\(\Delta=\left|\begin{array}{ll} 3 & -4 \\ 5 & -3 \end{array}\right|=-9-(-20)=11\)
\(\Delta_{x}=\left|\begin{array}{cc} 6 & -4 \\ -1 & -3 \end{array}\right|=-18-4=-22\)
\(\Delta_{y}=\left|\begin{array}{cc} 3 & 6 \\ 5 & -1 \end{array}\right|=-3-30=-33\)
\( x=\frac{\Delta_{x}}{\Delta}=\frac{-22}{11}=-2 \)
\( y=\frac{\Delta_{y}}{\Delta}=\frac{-33}{11}=-3\)
(x, y) = (-2, -3)

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

Let x represent the number of question with correct answer and y represent the number of questions with wrong answers.
By the given data, x + y 100 ............... (1)
x - \(\frac { 1 }{ 4 } \)y = 80
Multiplying by 4 we get we get
4x - y = 320...............(2)
From (1) and (2)
Δ = \(\\ \left| \begin{matrix} 1 & 1 \\ 4 & -1 \end{matrix} \right| \)= -1 - 4 = -5
Δ₁ = \(\left| \begin{matrix} 100 & 1 \\ 320 & -1 \end{matrix} \right| \) = -100 - 320 = -420
Δ₂ = \(\left| \begin{matrix} 1 & 100 \\ 4 & 320 \end{matrix} \right| \) = 320 - 400 = -80
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -720 }{ -5 } \) = +84
and y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -80 }{ -5 } \) = 16
Hence, the number of questions with correct answer is 84 and wrong question is 16.

5x − 2y + 16 = 0, x + 3y − 7 = 0
Given Δ = \(\left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| \) = 15+2 = 17
Δ₁ = \(\left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| \) = -48+14 = -34
Δ₂ = \(\left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| \) = 35+16 = 51
∴ x = \(\frac { \triangle _{ 1 } }{ \triangle } =\frac { -34 }{ 7 } \) = -2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 51 }{ 17 } \)
∴ Solution set is {-2, 3}

Giveny = ax² + bx + c .............(1)
(-6, 8) lies on (1)
⇒ 8 = a(-6)²+b(-6)+c
⇒ 8 = 36z-6b+c ..........(2)
(-2,12) lies on (1)
⇒ -12 = a(-2)²+b(-2)+c
⇒ -12 = 4a-2b+c ...........(3)
Also (3, 8) lies on (1)
⇒ 8 = a(3)²+b(3)+c
⇒ 8 = 9a+3b+c ............(4)
Reducing the augment matrix to an equivalent row-echelon form by using elementary. row operations, we get,
\(\left[ \begin{matrix} 36 & -6 & 1 \\ 4 & -2 & 1 \\ 0 & 3 & 1 \end{matrix}|\begin{matrix} 8 \\ -12 \\ 8 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { 9R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow 4{ R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -12 & 8 \\ 0 & 18 & 3 \end{matrix}|\begin{matrix} 8 \\ -116 \\ 24 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 4\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 3 }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -50 \end{matrix} \right] \)
Writing the equivalent equation from the row echelon matrix, we get
36a - 6b + c = 8 ........(1)
-3b+2c = -29 ....(2)
5c = -50
⇒ c = \(\frac{-50}{5}\) = -10
Substituting c = -10 in (2) we get,
-3b+2(-10)= -29
⇒ -3b+2(-10) = -29
⇒ -3b-20 = -29
⇒ -3b = -9
⇒ b = \(\frac{-9}{-3}\) = 3
Substituting b = 3 and c = -10 in (1) we get,
36a-6(3)-10 = 8
⇒ 36a-18-10 = 8
⇒ 36a-28 = 8
⇒ 6a = 8+28 = 36
⇒ a = \(\frac{36}{36}\) = 1
∴ a = 1, b = 3, c = -10
Hence the path of the boy is
y = 1(x²)+3(x)-10
⇒ y = x²+3x-10
Since his friend is at P(7, 60),
60 = (7)²+3(7)-10
⇒ 60 = 49+21-10
⇒ 60 = 70-10 = 60
⇒ 60 = 60
Since (7, 60) satisfies his path, he can meet his friend who is at P(7, 60)

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.

Let P(x) = ax²+ bx + c
Given P(-3) = 21
[∵ P(x) ÷ x + 3, the remainder is 21]
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a - 3b + c = 21
Also, P(5) = 61
⇒ a(5)2 + b(5) + c = 61
[using remainder theorem]
⇒ 25a +5b + c = 61..........(2)
and P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 ............(3)
Reducing the augment matrix to an equivalent row-echelon form using elementary row operations, we get
\(\left[ \begin{matrix} 9 & - & 1 \\ 25 & 5 & 1 \\ -1 & 1 & 1 \end{matrix}|\begin{matrix} 21 \\ 61 \\ 9 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 25 & 5 & 1 \\ 9 & -3 & 1 \end{matrix}|\begin{matrix} 9 \\ 61 \\ 21 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-9{ R }_{ 1 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }-25{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -20 & -24 \\ 0 & -12 & -8 \end{matrix}|\begin{matrix} 9 \\ -164 \\ -60 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & -3 & -2 \end{matrix}|\begin{matrix} 9 \\ -41 \\ -15 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 3 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & \frac { 8 }{ 5 } \end{matrix}|\begin{matrix} 9 \\ -41 \\ \frac { 48 }{ 5 } \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow 5{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & 8 \end{matrix}|\begin{matrix} 9 \\ -41 \\ 48 \end{matrix} \right] \)
Writing the equivalent equations from the row-echelon matrix we get,
a + b + c = 9 ............(1)
5b + 6c = 41 ................(2)
-8c = -48
⇒ c = 6
Substituting c = 6 
⇒  5b + 36 = 41
⇒ 5b = 5  
b = 1
Substituting b = 1, c = 6
a + 1 + 6 = 9
⇒ a + 7 = 9
⇒ a = 9 - 7
⇒ a = 2
∴ a = 2, b = 1, and c = 6

Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 25 \\ 13 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 4 & 3 & 6 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 13 \\ 25 \\ 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & -1 & -13 \end{matrix}|\begin{matrix} 13 \\ -27 \\ -25 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 1 & 13 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 25 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 17{ R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 0 & 199 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 398 \end{matrix} \right] \).
The equivalent system is written by using the echelon form:
x + 5y + 7 = 13, … (1)
17y + 22z = 27, … (2)
199z = 398..... (3)
From (3), we get z = \(\frac { 398 }{ 199 } \) = 2.
Substituting z = 2 in (2), we get y = \(\frac { 27-22\times 2 }{ 17 } =\frac { -17 }{ 17 } \) = -1
Substituting z = 2, y = -1, in (1), we get x = 13 - 5 x (-1) - 7 \(\times\) 2 = 4.
So, the solution is (x = 4, y = -1, z = 2).

Let the pump A can fill the tank in x minutes, and the pump B can fill. the tank in y minutes In 1 minute A can fill \(\frac { 1 }{ x } \) units and in 1 minute B can fill \(\frac { 1 }{ y } \) units.
∴ \(\frac { 1 }{ x } +\frac { 1 }{ y } \) = 10
and \(\frac { 1 }{ x } -\frac { 1 }{ y } \) = 30
Put \(\frac { 1 }{ x } \) = a and \(\frac { 1 }{ y } \) = b
⇒ a + b = \(\frac { 1 }{ 10 } \) ..............(1)
and a - b = \(\frac { 1 }{ 30 } \) .............(2)
Δ = \(\left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right| \)= -1-1= -2
Δ₁ = \(\left| \begin{matrix} \frac { 1 }{ 10 } & 1 \\ \frac { 1 }{ 30 } & -1 \end{matrix} \right| =\frac { -1 }{ 10 } -\frac { 1 }{ 30 } \)
= \(\frac { -3-1 }{ 30 } =\frac { -4 }{ 30 } =\frac { -2 }{ 15 } \)
Δ₂ = \(\left| \begin{matrix} 1 & \frac { 1 }{ 10 } \\ 1 & \frac { 1 }{ 30 } \end{matrix} \right| =\frac { 1 }{ 30 } -\frac { 1 }{ 10 } =\frac { 1-3 }{ 30 } \)
= \(\frac { -2 }{ 30 } =\frac { -1 }{ 15 } \)
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -2 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 15 } \Rightarrow \)= x = 15
b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -1 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 30 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 30 } \Rightarrow \)y = 30
Hence the pump A can fill the tank in 15 minutes and the pump B can fill the tank in 30 minutes.