Given C₂H₆ + O₂ ⟶ H₂O + CO₂
We have to find positive integers x₁ x₂, x₃ and x₄ such that
x₁C₂H₆ + x2O₆ ⟶ x₃H₂O + x₄CO₂ .....(1)
The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1)
∴ 2x₁ = 1x₄
⇒ 2x₁-x₄ = 0 ...(2)
Considering hydrogen atoms we get,
6x₁ = 2x₃
⇒ 6x₁- 2x₃ = 0
⇒ 3x₁- x₃ = 0....(3)
Also, considering oxygen atoms we get,
2x₂ = 1x₃ + 2x₄
⇒ 2x₂ - x₃ - 2a₄ = 0 ....(4)
Equations (2), (3) and (4) forni a homogeneous system of linear equations in 4 unknowns
∴ The augmented matrix [A|0] is
\(\left[ \begin{matrix} 2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ 0 \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ 0 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ \frac { -7 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { -7 }{ 2 } \\ \frac { 3 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow 2{ R }_{ 2 }\\ R_{ 2 }\rightarrow 2{ R }_{ 2 } }{ \underset { { R }_{ 2 }\rightarrow 2{ R }_{ } }{ \longrightarrow } } \left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 4 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -2 \end{matrix}\begin{matrix} -1 \\ -7 \\ 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1, the number of unknowns
∴ The system is consistent with one parameter family of solutions, so let x₄ = t
Writing the equations, from the row-echelon form we get
2x₁-x₄ = 0
⇒ 2x₁ = x₄
⇒ 2x = t
⇒ x₁ = \(\frac { t }{ 2 } \)
4x₂ - 7x₄ = 0
⇒ 4x₂ = 7t
⇒ x₂ = \(\frac { 7t }{ 4 } \)
-2x₃ + 3x₄ = 0
⇒ 2x₃ = 3x₄
⇒ x₃ = \(\frac { 3t }{ 4 } \)
Since x₁, x₂, x₃ and x₄ are positive integers, let us choose t = 4
∴ x₁ = \(\frac{4}{2}\) = 2

So, the balanced equation is
2C₂H₆ + 7O₂ ⟶ 6H₂O + 4CO₂

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

\(\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)
Let A = \(\left( \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right) \)
adj A = \(\left( \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right) \)
[Interchange the elements in the leading diagonal and change the sign of the elements in off diagonal]

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

We first find adj A. By definition, we get adj A = \({ \left[ \begin{matrix} +{ M }_{ 11 } & -{ M }_{ 12 } \\ -{ M }_{ 21 } & +{ M }_{ 22 } \end{matrix} \right] }^{ T }={ \left[ \begin{matrix} d & -c \\ -b & a \end{matrix} \right] }^{ T }=\left[ \begin{matrix} d & -c \\ -c & a \end{matrix} \right] \).
Since A is non-singular, |A| = ad - bc ≠ 0.
As \({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } \) adj A, we get A^-1 = \(\frac { 1 }{ ad-bc } \left[ \begin{matrix} d & -b \\ -c & a \end{matrix} \right] \).

The matrix form of the system is AX = B , where A = \(\left[ \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] \)
We find |A| = \(\left| \begin{matrix} 5 & 2 \\ 3 & 2 \end{matrix} \right| \) = 10 - 6 = 4 ≠ 0. So, A⁻¹ exists and A⁻¹ = \(\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \)
Then, applying the formula X = A⁻¹B, we get
\(\left[ \begin{matrix} x \\ y \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 2 & -2 \\ -3 & 5 \end{matrix} \right] \left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} -4 \\ 16 \end{matrix} \right] =\left[ \begin{matrix} \frac { -4 }{ 4 } \\ \frac { 16 }{ 4 } \end{matrix} \right] =\left[ \begin{matrix} -1 \\ 4 \end{matrix} \right] \).
So the solution is (x = −1, y = 4).

\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 7 \end{matrix} \end{matrix}\begin{matrix} 3 \\ \begin{matrix} 4 \\ 11 \end{matrix} \end{matrix} \right] \)
A =\(\left[ \begin{matrix} 1 \\ 2 \\ 5 \end{matrix}\begin{matrix} 1 \\ -1 \\ -1 \end{matrix}\begin{matrix} 1 \\ 3 \\ 7 \end{matrix}\begin{matrix} 3 \\ 4 \\ 11 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { { R } }_{ 2 }-2{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 5 \end{matrix}\begin{matrix} 1 \\ -3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 7 \end{matrix}\begin{matrix} 3 \\ -2 \\ 11 \end{matrix} \right] \)
\(\overset { { { R } }_{ 3 }\rightarrow { { R } }_{ 3 }-5{ { R } }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -6 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix}\begin{matrix} 3 \\ -2 \\ -4 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { { R } }_{ 3 }-2{ { R } }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix}\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} 3 \\ -2 \\ 0 \end{matrix} \right] \)
The last equivalent matrix is in row echelon form it ha two non-zero row \(\rho \)(A) = 2

Given A\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
Let B =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \) and 
C = \(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
∴ AB = C
Post multiply by B^-1 we get
A(BB^-1) = CB^-1
⇒ A = CB^-1     [∵ BB^-1 = 1]
|B| = \(\left| \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right| \)
= -10 + 3 = -7 ≠ 0
∴ B^-1 exists
B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { -1 }{ 7 } \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
A = CB^-1
=\(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(7\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} -4+1 & -6+5 \\ -2+1 & -3+5 \end{matrix} \right] =-\left[ \begin{matrix} -3 & -1 \\ -1 & 2 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 3 & 1 \\ 1 & -2 \end{matrix} \right] \).

Given A =\(\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
adj A =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of the elements in the off diagonal]
|A| = 24 - 20 = 4
∴ A(adj A) =\(\\ \left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & 32-32 \\ -15+15 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) ....(1)
(adj A)(A) =\(\left[ \begin{matrix} 3 & 4 \\ 5 & 8 \end{matrix} \right] =\left[ \begin{matrix} 8 & -4 \\ -5 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 24-20 & -12+12 \\ 40-40 & -20+24 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \)...(2)
|A|I₂ = 4\(\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 4 & 0 \\ 0 & 4 \end{matrix} \right] \) .....(3)
From (1), (2) and (3), it is proved that
A (adj A) = (adj A) A = |A|I₂

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution.
So, we have \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0, Applying R₂ ➝ R₂ - R₁ and R₃ ➝ R₃ - R₁ in the above equation,
we get \(\left| \begin{matrix} p & b & c \\ a-p & q-b & c \\ a-p & b & r-c \end{matrix} \right| \) = 0. That is,  \(\left| \begin{matrix} p & b & c \\ -\left( p-a \right) & q-b & c \\ -\left( p-a \right) & b & r-c \end{matrix} \right| \) = 0.
Since p ≠ a, q ≠ b, r ≠ c, we get (p - a)(q - b)(r - c) \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
So, we have \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
Expanding the determinant, we get \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 0.
That is, \(\frac { p }{ p-a } +\frac { q-\left( q-b \right) }{ q-b } +\frac { r-\left( r-c \right) }{ r-c } \) = 0
⇒ \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 2.

kx-2y+z = 1, -2ky+z = -2, x-2y+k = 1 
The matrix form of the system is AX = B where
\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
Applying elementary row operation on the augment matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} k & -2 & 1 \\ 1 & -2k & 1 \\ 1 & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k & 1 \\ k & -2 & k \end{matrix}|\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 1 & -2k+2 & k \\ 0 & -2+2k & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ - \\ 1-k \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & k \\ 0 & 0 & 1-k^{ 2 } \end{matrix}|\begin{matrix} 1 \\ -3 \\ 1-k \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & { k }^{ 2 }-k+2 \end{matrix}\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \)
\(\rightarrow \left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & (k+2)(1-k) \end{matrix}|\begin{matrix} 1 \\ -3 \\ -k-2 \end{matrix} \right] \).........(1)
Case (i): when k = 1
\([A|B]\rightarrow \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ -3 \end{matrix} \right] \overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -2 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B] ⇒ The system has no solution
Case (ii): when k ≠ 2, k ≠ -2 
\(\left[ \begin{matrix} 1 & -2 & k \\ 0 & -2k+2 & 1-k \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 1 \\ -3 \\ not\quad zero \end{matrix} \right] \)
⇒ \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
so, \(\rho \)(A) =\(\rho \)[A|B] = 3 = the number of unknowns Hence, the system has unique solution.
Case (iii): when k = -2
\(\rho [A|B]\rightarrow \left[ \begin{matrix} 1 \\ 1 \\ 0 \end{matrix}\begin{matrix} -2 \\ 6 \\ 0 \end{matrix}\begin{matrix} -2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 1 \\ -3 \\ 0 \end{matrix} \right] \)
Here \(\rho \) (A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2<3 the number of unknowns so the system is consistent with infinitely many solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where A = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

Since v(3) = 64, v(6) = 133,and v(9) = 208 , we get the following system of linear equations
9a + 3b + c = 64 ,
36a + 6b + c = 133 ,
81a + 9b + c = 208 .
We solve the above system of linear equations by Gaussian elimination method.
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row
operations, we get
[A | B] = \(\left[ \begin{matrix} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{matrix}|\begin{matrix} 64 \\ 133 \\ 208 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & -6 & -3 \\ 0 & -18 & -8 \end{matrix}|\begin{matrix} 64 \\ -123 \\ -368 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -3 \right) ,{ R }_{ 3 }\div \left( -2 \right) }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 9 & 4 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 184 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow 2{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 18 & 8 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 368 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ -1 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 1 \end{matrix} \right] \).
Writing the equivalent equations from the row-echelon matrix, we get
9a + 3b + c = 64, 2b + c = 41, c = 1.
By back substitution, we get c = 1, b = \(\frac { \left( 41-c \right) }{ 2 } =\frac { \left( 41-1 \right) }{ 2 } \) = 20, a = \(\frac { 64-3b-c }{ 9 } =\frac { 64-60-1 }{ 9 } =\frac { 1 }{ 3 } \).
So, we get v(t) = \(\frac { 1 }{ 3 }\)t² + 20t + 1. Hence, v(15) = \(\frac { 1 }{ 3 }\) (225) + 20(15) + 1 = 75 + 300 + 1 = 376.

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.