇒ \(\frac { dy }{ dx } \) - y = ex (−a sin x + bcos x) (2)
Again differentiating, we get
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \) = e^x (−a sin x + b cos x) + e^x (−a cos x − b sin x)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \)= e^x (−a sin x + bcos x) − e^x (a cos x + b sin x)
\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \) = \(​​\left( \frac { dy }{ dx } -y \right) -y\) (from (1) and (2))
⇒\(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -\frac { dy }{ dx } \)+2y = 0, which is the requited differential equation.

Since the number of years is odd(seven), we will leave the middle year’s production value and obtain the averages of first three years and last three years.

Year	Production	Average
2000	105	\(\frac{105+115+120}{3}=113.33\)
2001	115
2002	120
2003	100(left out)
2004	110	\(\frac{110+125+135}{3}=123.33\)
2005	125
2006	135

Let I = ഽ(x +1)² log x dx
Let u = logx dv = (x+1)²dx
\(du=\cfrac { 1 }{ x } dx\quad v=\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \)
\(\therefore\) Using integration by parts,
\(I=\int { udv } =uv-\int { vdu } \)
\(I=\int { \left( x+1 \right) ^{ 2 }\log xdx } \)
= \(\cfrac { (x+1)^{ 3 } }{ 3 } \log x-\int { \cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } .\cfrac { 1 }{ x } dx } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \int { \cfrac { { x }^{ 3 }+{ 3x }^{ 2 }+3x+1 }{ x } } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \int { \left( { x }^{ 2 }+3x+3+\cfrac { 1 }{ x } \right) dx } \)
= \(\cfrac { \left( x+1 \right) ^{ 3 } }{ 3 } \log x-\cfrac { 1 }{ 3 } \left[ \cfrac { { x }^{ 3 } }{ 3 } +\cfrac { { 3x }^{ 2 } }{ 2 } +3x+\log|x| \right] +c\)

= \(\left[ \left( x+1 \right) ^{ 3 }\log x-\cfrac { { x }^{ 3 } }{ 3 } -\cfrac { { 3x }^{ 2 } }{ 2 } -3x-log|x| \right] +c\)

If the occurrence of 3 or 4 on the die is called a success, then
Sample size = 9000; Number of Success = 3240
Sample proportion =\(P=\frac { 3240 }{ 9000 } =0.36\)
Population proportion (P) = Prob(getting 3 or 4 when a die is thrown)
\(=\frac { 1 }{ 6 } +\frac { 1 }{ 6 } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } =0.3333\)
Thus P = 0.3333 and Q = 1–P = 1– 0.3333 = 0.6667
The S.E for sample proportion is given by
\(S.E=\sqrt { \frac { PQ }{ N } } =\sqrt { \frac { (0.3333)(0.6667) }{ 9000 } } =0.00496\)
Hence the standard error for sample proportion is S.E = 0.00496.

Given  \(\begin{cases} cx, \\ 0, \end{cases}\begin{matrix} 0 < x < 1 \\ \text{otherwise} \end{matrix}\)
Also, Given \(\int _{ 0 }^{ 1 }{ f\left( x \right) } dx=2\)
\(\Rightarrow \int _{ 0 }^{ 1 }{ cx } dx=2\)
\(\Rightarrow c{ \left[ \frac { { x }^{ 2 } }{ 2 } \right] }_{ 0 }^{ 1 }=2\)
\(\Rightarrow \frac { c }{ 2 } \left[ { 1 }^{ 2 }-{ 0 }^{ 2 } \right] =2\)
\(\Rightarrow \frac { c }{ 2 } \left( 1-0 \right) =2\)
\(\Rightarrow \frac { c }{ 2 } =2\)
⇒ c = 2

Given μ = 3500, σ = 500
P(X<3100)
When X = 3100,
Z = \(\frac { X-\mu }{ \sigma } =\frac { 3100-3500 }{ 500 } \)
= -0.8
∴ P(X<3100) = P(Z<-0.8)


= 0.5-P(0 [By symmetry]
= 0.5-0.2881= 0.2119
Hence, the probability that a baby is born with weight less than 3100g is 0.2119.

Let p be the probability of mice recovering from vitamin A deficiency
Given p = 0.73 ⇒ q = 1 - 0.73 = 0.27 and n = 5
∴  P (atleast 3 mice recover) = P(X ≥ 3)
P(X≥3) = 1-P(X<3)
= 1 - [P(X = 0) + P(X = 1) + P (X = 2)]
= 1-[⁵C₀ (0.73)⁰ (0.27)⁵ + ⁵C₁ (0.73)¹ (0.27)⁴ + ⁵C₂ (0.73)² (0.27)³ ]
= 1-[(0.27)⁵ + 5(0.73)(0.27)⁴ + \(\frac { 5\times 4 }{ 2\times 1 } \)+(0.73)² (0.27)³ ]
= 1 - [0.00143 + 0.01939 + 0.10489]
= 1 - 0.1257
P(X≥3) = 0.8743

\(\int _{ 0 }^{ 1 }{ [{ e }^{ a \log x }+{ e }^{ x \log a }] } dx=\int _{ 0 }^{ 1 }{ ({ x }^{ a }+{ a }^{ x }) } dx\)
\(={ \left[ \frac { { x }^{ a+1 } }{ a+1 } +\frac { { a }^{ x } }{ \log a } \right] }_{ 0 }^{ 1 }\)
\(=\left( \frac { 1 }{ a+1 } +\frac { a }{ \log a } \right) -\left( 0+\frac { 1 }{ \log a } \right) \)
\(=\frac { 1 }{ a+1 } +\frac { a }{ \log a } -\frac { 1 }{ \log a } \)
\(=\frac { 1 }{ a+1 } +\frac { (a-1) }{ \log a } \)

\(\int { \frac { dx }{ { x }^{ 2 }-x-2 } } \)
\(=\int { \frac { dx }{ { x }^{ 2 }-x+\frac { 1 }{ 4 } -\frac { 1 }{ 4 } -2 } } \)
Adding and subtracting
\(\frac { 1 }{ 2 } { \left[\text{ co-effective of x }\right] }^{ 2 }\)
\(={ \left[ \frac { 1 }{ 2 } (-1) \right] }^{ 2 }=\frac { 1 }{ 4 } \)
\(=\int { \frac { dx }{ { \left( { x }^{ 2 }-\frac { 1 }{ 2 } \right) }^{ 2 }-\frac { 9 }{ 4 } } } \)
\(=\int { \frac { dx }{ { \left( x-\frac { 1 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } } \)
\(\left[ \because \int { \frac { dx }{ { x }^{ 2 }-{ a }^{ 2 } } =\frac { 1 }{ 2a } \log { \left| \frac { x-a }{ x+a } \right| } +c } \right] \)
\(=\frac { 1 }{ 2\left( \frac { 3 }{ 2 } \right) } \log { \left| \frac { x-\frac { 1 }{ 2 } -\frac { 3 }{ 2 } }{ x-\frac { 1 }{ 2 } +\frac { 3 }{ 2 } } \right| } +c\)
\(=\frac { 1 }{ 3 } \log { \left| \frac { x-2 }{ x+1 } \right| } +c\)

Probability p of guessing an answer correctly is p = \(\frac{1}{2}\)
⇒ q = \(\frac{1}{2}\)
Probability of guessing correctly x answers in 10 questions
\(P(X=x)=p(x)^{ n }{ C }_{ x }{ q }^{ n-x }=10Cx\left( \frac { 1 }{ 2 } \right) \left( \frac { 1 }{ 2 } \right) \)
The required probability P(X ≥ 6) = P(6) + P(7) + P(8) + P(9) + P(10)
\(={ \left( \frac { 1 }{ 2 } \right) }^{ 10 }\left[ { 10C }_{ 6 }+{ 10C }_{ 7 }+{ 10C }_{ 8 }+{ 10C }_{ 9 }+{ 10C }_{ 10 } \right] \)
\(=\left[ \frac { 1 }{ 1024 } \right] [210+120+45+10+1]\)
\(=\frac { 193 }{ 512 } \)

\(Let\ I=\int { \frac { { x }^{ e-1 }+{ e }^{ x-1 } }{ { x }^{ e }+{ e }^{ x } } } dx\)
\(put\ t={ x }^{ e }+{ e }^{ x }\)
\(\Rightarrow dt=\left( e{ x }^{ e-1 }+{ e }^{ x } \right) dx\)
\(dt=e\left( { x }^{ e-1 }+{ e }^{ x-1 } \right) dx\)
\(\Rightarrow \frac { dt }{ e } =\left( { x }^{ e-1 }+{ e }^{ x-1 } \right) dx\)
\(\therefore I=\int { \frac { dt }{ e(t) } } =\frac { 1 }{ e } \int { \frac { dt }{ t } } \)
\(=\frac { 1 }{ e } \log { \left| t \right| } +c\)
\(=\frac { 1 }{ e } \log { \left| { x }^{ e }+{ e }^{ x } \right| } +c\quad \left[ \because t={ x }^{ e }+{ e }^{ x } \right] \)

When a coin is tossed, sample space S = {H, T}
Since he is receiving Rs. 4 for a head and pays Rs. 2 for a tail,
∴ X take values 4 and -2.
∴ Probability for getting a head is \(\frac{1}{2}\) and probability for getting a tail is \(\frac{1}{2}\).
The probability mass function is

X = x	4	-2
P(X = x)	\(\frac{1}{2}\)	\(\frac{1}{2}\)

∴ Expectation E(X) = Σxp(x)
= 4(\(\frac{1}{2}\)) - 2(\(\frac{1}{2}\))
= 2-1 = 1
∴ His expectation is Rs. 1
E(x²) = ∑x²p(x)
= 4²(\(\frac{1}{2}\)) + (-2)²(\(\frac{1}{2}\))
= 16(\(\frac{1}{2}\)) + 4(\(\frac{1}{2}\))
= 8 + 2 = 10
Variance (X) = E(X²)-[E(X)]²
= 10-1² = 9
∴ Variance of his gains Rs. 9

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 demand function P_d = 1600 - x² and
Supply function P_s = 2x² + 400
Under perfect competition p_d = p_s
⇒ 1600 - x² = 2x² + 400
⇒ 1600-400 = 2x² + x²
⇒ 1200 = 3x²
\(\Rightarrow \frac { 1200 }{ 3 } ={ x }^{ 2 } \Rightarrow { x }^{ 2 }=400\)
\(\Rightarrow x=\pm \sqrt { 400 } =+20\quad or-20\)
Since x cannot be negative, x₀ = 20
∴ p₀ = 1600 - (20)² = 1600 - 400
= 1200
∴ p₀x₀ = (1200) (20) = 24000
Producer's Surplus PS = P_ox_o -\(\int _{ 0 }^{ x }{ g(x)dx } \)
\(=24000-\int _{ 0 }^{ 20 }{ ({ 2x }^{ 2 }+400)dx } \)
\(=24000-{ \left[ \frac { { 2x }^{ 3 } }{ 3 } +400x \right] }_{ 0 }^{ 20 }\)
\(=24000-\left[ \frac { { 2(20) }^{ 3 } }{ 3 } +400(20) \right] \)
\(=24000-\left[ \frac { 16000 }{ 3 } +8000 \right] \)
\(=24000-\left[ \frac { 16000+24000 }{ 3 } \right] \)
\(=24000-\left[ \frac { 40000 }{ 3 } \right] \)
\(=\frac { 72000-40000 }{ 3 } \)
PS = \( \frac{32000}{3}\)units

Given marginal cost function
\(MC=\frac { dc }{ dx } =100-10x+0.1{ x }^{ 2 }\)
\(\Rightarrow C=\int { (100-10x+0.1{ x }^{ 2 })dx } \)
\(\Rightarrow C=100x-\frac { { 10x }^{ 2 } }{ 2 } +\frac { { 0.1x }^{ 3 } }{ 3 } +k\) ...(1)
Given fixed cost is Rs. 500 ⇒ k = 500
∴ (1) becomes,
\(C=100x-5{ x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } +500\)
Average cost function
\(AC=\frac { C }{ x } =\frac { 100x-{ 5x }^{ 2 }+\frac { { 0.1x }^{ 3 } }{ 3 } }{ x } +500\)
\(AC=100-5x+\frac { { 0.1x }^{ 2 } }{ 3 } +\frac { 500 }{ x } \)

We know that
\(\int _{ -\infty }^{ \infty }{ f(x)dx=1 } \)
\(\int _{ 0 }^{ -\infty }{ ax\quad dx } \Rightarrow a\int _{ 0 }^{ 1 }{ xdx=1 } \)
\(\Rightarrow a{ \left( \frac { { x }^{ 2 } }{ 2 } \right) }^{ 1 }=1\)
\(\Rightarrow \frac { a }{ 2 } (1-0)=1\)
\(\Rightarrow\)a = 2
\(P\left[ x\le \frac { 1 }{ 2 } \right] =\int _{ -\infty }^{ \frac { 1 }{ 2 } }{ f(x)dx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ axdx } \)
\(=\int _{ 0 }^{ \frac { 1 }{ 2 } }{ 2xdx } \)
\(=\frac { 1 }{ 4 } \)

Given MC = 2 + 5e^x
\(C=\int { MC } dx+k\)
\(=\int { (2+5{ e }^{ x })dx } +k\)
= 2x + 5e^x + k
x = 0 ⇒ C = 100,
100 = 2(0)+ 5(e⁰ )+ k
k = 95
C = 2x + 5 e^x + 95.

Let I = ∫ x logx dx
Let u = log x; dv = x dx
\(du=\frac { 1 }{ x } dx;v=\frac { { x }^{ 2 } }{ 2 } \)
Using integration by parts,
∫ udv = uv - ∫ vdu
\(\Rightarrow \int { x\ \log\ x\ dx=\log x\left( \frac { { x }^{ 2 } }{ 2 } \right) } -\int { \frac { { x }^{ 2 } }{ 2 } } \frac { 1 }{ x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x\frac { 1 }{ 2 } \int { x } dx\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x-\frac { 1 }{ 2 } \frac { { x }^{ 2 } }{ 2 } +c\)
\(=\frac { { x }^{ 2 } }{ 2 } \log x-\frac { { x }^{ 2 } }{ 4 } +c\)
\(=\frac { { x }^{ 2 } }{ 2 } \left( \log\ x-\frac { 1 }{ 2 } \right) +c\)

\(\int { \frac { { e }^{ 3x }+{ e }^{ 5x } }{ { e }^{ x }+{ e }^{ -x } } } dx\)
\(=\int { \left( \frac { { e }^{ 3x }+{ e }^{ 5x } }{ { e }^{ x }+\frac { 1 }{ { e }^{ x } } } \right) } dx\quad =\quad \int { \frac { { e }^{ 3x }+{ e }^{ 5x } }{ \frac { { e }^{ 2x }+1 }{ { e }^{ x } } } } dx\)
\(=\int { \frac { { e }^{ x }\left( { e }^{ 3x }+{ e }^{ 5x } \right) }{ { e }^{ 2x }+1 } } dx\)

\(=\int { { e }^{ 4x } } dx=\frac { { e }^{ 4x } }{ 4 } +c\)

By factorisation,
2x² −14x + 24 = (x − 3)(2x − 8)
\(\int { \frac { { 2x }^{ 2 }-14x+24 }{ x-3 } dx } =\int { \frac { \left( x-3 \right) \left( 2x-8 \right) }{ x-3 } } dx\)
\(=\int { \left( 2x-8 \right) } dx\)
\(={ x }^{ 2 }-8x+c\)

Given  A =\(\left( \begin{matrix} 4 & 5 & 2 \\ 3 & 2 & 1 \\ 4 & 4 & 8 \end{matrix}\begin{matrix} 2 \\ 6 \\ 0 \end{matrix} \right) \)
\(\left( \begin{matrix} 4 & 5 & 2 \\ 3 & 2 & 1 \\ 1 & 1 & 2 \end{matrix}\begin{matrix} 2 \\ 6 \\ 0 \end{matrix} \right) { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 4 & 5 & 2 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) { R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 1 & -6 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) \)\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 1 & -6 \end{matrix}\begin{matrix} 0 \\ 6 \\ 2 \end{matrix} \right) \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } \end{matrix}\)
\(\left( \begin{matrix} 1 & 1 & 2 \\ 0 & -1 & -5 \\ 0 & 0 & -11 \end{matrix}\begin{matrix} 0 \\ 6 \\ 8 \end{matrix} \right) { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 }\)
The last equivalent matrix is in echelon form and there are 3 non-zero rows
\(\therefore \rho (A)=3\)

Let the amount invested in the two accounts be Rs. x and Rs. y respectively
By the given data, x + y = 8600   ..(1)
\(4\cfrac { 3 }{ 4 } \times \cfrac { x }{ 100 } +6\cfrac { 1 }{ 2 } \times \cfrac { y }{ 100 } =431.25\) \(\left[ \therefore interest=\cfrac { PNR }{ 100 } \right] \)
\(\Rightarrow \cfrac { 19x }{ 400 } +\cfrac { 13y }{ 3200 } =431.25\)
\(\Rightarrow \cfrac { 19x+26y }{ 400 } =431.25\)
19x + 26y = 172500   ...(2)
\(\Delta =\left| \begin{matrix} 1 & 1 \\ 19 & 26 \end{matrix} \right| =1(26)-1(19)\)
= 26-19 =7
\({ \Delta }x=\left| \begin{matrix} 8600 & 1 \\ 172500 & 26 \end{matrix} \right| =8600(26)-1(172500)\)
= 223600 - 172500 = 51100
\(\Delta y=\left| \begin{matrix} 1 & 8600 \\ 19 & 172500 \end{matrix} \right| =1(172500)-19(8600)\)
= 172500 - 163400 = 9100
\(x=\cfrac { \Delta x }{ \Delta } -\cfrac { 51100 }{ 7 } =7300\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 9100 }{ 7 } =1300\)
\(\therefore\) Investment in the interest of \(4\frac { 3 }{ 4 } \) % account is Rs. 7300 and investment in the rate of \(6\frac { 1 }{ 2 } \) account is Rs.1300. 

The order of A is 3 \(\times\) 4.
 \(\therefore \) \(\rho (A)\le 3.\)
Let us transform the matrix A to an echelon form

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} 1 & 1 & 1 \\ 3 & 4 & 5 \\ 2 & 3 & 4 \end{matrix}\begin{matrix} 1 \\ 2 \\ 0 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ -1 \\ -2 \end{matrix} \right) \) \(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -2 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ 3R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

The number of non zero rows is 3.
\(\therefore \) \(\rho (A)=3.\)