\(\Delta =\left| \begin{matrix} 5 & 3 \\ 3 & 7 \end{matrix} \right| =5(7)-3(3)\)
= 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.

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
\(\therefore \rho (A)=3\)

Given A =\(\left( \begin{matrix} -2 & 1 & 3 \\ 0 & 1 & 1 \\ 1 & 3 & 4 \end{matrix}\begin{matrix} 4 \\ 2 \\ 7 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ -2 & 1 & 3 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 1 }\rightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 7 & 11 \end{matrix}\begin{matrix} 7 \\ 2 \\ 18 \end{matrix} \right) { R }_{ 2 }\rightarrow { { R_{ 2 }+2{ R }_{ 1 } } }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 3 }\rightarrow { R_{ 3 }-{ 7R }_{ 2 } }\)
The last equivalent matrix is in echelon form and there are 3 non - zero rows.
\(\therefore \rho (A)=3\)

Transition probability matrix

Where A represents the percentage of subscribers and B represents the percentage of non - subscribers.
A 40% = ·40
By the given data, 40% received the order of subscription = 60% are non-subscribers.
A 40% = ·40
and B 60% = ·60

((-40)(-45) + (·60)(-30) (-40)(.55) + (-60)(.70))
(-18+·18 ·22+42)
(·36 ·64)
\(\Rightarrow\) 36% of those receiving the current letter can be expected to order a subscription

The Transition probability matrix is T = \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \)
At equilibrium, (S F) \(\left( \begin{matrix} 0.25 & 0.75 \\ 0.35 & 0.65 \end{matrix} \right) \) = (S F) 
where S + F = 1
0.25 S + 0.35 F = S
0.25 S + 0.35 (1 – S) = S
On solving this, we get S = \(\frac { 0.35 }{ 1.10 } \)
⇒ S = 0.318 and F = 0.682 
\(\therefore \)  Akash’s batting average is 31.8%

The transition porbability matrix is T = \(\left( \begin{matrix} \frac { 1 }{ 3 } & \frac { 2 }{ 3 } \\ \frac { 4 }{ 5 } & \frac { 1 }{ 5 } \end{matrix} \right) \)
At equilibrium, (S H) \(\left( \begin{matrix} \frac { 1 }{ 3 } & \frac { 2 }{ 3 } \\ \frac { 4 }{ 5 } & \frac { 1 }{ 5 } \end{matrix} \right) \) = (S H)
where S + H = 1
\(\frac { 4 }{ 5 } S+\frac { 1 }{ 3 } H=S\)
\(\frac { 1 }{ 3 } (1 - H)+\frac { 4 }{ 5 } H= 1- H\)
On solving this, we get S = \(\frac { 6 }{ 11 } \) and H = \(\frac { 5 }{ 11 } \)
In the long run, on a randomly selected day, his chances of being happy is \(\frac { 5 }{ 11 } \)

Transition probability matrix
T = \(_{ B }^{ A }\left( \begin{matrix} \overset { A }{ 0.9 } & \overset { B }{ 0.1 } \\ 0.3 & 0.7 \end{matrix} \right) \)
At equilibrium, (A B) T = (AB) where A + B = 1
(A B) \(\left( \begin{matrix} 0.9 & 0.1 \\ 0.3 & 0.7 \end{matrix} \right) \) = (A B)
0.9A + 0.3B = A
0.9A + 0.3(1−A) = A
0.9A−0.3A + 0.3 = A
0.6A + 0.3 = A
0.4A = 0.3
A = \(\frac { 0.3 }{ 0.4 } =\frac { 3 }{ 4 } \)
B = 1-\(\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
Hence the market share of brand A is 75% and the market share of brand B is 25%

Let the hourly charge for horse riding be Rs. x and the hourly charge for quad bike be Rs. y from the given data,
3x + 4y = 780
2x+ 3y = 560
\(\Delta =\left| \begin{matrix} 3 & 4 \\ 2 & 3 \end{matrix} \right| =3(3)-2(4)=9-8=1\neq 0\)
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied
\(\Delta x=\left| \begin{matrix} 780 & 4 \\ 560 & 3 \end{matrix} \right| =780(30-4(560)\)
= 2340 - 2240 = 100
\(\Delta y=\left| \begin{matrix} 3 & 780 \\ 2 & 560 \end{matrix} \right| =39560)-2(780)\)
= \(1680 - 1560\) = 120
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { 100 }{ 1 } =100\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 120 }{ 1 } =120\)
\(\therefore\)  Hourly charges for the two rides are Rs. 100 and Rs. 120 respectively.

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. 

Let Rs. x represents the cost per unit of labour and Rs. y represents the cost per unit of capital
Given
3x + 2y = 62
4x +y = 56
\(\Delta =\left| \begin{matrix} 3 & 2 \\ 4 & 1 \end{matrix} \right| =3(1)-4(2)=3-8=-5\)
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied.
\(\Delta x=\left| \begin{matrix} 62 & 2 \\ 56 & 1 \end{matrix} \right| =62(1)-56(2)=62-112=-50\)

\({ \Delta }_{ y }=\left| \begin{matrix} 3 & 62 \\ 4 & 56 \end{matrix} \right| =3(56)-4(62)\)
= 168 - 248 = -80

\(\therefore\) Cost per unit oflabour is Rs. 10 and the cost per unit of capital is Rs. 16.

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =10-9=1\neq 0\)
Since \(\Delta \neq 0\) 
we can apply Cramer's rule and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =7(5)-9(3)\)
= 35 - 27 = 8
\(\Delta y=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =2(9)-3(7)\)
= 18 - 21 = -3
 \(\therefore\) \(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 8 }{ 1 } =8\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
\(\therefore\) Solution set is (8, -3)

Let ‘x’ be the cost of a pencil
Let ‘y’ be the cost of an eraser
\(\therefore \) By given data, we get the following equations
11x + 3y = 64
8x + 3y = 49
\(\triangle =\left| \begin{matrix} 11 & 3 \\ 8 & 3 \end{matrix} \right| =9\neq 0,\) It has unique solution.
\({ \triangle }_{ x }\left| \begin{matrix} 64 & 3 \\ 49 & 3 \end{matrix} \right| =45\)
\({ \triangle }_{ y }\left| \begin{matrix} 11 & 64 \\ 8 & 49 \end{matrix} \right| =27\)
\(\therefore \) By Cramer’s rule
\(x={ \frac { \triangle x }{ \triangle } =\frac { 45 }{ 9 } =5 }\)
\(y={ \frac { \triangle y }{ \triangle } =\frac { 27 }{ 9 } =3 }\)
\(\therefore \) The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Let the price of one share of A be x
Let the price of one share of B be y
\(\therefore \) By given data, we get the following equations
10x + 5y = 125
9x + 12y = 150
\(\triangle =\left| \begin{matrix} 10 & 5 \\ 9 & 12 \end{matrix} \right| =75\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 125 & 5 \\ 150 & 12 \end{matrix} \right| =750\)
\({ \triangle }_{ y }=\left| \begin{matrix} 10 & 125 \\ 9 & 150 \end{matrix} \right| =375\)
\(\therefore \) Cramer’s rule 
\(x=\frac { \triangle x }{ \triangle } =\frac { 750 }{ 75 } =10\)
\( y=\frac { \triangle y }{ \triangle } =\frac { 375 }{ 75 } =5\)
The price of the share A is Rs10 and the price of the share B is Rs. 5.

The equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

Given non-homogeneous equations are
x + y + z = 3
x + 2y + 3z = 4
x + 4y + 9z = 6
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 11 & 4 & 9 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 4 \\ 6 \end{matrix} \right) \)
AX = B

Clearly the last equivalent matrix is in echelon form and it has three non-zero rows
\(\therefore \rho (A)=3\quad \rho \left( \left[ A,B \right] \right) =3\)
\(\rho (A)=\rho \left( \left[ A,B \right] \right) =3\)
\(\therefore\) The given system is consistent and has unique solution.
To find the solution, let us rewrite the above echelon form into the matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ 1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow\) x + y + z = 3
y + 2z = 1
\((3)\Rightarrow 2x=0\Rightarrow z=\cfrac { 0 }{ 2 } =0\)
\((2)\Rightarrow y+2(0)=1\Rightarrow y+0=1\Rightarrow y=1-0=1\)
\(\left( 1 \right) \Rightarrow x+1+0=3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)
\(\therefore\) Solution set [2, 1, 0]

Given 
A = \(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B = \(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
\(AB=\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-2+5 & -2+4-1 & 3-6+1 \\ 2+6+20 & -4-12+45 & 6+18-4 \\ 3+4+15 & -6-8+3 & 9+12-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) =\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (AB)=2\)
Now  \(BA=\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-4+9 & 1+6-6 & -1-8+9 \\ -2+8-18 & -2-12+12 & 2+16-18 \\ 5+2-3 & 5-3+2 & -5+4-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)

The number of non-zero rows is 2.
\(\therefore \rho (BA)=2\)

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 3 & -2 \\ 6 & -4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \)
AX = B

\(\therefore \)\(\rho ([A,B])=2\), \(\rho (A)=1\)
\(\rho (A)\neq \rho \left( [A,B] \right) \)
\(\therefore \) The given system is inconsistent and has no solution.

The matrix equation corresponding to the system is
\(\left( \begin{matrix} 2 & 1 \\ 4 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 10 \end{matrix} \right) \)
A X = B

\(\rho (A)=\rho ([A,B])=1<\)number of unknowns
\(\therefore \)The given system is consistent and has infinitely many solutions.
Now, the given system is transformed into the matrix equation.
\(\left( \begin{matrix} 2 & 1 \\ 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 0 \end{matrix} \right) \)
\(\Rightarrow 2x+y=5\)
Let us take y = k,k \(\in \)R
\(\Rightarrow 2x+k=5\)
\(x=\frac { 1 }{ 2 } (5-k)\) 
\(x=\frac { 1 }{ 2 } (5-k),y=k\ for\ all\ k\in R\)
Thus by giving different values for k, we get different solution.
Hence the system has infinite number of solutions.

The matrix equation corresponding to the given system is
\(\begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix}\left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ 8 \end{matrix} \right) \)
A X = B

Number of non-zero rows is 2.
\(\rho (A)=\rho ([A,B])=2=\) Number of unknowns.
The given system is consistent and has unique solution.
Now, the given system is transformed into
\(\left( \begin{matrix} 0 & 1 \\ 0 & -1 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 5 \\ -2 \end{matrix} \right) \)
x + y = 5
y = 2
\(\therefore (1)\Rightarrow x+2=5\)
x = 3
Solution is x = 3, y = 2

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

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

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

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

The order of A is 3 \(\times\) 3.
\(\therefore \rho (A)=\le 3.\)
Let us transform the matrix A to an echelon form by using elementary transformations.

The number of non zero rows is 2
∴ Rank of A is 2.
\(\rho (A)=2.\)

Let A = \(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
Order of A is 3 \(\times\) 4
∴\(\rho \)(A)\(\le \)3
Consider the third order minors
\(\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix} \right| =\) 0, \(\left| \begin{matrix} 1 & -1 & 3 \\ 2 & 1 & -2 \\ 3 & 3 & -7 \end{matrix} \right| =0\) 
\(\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & -2 \\ 3 & 6 & -7 \end{matrix} \right| =0,\) \(\left| \begin{matrix} 2 & -1 & 3 \\ 4 & 1 & -2 \\ 6 & 3 & -7 \end{matrix} \right| =0\)
Since all third order minors vanishes, \(\rho (A)\neq 3\)
Now, let us consider the second order minors,
Consider one of the second order minors \(\\ \\ \left| \begin{matrix} 2 & -1 \\ 4 & 1 \end{matrix} \right| =6\neq 0\)
There is a minor of order 2 which is not zero.
\(\therefore \rho (A)=2\)

Let A= \(\left( \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3.
∴\(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right| =0\)
Since the third order minor vanishes, therefore \(\rho (A)\neq 3\)
Consider a second order minor \(\left| \begin{matrix} 5 & 3 \\ 1 & 2 \end{matrix} \right| =7\neq 0\)
There is a minor of order 2, which is not zero.
\(\therefore \rho (A)=2\)

Let A =\(\left( \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3  
∴ \(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 0 & -1 & 5 \\ 2 & 4 & -6 \\ 1 & 1 & 5 \end{matrix} \right| =6\neq 0\)
There is a minor of order 3, which is not zero 
\(\therefore \rho (A)=3\)