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Questions:

Generic 5

5 x 5 = 25

A new transit system has just gone into operation in a city. Of those who use the transit system this year, 10% will switch over to using their own car next year and 90% will continue to use the transit system. Of those who use their cars this year, 80% will continue to use their cars next year and 20% will switch over to the transit system. Suppose the population of the city remains constant and that 50% of the commuters use the transit system and 50% of the commuters use their own car this year,
(i) What percent of commuters will be using the transit system after one year?
(ii) What percent of commuters will be using the transit system in the long run?

Using determinants, find the quadratic defined by f(x) = ax² + bx + c if
f(1) = 0,
f(2) = - 2 and
f(3) = -6.

For what values of k, the system of equations kx+ y+z = 1, x+ ky+z= 1, x+ y+kz = 1 have
(I) Unique solution
(ii) More than one solution
(iii) no solution

A mixture is to be made of three foods A, B, C. The three foods A, B, C contain nutrients P, Q, R as shown below

Ounces per pound of Nutrient
Food	P	Q	R
A	1	2	5
B	3	1	1
C	4	2	1

How to form a mixture which will have 8 ounces of P, 5 ounces of Q and 7 ounces of R? (Cramer's rule).

The sum of three numbers is 6. If we multiply the third number by 2 and add the first number to the result we get 7. By adding second and third numbers to three times the first number we get 12. Find the numbers using rank method

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Answers Key:

Generic 5

5 x 5 = 25

Let A represents the percent of commuters who use the transit system and B represents the percent of commuters who use their own car. Transition probability matrix

Given 50% of commuters use the transit system and 50% of the commuters use their own car this year.
(i) Percentage of commuters after one year
\(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·9+·5\(\times\).2 ·5\(\times\)·1+·5\(\times\)·8)
= (-45 + ·10 ·05 +.40)
= (-55 - 45)
A = 55% and B = 45%
(ii) Equilibrium will be reached in the long run at equilibrium, we must have
(A B)T = (A B) wher A+B = 1
\(\Rightarrow \left( \begin{matrix} A & B \end{matrix} \right) \left( \begin{matrix} \cdot 9 & \cdot 1 \\ \cdot 2 & \cdot 8 \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
\(\left( \begin{matrix} \cdot 9A+\cdot 2B & \cdot 1A+8B \end{matrix} \right) =\left( \begin{matrix} A & B \end{matrix} \right) \)
Equating the corresponding entries on both sides we get,
\(\cdot 9A+\cdot 2B=A\Rightarrow \cdot 9A+\cdot 2(1-A)=A\)
[Since A + B = 1, B = 1 -A]
\(\Rightarrow \cdot 9A+\cdot 2-\cdot 2A=A\)
\(\Rightarrow \cdot 2=A-\cdot 9A+\cdot 2A\)
\(\Rightarrow \cdot 2=A(1-\cdot 9+\cdot 2)\)
\(\Rightarrow \cdot 2=A=(\cdot 3)\)
\(\therefore\) 67% of the commuters will be using the transit system in the long run.

fix) = ax² + bx + c
\(f(1)=0\Rightarrow a\left( 1 \right) ^{ 2 }+b(1)+c=0\Rightarrow a+b+c=0\) ...(1)
\(f(2)=-2\Rightarrow a\left( { 2 }^{ 2 } \right) +b(2)+c=-2\Rightarrow 4a+2b+c=2\)..(2)
\(f(3)-6\Rightarrow a(3^{ 2 })+b(3)+c=-6\Rightarrow 9a+3b+c=-6\)
Now \(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 9 & 3 & 1 \end{matrix} \right| \)
= 1(2 - 3) - 1(4 - 9) + 1(12 - 18)
= \(-1+5-6=-2\neq 0\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system has unique solution
\(\Delta a=\left| \begin{matrix} 0 & 1 & 1 \\ -2 & 2 & 1 \\ -6 & 3 & 1 \end{matrix} \right| \)
= 0-1(-2+6)+ 1(-6+ 12)
= -4 + 6 = 2
\(\Delta b=\left| \begin{matrix} 1 & 0 & 1 \\ 4 & -2 & 1 \\ 9 & -6 & 1 \end{matrix} \right| \)
= 1(-2+6)+0+1(-24+ 18)
= 4 - 6 = -2
\(\Delta c=\left| \begin{matrix} 1 & 1 & 0 \\ 4 & 2 & -2 \\ 9 & 3 & -6 \end{matrix} \right| \)
= 1 (-12 + 6) - 1( - 24 + 18) + 0
= -6 + 6 = 0

f(n) = (-1)x² + 1(x) + 0
f(x) = x²+ x.

The given non-homogeneous equations can be written as
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

Augmented matrix [A, B]	Elementary Transformation
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 1 & k & 1 \\ k & 1 & 1 \end{matrix}\begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 0 & k-1 & 1-k \\ 0 & 1-k & 1-{ k }^{ 2 } \end{matrix}\begin{matrix} 1 \\ 0 \\ 1-k \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-R_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-k{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & k \\ 0 & k-1 & 1-k \\ 0 & 0 & 2-k-{ k }^{ 2 } \end{matrix}\begin{matrix} 1 \\ 0 \\ 1-k \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }{ +R_{ 2 } }\)

Case (i):
When \(k\neq 1\) and \(k\neq 2\)
\(\rho (A)=\rho (A,B)=3=\) Number of unknowns
\(\therefore \) The system has unique solution
Case (ii):
When k = 1
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right) \)
\(\rho (A)=\rho\) (A, B) = 1
\(\therefore \) The system is consistent and has infinitely many solutions.
Case (iii):
When k = - 2
\(\left[ A,B \right] \sim \left( \begin{matrix} 1 & 1 & -2 \\ 0 & -3 & 3 \\ 0 & 0 & 0 \end{matrix}\begin{matrix} 1 \\ 0 \\ -3 \end{matrix} \right) \)
\(\rho (A)=2\rho (A,B)\)= 3
\(\Rightarrow \rho (A)\neq 2\rho (A,B)\)
\(\therefore\) The system is inconsistent and has no solution.

Let x pounds of food A, y pounds of food B and z pounds of food C be needed to form the mixture.
Given x + 3y + 4z = 8
2x + y + 2z = 5
5x + y + z = 7
\(\Delta =\left| \begin{matrix} 1 & 3 & 4 \\ 2 & 1 & 2 \\ 5 & 1 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +4\left| \begin{matrix} 2 & 1 \\ 5 & 1 \end{matrix} \right| \)
= 1(1 - 2) - 3(2 - 10) + 4(2 - 5)
= 1(-1)-3(-8)+4(-3)
= - 1 + 24 - 12 = 11
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system has unique solution.
\(\Delta x=\left| \begin{matrix} 8 & 3 & 4 \\ 5 & 1 & 2 \\ 7 & 1 & 1 \end{matrix} \right| \)
= \(8\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| -3\left| \begin{matrix} 5 & 2 \\ 7 & 1 \end{matrix} \right| +4\left| \begin{matrix} 5 & 1 \\ 7 & 1 \end{matrix} \right| \)
= 8(1 - 2) - 3(5 - 14) + 4(5 - 7)
= 8 (- 1) - 3 (- 9) + 4 (- 2)
= 8 + 27 - 8 = 11
\(\Delta y=\left| \begin{matrix} 1 & 8 & 4 \\ 2 & 5 & 2 \\ 5 & 7 & 1 \end{matrix} \right| =1\left| \begin{matrix} 5 & 2 \\ 7 & 1 \end{matrix} \right| -8\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +4\left| \begin{matrix} 2 & 5 \\ 5 & 7 \end{matrix} \right| \)
= 1(5 - 14) - 8(2 - 10) + 4(14 - 25)
= 1 (- 9) - 8 (- 8) + 4 (- 11)
= 9 + 64 - 44 = 11
\(\Delta z=\left| \begin{matrix} 1 & 3 & 8 \\ 2 & 1 & 5 \\ 5 & 1 & 7 \end{matrix} \right| =1\left| \begin{matrix} 1 & 5 \\ 1 & 7 \end{matrix} \right| -3\left| \begin{matrix} 2 & 5 \\ 5 & 7 \end{matrix} \right| +8\left| \begin{matrix} 2 & 1 \\ 5 & 1 \end{matrix} \right| \)
= 1(7 - 5) - 3(14 - 25) + 8(2 - 5)
= 1 (2) - 3 (- 11) + 8 (- 3)
= 2 + 33 - 24
= 11
\(\therefore\ x=\cfrac { \Delta x }{ \Delta } =\cfrac { 11 }{ 11 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 11 }{ 11 } =1\)
\(z=\cfrac { \Delta z }{ \Delta } =\cfrac { 11 }{ 11 } =1\)
Hence, the mixture is formed by mixing one pound of each of the foods A, B and C.

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.