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 \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

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

Let A =\(\begin{pmatrix} 1 & 5 \\ 3 & 9 \end{pmatrix}\)
Order of A is 2 \(\times\) 2
∴ \(\rho \) (A) \(\le \) 2
Consider the second order minor
\(\begin{vmatrix} 1 & 5 \\ 3 & 9 \end{vmatrix}=-6\neq 0\)
There is a minor of order 2, which is not zero.
∴ \(\rho \) (A) \(\le \) 2

The order of A is 3 x 4
\(\therefore \rho (A)\le min\left( 3,4 \right) \)
\(\rho (A)\le 3\)
Consider the third order minor
\(\left| \begin{matrix} 1 & 2 & -4 \\ 2 & -1 & 3 \\ 8 & 1 & 9 \end{matrix} \right| =1\left| \begin{matrix} -1 & 3 \\ 1 & 9 \end{matrix} \right| -2\left| \begin{matrix} 2 & 3 \\ 8 & 9 \end{matrix} \right| -4\left| \begin{matrix} 2 & -1 \\ 8 & 1 \end{matrix} \right| \)
= 1(- 9 - 3) - 2(18 - 24) - 4(2 + 8)
= 1 (-12) - 2 (- 6) - 4 (10)
= - 12 + 12 - 40
= - 40::\(\neq \) 0.
There is a minor of order 3, which is not zero
\(\therefore \rho (A)=3\)

The order of A is 3 x 3
\(\therefore \rho (A)\le min(3,3)\)
\(\Rightarrow \rho (A)\le 3\)

Matrix	Elementary Transformation
\(\left( \begin{matrix} 2 & 4 & 5 \\ 4 & 8 & 10 \\ -6 & -12 & -15 \end{matrix} \right) \)
\(-\left( \begin{matrix} 1 & 1 & 1 \\ 2 & 2 & 2 \\ -3 & -3 & -3 \end{matrix} \right) \)	\({ C }_{ 1 }\rightarrow { C }_{ 1 }\div 2\) \({ C }_{ 2 }\rightarrow { C }_{ 2 }\div 4\) \({ C }_{ 3 }\rightarrow { C }_{ 3 }\div 5\)
\(-\left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 1 }-{ 2R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }+3{ R }_{ 1 }\)

The last equivalent matrix is in echelon form and it has one non-zero row
\(\therefore \rho (A)=1\)

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

Matrix (AB)	Elementary Transformation
\(AB=\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \)	\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ -12 & 28 & 20 \\ -11 & 22 & 18 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+12{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & -44 & -4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }+11{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -6 & -2 \\ 0 & -44 & -4 \\ 0 & 0 & 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

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

Matrix (BA)	Elementary Transformation
\(BA=\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ -2 & -12 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)	\({ C }_{ 1 }\leftrightarrow { C }_{ 2 }\)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 6 & 0 \\ 0 & 0 & 0 \\ 0 & -20 & -4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-4R_{ 1 }\)

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

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 
\(\therefore \rho (A)=3\)

Let the three numbers be x, y and z respectively
Given
x + y + z = 6
x + 2z = 7
3x + y + z = 12

Augmented matrix [A, B]	Elementary Transformation'
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix}\begin{matrix} 6 \\ 7 \\ 12 \end{matrix} \right) \)
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & -2 & -2 \end{matrix}\begin{matrix} 6 \\ 1 \\ -6 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -4 \end{matrix}\begin{matrix} 6 \\ 1 \\ -8 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 2 }\)

The last equivalent matrix is in echelon form
\(\rho (A)=3\) and \(\rho (A,B)=3\)
\(\therefore \rho (A)=\rho (A,B)=3=Numberofunknowns\)
\(\therefore\) The system is consistent and has unique solution.
To find the solutions, let us rewrite the echelon form into matrix form
\(\left( \begin{matrix} 1 & 1 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & -4 \end{matrix}\begin{matrix} x \\ y \\ z \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 1 \\ -8 \end{matrix} \right) \)
x + y + z = 6    ..(1)
-y + z = 1
-4z = -8
From (3),\(-4z=-8\Rightarrow z=\cfrac { -8 }{ -4 } =2\)
Substituting z = 2 in (2) we get
\(-y+2=1\Rightarrow -y=1-2\Rightarrow -y=-1\)
\(\Rightarrow y=1\)
Substitutingy = 1 andz = 2 in (1) we get
\(x+1+2=6\Rightarrow x+3=6\Rightarrow x=6-3\)
\(\Rightarrow x=3\)
Hence, the numbers are 3, 1, 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}