\(\Delta=\left|\begin{array}{lll} 1 & 2 & 5 \\ 3 & 1 & 4 \\ 6 & 1 & 7 \end{array}\right|\)
\(=1(7-4)-2(21-24)+5(3-6)\)
\(=3+6-15=-6 \neq 0\)
\(\Delta_{x}=\left|\begin{array}{lll} 23 & 2 & 5 \\ 26 & 1 & 4 \\ 47 & 1 & 7 \end{array}\right|\)
\(=23(7-4)-2(182-188)\) + 5 (26 - 47)
= 69 +12-105 = -24
\(\Delta_{y}=\left|\begin{array}{lll} 1 & 23 & 5 \\ 3 & 26 & 4 \\ 6 & 47 & 7 \end{array}\right|\)
= 1(182 -188) -23(21 - 24) +5(141 -156)
= -6+69-75 = -12
\(\Delta_{z}=\left|\begin{array}{lll} 1 & 2 & 23 \\ 3 & 1 & 26 \\ 6 & 1 & 47 \end{array}\right|\)
= 1(47 - 26)-2(141 - 156) +23(3 - 6) + 23(3 - 6)
 = 21+ 30-69 = -18
\( x=\frac{\Delta_{x}}{\Delta}=\frac{-24}{-6}=4 \)
\( \mathrm{y}=\frac{\Delta_{y}}{\Delta}=\frac{-12}{-6}=2 \)
\( \mathrm{z}=\frac{\Delta_{z}}{\Delta}=\frac{-18}{-6}=3 \)
\( x=4, \mathrm{y}=2, \mathrm{z}=3\)
(ii) \(\Delta=\left|\begin{array}{lll} 2 & 2 & -1 \\ 1 & 1 & -1 \\ 3 & 2 & -3 \end{array}\right|\)
= 2(-3+2)-2(-3 +3)- 1(2-3)
= -2 +0+1= -1\(\neq\)0
\(\Delta_{x}=\left|\begin{array}{lll} 1 & 2 & -1 \\ 0 & 1 & -1 \\ 1 & 2 \cdot & -3 \end{array}\right|\)
= 1(-3 +2)-2(0 + 1)-1(0- 1)
= -1-2+1 = -2
\(\Delta_{y}=\left|\begin{array}{lll} 2 & 1 & -1 \\ 1 & 0 & -1 \\ 3 & 1 & -3 \end{array}\right|\)
= 2(0+1)-1(-3+3)-1 (1-0)
= 2+0-1 = 1
\(\Delta_{z}=\left|\begin{array}{lll} 2 & 2 & 1 \\ 1 & 1 & 0 \\ 3 & 2 & 1 \end{array}\right|\)
= 2(1-0)-2(1-0) +1 (2-3)
= 2 - 2 - 1 = -1
\( x=\frac{\Delta_{x}}{\Delta}=\frac{-2}{-1}=2 \)
\(y=\frac{\Delta_{y}}{\Delta}=\frac{1}{-1}=-1 \)
\( z=\frac{\Delta_{z}}{\Delta}=\frac{-1}{-1}=1 \)
x = 2, y = -1, z = 1 is the solution.
(iii) \(\frac{1}{x}+\frac{2}{y}-\frac{1}{z}=1, \frac{2}{x}+\frac{4}{y}+\frac{1}{z}=5 ; \frac{3}{x}-\frac{2}{y}-\frac{2}{z}=0\)
\(\text { Let } \frac{1}{x}=a, \frac{1}{y}=b, \frac{1}{z}=c\)
a+ 2b-c =1
2a +4b+ c = 5
3a-2b-2c = 0
\(\Delta=\left|\begin{array}{ccc} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & -2 & -2 \end{array}\right|\)
1(-8 +2)-2(-4-3)- 1(-4-12)
= -6+ 14+ 16 24 \(\neq\)0
\(\Delta_{a}=\left|\begin{array}{ccc} 1 & 2 & -1 \\ 5 & 4 & 1 \\ 0 & -2 & -2 \end{array}\right|\)
= 1(-8+2)-2(-10-0)-1-10-0)
= -6+20+10 = 24
\(\Delta_{b}=\left|\begin{array}{ccc} 1 & 1 & -1 \\ 2 & 5 & 1 \\ 3 & 0 & -2 \end{array}\right|\)
= 1(-10-0)-1(-4-3)-1 (0-15)
-10+7 + 15 = 12
\(\Delta_{c}=\left|\begin{array}{ccc} 1 & 2 & 1 \\ 2 & 4 & 5 \\ 3 & -2 & 0 \end{array}\right|\)
\( =1(0+10)-2(0-15)+1(-4-12) \)
= 10+30-16 = 24
\( \mathrm{a}=\frac{\Delta_{a}}{\Delta}=\frac{24}{24}=1 \Rightarrow x=1 \)
\( \mathrm{~b}=\frac{\Delta_{b}}{\Delta}=\frac{12}{24}=\frac{1}{2} \Rightarrow \mathrm{y}=2 \)
\( \mathrm{c}=\frac{\Delta_{c}}{\Delta}=\frac{24}{24}=1 \Rightarrow \mathrm{z}=1\)

(i) In matrix form
\(\left(\begin{array}{lll} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 25 \\ 13 \\ 1 \end{array}\right)\)
AX = B
The augmented matrix is
\((A, B)=\left(\begin{array}{cccc} 4 & 3 & 6 & 25 \\ 1 & 5 & 7 & 13 \\ 2 & 9 & 1 & 1 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 5 & 7 & 13 \\ 4 & 3 & 6 & 25 \\ 2 & 9 & 1 & 1 \end{array}\right) R_{1} \leftrightarrow R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & 5 & 7 & 13 \\ 0 & -17 & -22 & -27 \\ 0 & -1 & -13 & -25 \end{array}\right) R_{3} \rightarrow R_{3}-2 R_{1}\)
\(\sim\left(\begin{array}{cccc} 1 & 5 & 7 & 13 \\ 0 & -17 & -22 & -27 \\ 0 & 0 & 199 & 398 \end{array}\right) R_{3} \rightarrow-17 R_{3}+R_{2}\)
\(\rho(A, B)=\rho(A)=3\)
System is consistent and has unique solution
In matrix form
\(\left(\begin{array}{ccc} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & 0 & 199 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 13 \\ -27 \\ 398 \end{array}\right)\)
199z = 398
z = 2
-17y - 22z = -27
-17y = 44-27
-17y = 17
y = -1
x+ 5y+7z = 13
\(x-5+14=13 \Rightarrow x=4\)
Solution is x = 4, y = -1, z = 2
(ii) In matrix form
\(\left(\begin{array}{ccc} 1 & -3 & -8 \\ 3 & 1 & -4 \\ 2 & 5 & 6 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -10 \\ 0 \\ 13 \end{array}\right)\)
AX = B
Augumented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 3 & 1 & -4 & 0 \\ 2 & 5 & 6 & 13 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 10 & 20 & 30 \\ 0 & 11 & 22 & 33 \end{array}\right) \begin{gathered} R_{2} \rightarrow R_{2}-3 R_{1} \text {, } \\ R_{3} \rightarrow R_{3}-2 R_{1} \end{gathered}\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2} / 10 \\ &R_{3} \rightarrow R_{3} / 11 \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\(\rho(A, B)=\rho(A)=2<3\)
It is consistent and has infinitely many solution.
In matrix form
\(\left(\begin{array}{ccc} 1 & -3 & -8 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -10 \\ 3 \\ 0 \end{array}\right)\)
x- 3y - 8z = -10 
y + 2z = 3
Let z = k, \(k \in R\)
y = 3-2k
x - 9+ 6k- 8k = -10
x = 2k-1
Solution set : x = 2k-1, y = 3-2k, z = \(k \in R\)

In matrix form 
\(\left(\begin{array}{rrr} k & -1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right)\)
AX = B
The augmented matrix
\((\mathrm{A}, \mathrm{B})=\left(\begin{array}{llll} k & 1 & 1 & 1 \\ 1 & k & 1 & 1 \\ 1 & 1 & k & 1 \end{array}\right)\)
\(\sim\left(\begin{array}{llll} 1 & 1 & k & 1 \\ 1 & k & 1 & 1 \\ k & 1 & 1 & 1 \end{array}\right) R_{1} \leftrightarrow R_{3}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & k & 1 \\ 0 & k-1 & 1-k & 0 \\ 0 & 1-k & 1-k^{2} & 1-k \end{array}\right) \begin{gathered} R_{2} \rightarrow R_{2}-R_{1} \\ R_{3} \rightarrow R_{3}-k R_{1} \end{gathered}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & k & 1 \\ 0 & k-1 & 1-k & 0 \\ 0 & 0 & 2-k-k^{2} & 1-k \end{array}\right) \quad R_{3} \rightarrow R_{3}+R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & k & 1 \\ 0 & k-1 & 1-k & 0 \\ 0 & 0 & (k+2)(1-k) & 1-k \end{array}\right)\)
Case (i):
when k\(\neq\)1 
\(\rho(A, B)=\rho(A)=3\)
The system has unique solution.
Case (ii):
If k = 1
\(\rho(A, B)=\rho(A)=1<3\)
The system is consistent and has infinitely many solution.
Case (iii):
If k = -2
\(\rho(A, B)=3, \rho(A)=2\)
\(\rho(A, B) \neq \rho(A)\)
The system is 2 = 2 inconsistent. Hence no solution.

In matrix form
\(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 1 & -2 \\ \lambda & 1 & 4 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 2 \\ 2 \\ 2 \end{array}\right)\)
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 2 & 1 & -2 & 2 \\ \lambda & 1 & 4 & 2 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & -2 \\ 0 & 1-\lambda & 4-\lambda & 2-2 \lambda \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-2 R_{1} \\ &R_{3} \rightarrow R_{3}-\lambda R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & -2 \\ 0 & -\lambda & -\lambda & -2 \lambda \end{array}\right) R_{3} \rightarrow R_{j}+R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 2 \\ 0 & -1 & -4 & -2 \\ 0 & 0 & 3 \lambda & 0 \end{array}\right) \quad R_{3} \rightarrow R_{3}-\lambda R_{2}\)
Case (i) : When \(\lambda \neq 0\)
\(\rho(A, B)=\rho(A)=3\)
The system has unique solution
Case (ii) : When \(\lambda =0\)
\(\rho(A, B)=\rho(A)=2\)
The system is consistent and has infinitely many solution.

In matrir form
\(\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & \lambda \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6 \\ 10 \\ \mu \end{array}\right)\)
AX = B
The augmented matrix is
\((A, B)=\left(\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 10 \\ 1 & 2 & \lambda & \mu \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 4 \\ 0 & 0 & \lambda-3 & \mu-10 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-R_{2} \end{aligned}\)
Case 1 :
\( \lambda-3=0 \text { and } \mu \neq 10 \)
\( \rho(A)=2, \rho(A, B)=3 \)
\( \therefore \rho(A) \neq \rho(A, B)\)
System is inconsistent and has no solution.
Case 2 :
\(\lambda \neq 3\) and \(\mu\) cantakeany value in R
\( \rho(A)=3, \rho(A, B)=3 \)
\(\rho(A)=\rho(A, B)=3 \)
System is consistent and "has, unique solution
Case 3 :
\(\lambda=3, \mu=10\)
\(\rho(A)=\rho(A, B)=2<3\)
The system is consistent and has infinite number of solution.

In matrix form 
\(\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 6 \\ 14 \\ 30 \end{array}\right)\)
AX = B
The augnmented matrix is
\((A, B)=\left(\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 1 & 2 & 3 & 14 \\ 1 & 4 & 7 & 30 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ 0 & 3 & 6 & 24 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-R_{2} \\ &R_{3} \rightarrow R_{3}-R_{2} \end{aligned}\)
\(\sim\left(\begin{array}{llll} 1 & 1 & 1 & 6 \\ 0 & 1 & 2 & 8 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-3 R_{2}\)
\(\rho(A, B)=\rho(A)=2\)
System is consistent and has an infinite number of solution 
In matrix form
\(\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 6 \\ 8 \\ 0 \end{array}\right)\)
\( x+y+z =6 \)
\(y+2 z =8 \)
\(\text {Let } z =k, k \in R \)
\(y =8-2 \mathrm{k} \)
\(x =6-y-z \)
\(=6-8 +2 \mathrm{k}-\mathrm{k} \)
\(=\mathrm{k}-2 \)
Solution is \(x=\mathrm{k}-2, \mathrm{y}=\mathrm{s}-2 \mathrm{k}, \mathrm{z}=\mathrm{k}\)
\(k \in R\)

In matrix form
\(\left(\begin{array}{ccc} 2 & 5 & 7 \\ 1 & 1 & 1 \\ 2 & 1 & -1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 52 \\ 9 \\ 0 \end{array}\right)\)
AX = B
Augumented matrix
\((A, B)=\left(\begin{array}{cccc} 2 & 5 & 7 & 52 \\ 1 & 1 & 1 & 9 \\ 2 . & 1 & -1 & 0 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 9 \\ 2 & 5 & 7 & 52 \\ 2 & 1 & -1 & 0 \end{array}\right) R_{1} \leftrightarrow R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 9 \\ 0 & 3 & 5 & 34 \\ 0 & -1 & -3 & -18 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-2 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 9 \\ 0 & 3 & 5 & 34 \\ 0 & 0 & -4 & -20 \end{array}\right) R_{3} \rightarrow 3 R_{3}+R_{2}\)
\(\rho(A, B)=\rho(A)=3\)
It is consistent and has unique solution. In matrix form
\(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 3 & 5 \\ 0 & 0 & -4 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 9 \\ 34 \\ -20 \end{array}\right)\)
\(-4 z=-20 \Rightarrow z=5\)
3y + 5z = 34
\(3 y+25=34 \Rightarrow y=9 / 3=3\)
x + y + z = 9
x = 9-3-5 = 1
Solutions are x = 1, y = 3, z = 5.

In matrix form
\(\left(\begin{array}{lll} 1 & 1 & 1 \\ 1 & 3 & 2 \\ 1 & 5 & 3 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{l} 3 \\ 6 \\ k \end{array}\right)\)
AX = B
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 1 & 3 & 2 & 6 \\ 1 & 5 & 3 & k \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 0 & 2 & 1 & 3 \\ 0 & 4 & 2 & k-3 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-R_{1} \\ &R_{3} \rightarrow R_{3}-R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 0 & 2 & 1 & 3 \\ 0 & 0 & 0 & k-9 \end{array}\right) R_{3} \rightarrow R_{3}-2 R_{2}\)
The system is inconsistent \(\rho(A, B) \neq \rho(A)\)
\( \rho(A, B)=3 \text { as } \rho(A) =2 \)
\(\therefore k-9 \neq 0 \)
\(k \neq 9 \)

In matrix form
\(\left(\begin{array}{ccc} 1 & 2 & -3 \\ 3 & -1 & -2 \\ 2 & 3 & -5 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -2 \\ 1 \\ k \end{array}\right)\)
AX = B
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 3 & -1 & -2 & 1 \\ 2 & 3 & -5 & k \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -7 & 7 & 7 \\ 0 & -1 & 1 & K+4 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -1 & 1 & 1 \\ 0 & -1 & 1 & k+4 \end{array}\right) \quad R_{2} \rightarrow R_{2} / 7\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -3 & -2 \\ 0 & -1 & 1 & 1 \\ 0 & 0 & 0 & k+3 \end{array}\right) \quad R_{3} \rightarrow R_{3}-R_{2}\)
Since the system is consistent
\(\rho(A, B)=\rho(A)=2\)
k + 3 = 0
k = -3

In matrix form
\(\left(\begin{array}{ccc} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 7 \\ 13 \\ 0 \end{array}\right)\)
AX = B
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 2 & -1 & 1 & 7 \\ 3 & 1 & -5 & 13 \\ 1 & 1 & 1 & 0 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 3 & 1 & -5 & 13 \\ 2 & -1 & 1 & 7 \end{array}\right) R_{1} \leftrightarrow R_{3}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & -1 & 0 \\ 0 & -2 & -8 & 13 \\ 0 & -3 & -1 & 7 \end{array}\right) \begin{gathered} R_{2} \rightarrow R_{2}-3 R_{1} \\ R_{3} \rightarrow R_{3}-2 R_{1} \end{gathered}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 0 \\ 0 & -2 & -8 & 13 \\ 0 & 0 & 22 & -25 \end{array}\right) R_{3} \rightarrow 2 R_{3}-3 R_{2}\)
In matrix form
\(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & -2 & -8 \\ 0 & 0 & 22 \end{array}\right)\left(\begin{array}{c} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 0 \\ 13 \\ -25 \end{array}\right)\)
\(\rho(A ; B)=\rho(A)=3\)
It is consistent and has unique solution
\( 22 \mathrm{z} =-25 \)
\(\mathrm{z} =-\frac{25}{22} \)
\(-2 \mathrm{y}-8 \mathrm{z} =13 \)
\(-2 \mathrm{y}-8\left(-\frac{25}{22}\right) =13 \)
\(\Rightarrow \quad-2 \mathrm{y} =13-\frac{100}{11}=\frac{43}{11} \)
\(\mathrm{y} =-\frac{43}{22} \)
\(x+y+z =0 \)
\(x-\frac{43}{22}-\frac{25}{22} =0 \)
\(x =\frac{68}{22} \)

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
\(\left( \begin{matrix} k & 1 & 1 \\ 1 & k & 1 \\ 1 & 1 & k \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ 1 \\ 1 \end{matrix} \right) \)

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

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.