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 \(A=\left( \begin{matrix} i & -1 \\ 3 & -6 \end{matrix} \right) \)
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
\(\therefore \rho (A)=2\)

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 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.