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

\(\Delta =\left| \begin{matrix} 1 & 4 & 3 \\ 2 & -6 & 6 \\ 5 & -2 & 3 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -6 & 6 \\ -2 & 3 \end{matrix} \right| -4\left| \begin{matrix} 2 & 6 \\ 5 & 3 \end{matrix} \right| +3\left| \begin{matrix} 2 & -6 \\ 5 & -2 \end{matrix} \right| \)
= 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 \(\left\{ -1,\frac { 1 }{ 2 } ,\frac { 1, }{ 3 } \right\} \)

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

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.