\(\Delta =\left| \begin{matrix} 5 & 3 \\ 3 & 7 \end{matrix} \right| =5(7)-3(3)\)
= 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 =\(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -1 \\ -1 & 2 & 7 \end{matrix}\begin{matrix} 4 \\ -3 \\ 6 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho (A)\le \text{minimum} \ of(3,4)\)
\(\rho (A)\le 3\)
Let us transform the matrix A to an echelon form

The matrix is in echelon form and the number of non- zero rows is 2
\(\therefore \rho (A)=2\)

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

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

Let A =\(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
The order of A is 3 x 4
\(\therefore \rho \left( A \right) \le 3\) [Since minimum of (3, 3) is 3]
Let us transform the matrix to an echelon form.

The matrix is in echelon form and the number of non- zero rows is 2.
\(\therefore \rho (A)=2\)

Let A =  \(\left( \begin{matrix} -1 & 2 & -2 \\ 4 & -3 & 4 \\ -2 & 4 & -4 \end{matrix} \right) \)
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 to an echelon form

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (A)=2\)

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

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

Let A represents the percent of people who subscribe the magazine and B represents the percent of people who do not subscribe the magazine.
Given 60% of people subscribe again implies 40% of people do not subscribe. And 25% of people are going to subscribe implies 75% of people are not going to subscribe.
\(\therefore\) Transition probability matrix.

Also, it is given that 40% of those received the order of subscription implies 60% are not going to receive the order.
\(\left( \begin{matrix} \cdot 4 & \cdot 6 \end{matrix} \right) \left( \begin{matrix} \cdot 6 & \cdot 4 \\ \cdot 25 & \cdot 75 \end{matrix} \right) \)
= \(\left( \left( \begin{matrix} \cdot 4 & \cdot 6 \end{matrix} \right) +\left( \cdot 6 \right) \left( \cdot 25 \right) \left( \cdot 4 \right) \left( \cdot 4 \right) +\left( \cdot 6 \right) \left( \cdot 75 \right) \right) \)
= \(\left( \cdot 24+\cdot 15\quad \cdot 16+\cdot 45 \right) =\left( \cdot 39\quad \cdot 61 \right) \)
\(\therefore\) 39% of people who received the current letter can be expected to order a subscription.

Given non homogeneous equation are x +y += 7, x + 2y + 3z = 18, y + 2z = 6

Here \(\rho (A)=2\) and \(\rho (A,b)=3\)
Since \(\rho (A)\neq \rho (A,B)\) the given system is inconsistent and has no solution.

\(\Delta =\left| \begin{matrix} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(-18 + 12) - 4(6 - 30) +3 (- 4 +30)
= 1(- 6) - 4(- 24) + 3(26)
= - 6 + 96 + 78 = 168 \(\neq \) 0
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 2 & 4 & 3 \\ -3 & -6 & 6 \\ -5 & -2 & 3 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| +3\left| \begin{matrix} -3 & -6 \\ -5 & -2 \end{matrix} \right| \)
= 2 (- 18 + 12) - 4(- 9 +30) + 3(6 -30)
= 2(- 6) - 4(21) + 3(- 24)
= -12-84-72 =-168
\(\Delta y=\left| \begin{matrix} 1 & 2 & 3 \\ 2 & - & 6 \\ 5 & -5 & 3 \end{matrix} \right| =1\left| \begin{matrix} -3 & 6 \\ -5 & 3 \end{matrix} \right| -2\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| \)
= 1 (-9+30)-2(6-30)+3(- 10+ 15)
= 1(21) - 2(- 24) + 3(5)
= 21 + 48 + 15 = 84
\(\Delta z=\left| \begin{matrix} 1 & 4 & 2 \\ 2 & -6 & -3 \\ 5 & -2 & -5 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & -3 \\ -2 & -5 \end{matrix} \right| -4\left| \begin{matrix} 2 & -3 \\ 5 & -5 \end{matrix} \right| +2\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 1(30-6)-4(-10+ 15)+2(-4+30)
= 24 - 4(5) + 2(26)
= 24 - 20 + 52 = 56


Solution set is \(\left\{ -1,\frac { 1 }{ 2 } ,\frac { 1, }{ 3 } \right\} \)

\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 6 & -2 & -3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 3 & -1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 6 & 3 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 6 & -2 \end{matrix} \right| \)
= 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}

\(\Delta =\left| \begin{matrix} 2 & 1 & -1 \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{matrix} \right| =2\)
\(\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2 (-3+2) - 1 (-3 -1) -1 (-2-1)
= 2(-1) -1 (-4) -1 (-3)
= -2 + 4 + 3 = 5.
Since\(\Delta \neq 0\),
we can apply Cramer's rule and the system is consistent with unique solution.
\(x=\left| \begin{matrix} 3 & 1 & -1 \\ 1 & 1 & 1 \\ 4 & -2 & 3 \end{matrix} \right| =3\left| \begin{matrix} 1 & 1 \\ -2 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 4 & -2 \end{matrix} \right| \)
= 3(-3 + 2) -1(-3 -4) -1(-2 -4)
= 3 (-1) -1 (-7) -1 (-6)
= -3 + 7 + 6 = 10.
\(\Delta y=\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 1 & 4 & -3 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ 4 & -3 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 1 & -3 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| \)
= 2(-3 -4) -3 (-3 -1) -1 (4-1)
= 2 (-7) -3 (-4) -1(3)
= 14 + 12 - 3 = -5
\(\Delta z=\left| \begin{matrix} 2 & 1 & 3 \\ 1 & 1 & 1 \\ 1 & -2 & 4 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & 1 \\ -2 & 4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 1 \\ 1 & -2 \end{matrix} \right| \)
= 2(4 + 2) -1(4 -1) + 3(-2 -1)
= 2(6) -1(3) + 3(-3)
= 12 - 3 - 9
= 0

\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 0 }{ 5 } =0\)
\(\therefore\)Solution set is (2, -1, 0)
 

Let the rate of commission on the items A, B and C be x, y and z respectively.
By the given data, the non-homogeneous equations are
90x + 100y + 20z = 800
\(\Rightarrow\)9x + 10y + 2z = 80
130x + 50y + 40z = 900
\(\Rightarrow\) 13x + 5y + 4z = 90
60x + 100y + 30z = 850
\(\Rightarrow\) 6x + 10y + 3z = 85

\(\Delta =\left| \begin{matrix} 9 & 10 & 2 \\ 13 & 5 & 4 \\ 6 & 10 & 3 \end{matrix} \right| \)
\(9\left| \begin{matrix} 5 & 4 \\ 10 & 3 \end{matrix} \right| -10\left| \begin{matrix} 13 & 4 \\ 6 & 3 \end{matrix} \right| +2\left| \begin{matrix} 13 & 5 \\ 6 & 10 \end{matrix} \right| \)
= 9 (15 - 40) - 10 (39 - 24) + 2(130 - 30)
= 9 (- 25) - 10(15) + 2(100)
= - 225 - 150 + 200
= -175
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 80 & 10 & 2 \\ 90 & 5 & 4 \\ 85 & 10 & 3 \end{matrix} \right| \)
= \(80\left| \begin{matrix} 5 & 4 \\ 10 & 3 \end{matrix} \right| -10\left| \begin{matrix} 90 & 4 \\ 85 & 3 \end{matrix} \right| +2\left| \begin{matrix} 90 & 5 \\ 85 & 10 \end{matrix} \right| \)
= 80(15 - 40) - 10(270 - 340) + 2(900 - 425)
= 80 (- 25) - 10 (- 70) + 2 (475)
= - 2000 + 700 + 950
= -350
\(\Delta y=\left| \begin{matrix} 9 & 80 & 2 \\ 13 & 90 & 4 \\ 6 & 85 & 3 \end{matrix} \right| \)
= \(9\left| \begin{matrix} 90 & 4 \\ 85 & 3 \end{matrix} \right| -80\left| \begin{matrix} 13 & 4 \\ 6 & 3 \end{matrix} \right| +2\left| \begin{matrix} 13 & 90 \\ 6 & 85 \end{matrix} \right| \)
= 9(270 - 340) - 80(39 - 24) +2(1105 - 540)
= 9(- 70) - 80(15) + 2(565)
= - 630 - 1200 + 1130
= -700
\(\Delta z=\left| \begin{matrix} 9 & 10 & 80 \\ 13 & 5 & 90 \\ 6 & 10 & 85 \end{matrix} \right| \)
= \(9\left| \begin{matrix} 5 & 90 \\ 10 & 85 \end{matrix} \right| -10\left| \begin{matrix} 13 & 90 \\ 6 & 85 \end{matrix} \right| +80\left| \begin{matrix} 13 & 5 \\ 6 & 10 \end{matrix} \right| \)
= 9(425 - 900) - 10(1105 - 540) + 80(130 - 30)
= 9(- 475) - 10(565) + 80 (100)
= - 4275 - 5650 + 8000
= - 1925

\(\therefore\) The rate of commission on the items A, Band Care 2%, 4% and 11%

Let the cost of lkg of wheat be Rs. x, 1kg of sugar be Rs. y and lkg of rice be Rs. z.
By the given data,
2x + y = 100
x + z = 80
3x + 2y + z = 220
\(\Delta =\left| \begin{matrix} 2 & 1 & 0 \\ 1 & 0 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =2\left| \begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & 1 \end{matrix} \right| +0\)
= 2 (0 - 2) -1 (1 - 3) + 0
= 2(-2) - 1(- 2)
= - 4 + 2 = -2
\(\Delta x=\left| \begin{matrix} 100 & 1 & 0 \\ 80 & 0 & 1 \\ 220 & 2 & 1 \end{matrix} \right| =100\left| \begin{matrix} 0 & 1 \\ 2 & 1 \end{matrix} \right| -1\left| \begin{matrix} 80 & 1 \\ 220 & 1 \end{matrix} \right| +0\)
= 100(0 - 2) - 1 (80 - 220)
= 100(- 2) - 1(- 140)
= - 200 + 140 = - 60.
\(\Delta y=\left| \begin{matrix} 2 & 100 & 0 \\ 1 & 80 & 1 \\ 3 & 220 & 1 \end{matrix} \right| =2\left| \begin{matrix} 80 & 1 \\ 220 & 1 \end{matrix} \right| -100\left| \begin{matrix} 1 & 1 \\ 3 & 1 \end{matrix} \right| +0\)
= 2 (80 - 220) - 100 (1 - 3)
= 2 (- 140) - 100 (-2)
= - 280 + 200 = - 80.
\(\Delta z=\left| \begin{matrix} 2 & 1 & 100 \\ 1 & 0 & 80 \\ 3 & 2 & 220 \end{matrix} \right| \)
\(2\left| \begin{matrix} 0 & 80 \\ 2 & 220 \end{matrix} \right| -1\left| \begin{matrix} 1 & 80 \\ 3 & 220 \end{matrix} \right| +100\left| \begin{matrix} 1 & 0 \\ 3 & 2 \end{matrix} \right| \)
= 2(0 - 160) - 1(220 - 240) + 100(2 - 0)
= 2(- 160) - 1(- 20) + 100(2)
= - 320 + 20 + 200
= -100
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -60 }{ -2 } =30\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -80 }{ -2 } =40\)
\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { -100 }{ -2 } =50\)
\(\therefore\) The cost of 1 kg of wheat is Rs. 30
The cost of 1 kg sugar is Rs. 40 and The cost of 1 kg of rice is Rs. 50

\(\Delta =\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 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}

x +y + z = 1, 3x - y - z = 4, x + 5y + 5z = k

Here clearly \(\rho (A)=2\)
Since the given system is inconsistent \(\rho (A)\neq \rho (A,B)\)
This can take any value other than zero.
\(\therefore\) k can take any value other than zero.

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

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

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%

Given \(\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x+0+x \\ 0+y+0 \\ 0+0+z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x=1,y=-1,z=0\)

The matrix A is singular, if
\(\left| \begin{matrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{matrix} \right| =0\)
\(1\left| \begin{matrix} 2 & 1 \\ 2 & -3 \end{matrix} \right| +2\left| \begin{matrix} 1 & 1 \\ x & -3 \end{matrix} \right| +3\left| \begin{matrix} 1 & 2 \\ x & 2 \end{matrix} \right| =0\)
\(\Rightarrow\) (-8) -6 - 2x + 6 - 6x = 0
\(\Rightarrow\) -8-2x-6x = 0
\(\Rightarrow\) -8-8x = 0
\(\Rightarrow\) -8 = 8x
\(\Rightarrow\) \(x=\cfrac { -8 }{ 8 } =-1\)

Since A and B are non-singular,
|A| \(\neq \) 0, |B|\(\neq \) 0
Consider |AB| |A|·|B|
\(\neq \) 0 since |A|\(\neq \) 0 and |B|\(\neq \) 0.=? |AB| \(\neq \) 0
\(\therefore\) AB is non-singular.

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 4 & 6 \end{matrix} \right| =12-12=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\Delta x=\left| \begin{matrix} 4 & 3 \\ 8 & 6 \end{matrix} \right| =24-24=0\)
\(\therefore \Delta =\Delta x=\Delta y=0\)
\(\therefore \) 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

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

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

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

\(\Delta =\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| =1(1)-(1)(2)=1-2=-1\)
Since \(\Delta \neq O\)  Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| =3-4=-1\)
\(\Delta y=\left| \begin{matrix} 1 & 3 \\ 1 & 2 \end{matrix} \right| =2-3=-1\)
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
\(y=\cfrac { \Delta x }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
\(\therefore\) solution set is {1, 1}

Let A = \(\left( \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right) \)
The order of A is 2 \(\times\) 2
\(\rho (A)\le min(2,2)\)
\(\Rightarrow \rho (A)\le 2\)
\(\left| \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right| =4-4=0\)
Since the second order minor vanishes \(\rho (A)\neq 2\)
 We have to try for atleast one non-zero first order minor.
ie. atleast one non-zero element of A.
This is possible because A has non-zero element
\(\therefore \rho (A)-1\)

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
\(\therefore \rho (A)=2\)

Let A = \(\left( \begin{matrix} 1 & 4 \\ 2 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2 [Since minimum of (2,2) is 2]
Consider the second order minor \(\left| \begin{matrix} 1 & 4 \\ 2 & 8 \end{matrix} \right| \)
= 8-8
= 0
Since the second order minor vanishes \(\rho (A)\neq 2\)
Consider a first order minor \(\left[ 1 \right] \neq 0\)
There is a minor of order 1, which is not zero
\(\therefore \rho \left( A \right) =1\)

Let \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\) [Since minimum of (2, 2) is 2]
 Consider the second order minor
\(\left| \begin{matrix} 1 & -1 \\ 3 & -6 \end{matrix} \right| =-6-(-3)\)
= -6 + 3 = -3
\(\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)