\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 6 & -2 & -3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 3 & -1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 6 & 3 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 6 & -2 \end{matrix} \right| \)
= 1(-9 -2) -1(-6 +6) + 1(-4 -18)
= 1(-11) -1(0) +1(-22)
= -11 -22 = -33 \(\neq \)0
Since \(\Delta \neq 0\) 
Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 6 & 1 & 1 \\ 5 & 3 & -1 \\ -7 & -2 & -3 \end{matrix} \right| \)
= \(6\left| \begin{matrix} 3 & -1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 5 & -1 \\ -7 & -3 \end{matrix} \right| +1\left| \begin{matrix} 5 & 3 \\ -7 & -2 \end{matrix} \right| \)
= 6 (-9 -2) -1(-15 -7) + 1(-10 +21)
= 6 (-11) -1 (-22) + 1 (11)
= -66 + 22 + 11= - 33
\(\Delta y=\left| \begin{matrix} 1 & 6 & 1 \\ 2 & 5 & -1 \\ 6 & -7 & -3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 5 & -1 \\ -7 & -3 \end{matrix} \right| -6\left| \begin{matrix} 2 & -1 \\ 6 & -3 \end{matrix} \right| +1\left| \begin{matrix} 2 & 5 \\ 6 & -7 \end{matrix} \right| \)
= 1(-15 -7) -6(-6 +6) + 1(-14 -30)
= 1(-22) -6(0) + 1 (-44)
= -22 - 44 = - 66
\(\Delta z=\left| \begin{matrix} 1 & 1 & 6 \\ 2 & 3 & 5 \\ 6 & -2 & -7 \end{matrix} \right| \)
= \(=1\left| \begin{matrix} 3 & 5 \\ -2 & -7 \end{matrix} \right| -1\left| \begin{matrix} 2 & 5 \\ 6 & -7 \end{matrix} \right| +6\left| \begin{matrix} 2 & 3 \\ 6 & -2 \end{matrix} \right| \)

= 1(-21 +10) -1(-14 -30) +6 (-4 -18)
=1(-11) -1(-44) +6(-22)
= -11 + 44 - 132 = - 99

\(\therefore\) Solution set is {1, 2, 3}

\(\Delta =\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1(2 -2) - 2(-4 -2) + 1(2 + 1)
= 1(0)-2(-6)+1(3)
= 12 + 3 = 15\(\neq \)0.
Since \(\Delta \neq 0\) Cramer's rule can be applied and thesystem is consistent with unique solution.
\({ \Delta }x=\left| \begin{matrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{matrix} \right| \)
= \(7\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 4 & -1 \\ -1 & 1 \end{matrix} \right| \)
= 7 (2 -2) -2 (-8 + 2) + 1 (4 - 1)
= 7 (0) - 2(-6) + 1(3)
= 12 + 3 = 15
\(\Delta y=\left| \begin{matrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| -7\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| \)
= 1 (- 8 + 2) -7(-4 -2) + 1(-2 -4)
= 1 (-6) -7 (-6) + 1 (-6)
= - 6 + 42 - 6 = 30
\(\Delta z=\left| \begin{matrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 4 \\ 1 & -1 \end{matrix} \right| -2\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| +7\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1 (1 - 4) - 2(- 2 - 4) + 7(2 + 1)
= 1(-3)-2(-6)+7(3)
= - 3 + 12 + 21 = 30

\(\therefore\) Solution set is {1, 2, 2}

Given non-homogeneous equations are
2x + 3y - z = 5, 3x - y + 4z = 2, x + 7y - 6z = k

Augmented matrix	Elementary Transformation
\(\left( \begin{matrix} 2 & 3 & -1 \\ 3 & -1 & 4 \\ 1 & 7 & -6 \end{matrix}\begin{matrix} 5 \\ 2 \\ k \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 7 & -6 \\ 3 & -1 & 4 \\ 2 & 3 & -1 \end{matrix}\begin{matrix} k \\ 2 \\ 5 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\left( \begin{matrix} 1 & 7 & -6 \\ 0 & -22 & 22 \\ 0 & -11 & 11 \end{matrix}\begin{matrix} k \\ 2-3k \\ 5-2k \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 7 & -6 \\ 0 & -22 & 22 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} k \\ 2-3k \\ 2(5-2k)-(2-3k) \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 7 & -6 \\ 0 & -22 & 22 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} k \\ 2-3k \\ 10-4k-2+3k \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow 2{ R }_{ 3 }-{ R }_{ 2 }\)
\(\left( \begin{matrix} -1 & 7 & -6 \\ 0 & -22 & 22 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} k \\ 2-3k \\ 8-k \end{matrix} \right) \)

Here \(\rho (A)=2\)
Since the given system is consistent, \(\rho \)(A, B) must be equal to 2.
This can happen only when
8 - k = 0 \(\Rightarrow\) k = 8

Transition probability matrix
(A B) T = (A B)

Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought soap B.
By the given data
A = 15% = ·15
and B = 85% = ·85
Percentage after one year is
\(\left( \cdot 15\quad \cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share after one year is 48% and 52% At equilibrium,
\(\left( A\quad B \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) =(A\quad B)\)
(-65A + A5B ·35A +·55B) = (A B)
Equating the corresponding entries on both sides we get
\(\Rightarrow \cdot 65A+\cdot 45B=A\)
\(\Rightarrow \cdot 65A+\cdot 45(1-A)=A\)
[Since A+B = 1   B = 1-A]
\(\Rightarrow \cdot 65A+\cdot 45-\cdot 45A=A\)
\(\Rightarrow \cdot 45=A-\cdot 65A+\cdot 45A\)
\(\Rightarrow \cdot 45=A\left( \cdot 35+45 \right) \)
\(\Rightarrow \cdot 45=A(\cdot 35+45)\)
\(\Rightarrow \cdot 45=A(-8)\)
\(\Rightarrow A=\cfrac { \cdot 45 }{ \cdot 8 } =\cdot 5625=56.25\)
\(\therefore B=1-A=1-\cdot 5625=\cdot 4375\)
= 43.75%
\(\therefore\) Equilibrium is reached when A = 56.25% and B = 43.75%

Let the amount of investment in each bond be
Rs. x, Rs. y, Rs. z respectively.
Given x + y + z = 5000    ..(1)
Also \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } +\cfrac { 8z }{ 100 } =358\)
∴ Interes \(= \cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 6 }{ 100 } =\cfrac { 6x }{ 100 } \)
\(\Rightarrow \cfrac { 6x+7y+8z }{ 100 } =358\)
\(\Rightarrow 6x+7y++8z=35800\)
Given that \(\cfrac { 6x }{ 100 } +\cfrac { 7y }{ 100 } =70+\cfrac { 8z }{ 100 } \)
\(\Rightarrow 6x+7y=7008z\)
\(\Rightarrow 6x+7y-8z=7000\)
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right) \)

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 1 & 1 & 1 \\ 6 & 7 & 8 \\ 6 & 7 & -8 \end{matrix}\begin{matrix} 5000 \\ 35800 \\ 7000 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & -14 \end{matrix}\begin{matrix} 5000 \\ 5800 \\ -2300 \end{matrix} \right) \)	\(\ { R }_{ 2 }\rightarrow { R }_{ 2 }-6{ R }_{ 1 }\) \({ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ 6R }_{ 1 } }\)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -16 \end{matrix}\begin{matrix} 5000 \\ 5800 \\ -28800 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

The last equivalent matrix is in echelon form and ρ(A) = ρ([A,B]) = 3 = Number of unknowns
Thus, the given system is consistent with unique solution. To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & -16 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 5000 \\ 5800 \\ -28800 \end{matrix} \right) \)
\(\Rightarrow x+y+z=5000\)   ...(1)
\(\Rightarrow y+2z=5800\)      ..(2)
\(\Rightarrow -16z=-28800\)    ...(3)
\((3)\Rightarrow -16z=-28800\)
\(\Rightarrow z=\cfrac { 28800 }{ -16 } =1800\)
Substituting z = 1800 in (2) we get,
y + 2(1800) = 5800
\(\Rightarrow\) y + 3600 = 5800
\(\Rightarrow\)  y=5800-3600
\(\Rightarrow\) y = 2200
Substituting y = 2200 and z = 1800 in (1) we get
\(\Rightarrow\) x + 2200 + 1800 = 5000
\(\Rightarrow\)x + 4000 = 5000
\(\Rightarrow\)x = 5000 - 4000
\(\Rightarrow\)x = 1000
Hence, the amount of investment in each bond is Rs. 1000, Rs. 2200 and Rs. 1800 respectively

Given non-homogeneous equations are
\(3x-y+\lambda z=1\)
\(2x+y+z=2\)
\(x+2y-\lambda z=-1\)
The matrix equation corresponding to the given system is

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 3 & -1 & \lambda \\ 2 & 1 & 1 \\ 1 & 2 & - \end{matrix}\begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 2 & 1 & 1 \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 0 & -7 & 4\lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2\lambda }{ 3 } \\ 0 & - & \frac { 4\lambda }{ 7 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { 4 }{ 7 } \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 3\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 7\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2 }{ 3 } \\ 0 & 0 & \frac { -7-2\lambda }{ 21 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { -16 }{ 21 } \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

Since
\(\cfrac { 4\lambda }{ 7 } -\cfrac { 1+2\lambda }{ 3 } \)
= \(\cfrac { 12\lambda -7-14\lambda }{ 21 } =\cfrac { -7-2\lambda }{ 21 } \)
and \(\cfrac { 4 }{ 7 } -\cfrac { 4 }{ 3 } =\cfrac { 12-28 }{ 21 } \)
= \(\cfrac { -16 }{ 21 } \)
\(\therefore\) Since the system is fail to have unique solution either it can have infinitely many solution or it may be inconsistent.
This can happen only when \(\cfrac { -7-2\lambda }{ 21 } =0\)
\(\Rightarrow -7-2\lambda =0\)
\(\Rightarrow -7=2\lambda \)
\(\Rightarrow \lambda =\cfrac { -7 }{ 2 } \)

Given non-homogeneous equations are
x + yz = 9
2x + 5y + 7z = 52
2x - y - z = -6
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & -1 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 9 \\ 52 \\ -6 \end{matrix} \right) \)
Augmented matrix Elementary Transformation  \(\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 5 & 7 \\ 2 & -1 & -1 \end{matrix}\begin{matrix} 9 \\ 52 \\ -6 \end{matrix} \right) \)   \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 2 & -1 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ -1 & -6 \end{matrix} \right) \) \({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\) \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 0 & -3 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ -3 & -24 \end{matrix} \right) \) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }_{ 1 }\) \(\sim \left( \begin{matrix} 1 & 1 \\ 0 & 3 \\ 0 & 0 \end{matrix}\begin{matrix} 1 & 9 \\ 5 & 34 \\ 2 & 10 \end{matrix} \right) \) \({ R }_{ 3 }\rightarrow { R }_{ 3 }+{ R }_{ 2 }\) The last equivalent matrix is in echelon form andit has three non-zero rows.
\(\therefore \rho (A)=3\)
\(\rho \left( \left[ A,B \right] \right) =3\) Number of unknowns
\(\therefore\) The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon from into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 9 \\ 34 \\ 16 \end{matrix} \right) \)
\(\Rightarrow \) x + y + z = 9 .. (1)
3y + 5z = 34  ..(2)
2z = 16   ..(3)
From (3), \(2z=16\Rightarrow z=\cfrac { 16 }{ 2 } =8\)
Substituting Z = 8 in (2) we get,
3y + 5 (8) = 34
\(\Rightarrow\) 3y + 40 = 34
\(\Rightarrow\) 3y = 34 - 40
\(\Rightarrow\) 3y = -6
y = -2
Substituting y = -2 and z = 8 in (1) we
\(\Rightarrow\) x -2 + 8 = 9
\(\Rightarrow\) x + 6 = 9
\(\Rightarrow\)x = 9-6
\(\Rightarrow\)x = 3
\(\therefore\) Solution set is {3, -2, 8}

\(\Delta =\left| \begin{matrix} 5 & 3 \\ 3 & 7 \end{matrix} \right| =5(7)-3(3)\)
= 119 - 93 = 26
\(\Delta x=\left| \begin{matrix} 17 & 3 \\ 31 & 7 \end{matrix} \right| =17(7)-31(3)\)
= 119 - 93 = 2
\(\Delta y=\left| \begin{matrix} 5 & 17 \\ 3 & 31 \end{matrix} \right| =5(31)-17(3)\)
= 155 - 51 = 104

\(\therefore\) Solution set is (1, 4)

A =\(\left( \begin{matrix} 3 & 1 & -5 \\ 1 & -2 & 1 \\ 1 & 5 & -7 \end{matrix}\begin{matrix} -1 \\ -5 \\ 2 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho (A)\le \text{minimum} \ of(3,4)\)
\(\therefore \rho (A)\le 3\)
Let us transform the matrix A to an echelon form

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} 3 & 1 & -5 \\ 1 & -2 & 1 \\ 1 & 5 & -7 \end{matrix}\begin{matrix} -1 \\ -5 \\ 2 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 1 & -2 & 1 \\ 3 & 1 & -5 \end{matrix}\begin{matrix} 2 \\ -5 \\ -1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 3 & 1 & -5 \end{matrix}\begin{matrix} 2 \\ -7 \\ -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 0 & -14 & 16 \end{matrix}\begin{matrix} 2 \\ -7 \\ -7 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }+{ 3R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 5 & -7 \\ 0 & -7 & 8 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 2 \\ -7 \\ 7 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 2R }^{ 2 }\)

The matrix is in echelon form and the number of non-zero matrix is 3
\(\therefore \rho (A)=3\)

Let \(A=\left( \begin{matrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{matrix} \right) \)
The order of A is 3 x 3
\(\therefore \rho (A)\le 3\) [Since minimum of (3,3) is 3]
Let us transform the matrix A to an echelon form

Matrix A	Elementary Transformation
\(A=\left( \begin{matrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 3 & 1 & -5 \\ 2 & -1 & 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -2 & -8 \\ 0 & -3 & -1 \end{matrix} \right) \)	\({ { R }_{ 2 }\rightarrow { R }_{ 2 }3{ R }_{ 1 } }\) \({ R }_{ 3 }-{ R }_{ 3 }-2{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -4 \\ 0 & -3 & -1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 2\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & -4 \\ 0 & 0 & 11 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 2 }\)

This matrix is in echelon from and number of non zero rows is 3.
\(\therefore \rho (A)=3\)

Transition probability matrix

Given present market shares are 15% for A and 85% for B
\(\therefore\) Market shares after one year 
= \(\left( \cdot 15\cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((-15)(-65)+(-85)(-45) ·15x·35+·85x·55)
= (-0975 + 0.3825 .0525 + 4675)
= (0 .48 0.52)
\(\therefore\) Market shares after one year for A is 48% and for B is 52%

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right| =12-12=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\therefore \Delta =\Delta x=\Delta y=0\)
\(\therefore \) The system is consistent with infinite number of solutions
let y = k, \(k\epsilon R\)
\(\therefore 2x+3k=4\Rightarrow 2x=4-3k\)
\(\Rightarrow x=\cfrac { 1 }{ 2 } \left( 4-3k \right) ,k\epsilon R\)
\(\therefore \) Solution set is \(\left\{ \cfrac { 4-3k }{ 2 } ,k \right\} ,k\epsilon R\)

Given non-homogeneous equations are x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30

Augmented matrix	Elementary Transformation
\(\left( \begin{matrix} 1 & 2 & \begin{matrix} 1 & 6 \end{matrix} \\ 1 & 2 & \begin{matrix} 3 & 14 \end{matrix} \\ 1 & 4 & \begin{matrix} 7 & 30 \end{matrix} \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 3 & 6 \end{matrix}\begin{matrix} 6 \\ 8 \\ 24 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)
\(-\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 6 \\ 8 \\ 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }\)

Here \(\rho (A)=\rho (A,B) = 2\) 
\(\therefore\) The given system is consistent

The non-homogeneous equations are
x + 2y = 3, 2x + 4y = 6

Augmented matrix [A, b]	Elementary Transformation
\(\left( \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \end{matrix} \right) \)
\(-\left( \begin{matrix} 1 & 2 & 3 \\ 0 & 0 & 0 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\)

Here \(\rho (A)=1\) and \(\rho \left( \left[ A,B \right] \right) =1\)
Since \(\rho (A)=\rho \left[ \left( A,B \right) \right] =1<\) Number of unknowns, the given system is consistent with infinitely many solutions.
To find the solution, let us rewrite the above echelon form into the matrix form, we get
\(\left( \begin{matrix} 1 & 2 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 3 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x+2y=3\)
let \(y=k,k\varepsilon R\)
\((1)\Rightarrow x+2k=3\Rightarrow x=3-2k\)
\(\therefore\) Solution set is \(\left\{ 3-2k,k \right\} ,k\epsilon R\)
For different values of k; we get infinite number of solutions

Let \(A=\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] \)
The order of A is 2 x 2
\(\rho (A)\le min(2,2)\)
\(\left[ \begin{matrix} 7 & -1 \\ 2 & 1 \end{matrix} \right] =7-(-2)=7+29\neq 0\)
The highest order of non-vanishing minor of A is 2
\(\therefore \rho (A)=2\)