In matrix form
\(\left(\begin{array}{ccc} 1 & 1 & 1 \\ 3 & 1 & -2 \\ 2 & 4 & 7 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -3 \\ -2 \\ 7 \end{array}\right)\)
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & 1 & 1 & -3 \\ 3 & 1 & -2 & -2 \\ 2 & 4 & 7 & 7 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & -3 \\ 0 & -2 & -5 & 7 \\ 0 & 2 & 5 & 13 \end{array}\right) R_{2} \rightarrow R_{3}-2 R_{1}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & -3 \\ 0 & -2 & -5 & 7 \\ 0 & 0 & 0 & 20 \end{array}\right) \quad R_{3} \rightarrow R_{3}+R_{2}\)
\( \rho(A, B)=3, \rho(A)=2 \)
\(\rho(A, B) \neq \rho(A) \)
The system is inconsistent.

In matrix form
\(\left(\begin{array}{ccc} 4 & -5 & -2 \\ 5 & -4 & 2 \\ 2 & 2 & 8 \end{array}\right)\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} 2 \\ -2 \\ -1 \end{array}\right)\)
Augmented matrix
\((\Lambda, B)=\left(\begin{array}{cccc} 1 & -5 & -2 & 2 \\ 5 & -1 & 2 & -2 \\ 2 & 2 & 8 & -1 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -5 & -2 & 2 \\ 0 & 9 & 18 & -18 \\ 0 & 9 & 18 & -1 \end{array}\right) \begin{aligned} &R_{2} \rightarrow 4 R_{2}-5 R_{1} \\ &R_{3} \rightarrow 2 R_{3}-R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & -5 & -2 & 2 \\ 0 & 9 & 18 & -18 \\ 0 & 0 & 0 & 1.1 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\( \rho(A, B)=3, \rho(A)=2 \)
\( \rho(A, B) \neq \rho(A) \)
The system is inconsistent

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
Augmented matrix
\((A, B)=\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 3 & 1 & -4 & 0 \\ 2 & 5 & 6 & 13 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 10 & 20 & 30 \\ 0 & 11 & 22 & 33 \end{array}\right) R_{2} \rightarrow R_{2}-3 R_{1}\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2} / 10 \\ &R_{3} \rightarrow R_{3} / 11 \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & -3 & -8 & -10 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\(\rho(A, B)=\rho(A)=2<3\)
It is consistent and has infinite solution.

\(A+B=\left(\begin{array}{ccc} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{array}\right)+\left(\begin{array}{ccc} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{array}\right)\)
\(=\left(\begin{array}{ccc} 0 & -1 & -2 \\ 8 & 9 & 10 \\ 8 & 8 & 8 \end{array}\right)\)
\(\sim\left(\begin{array}{ccc} 1 & 1 & 1 \\ 8 & 9 & 10 \\ 0 & -1 & -2 \end{array}\right) \begin{aligned} &R_{1} \leftrightarrow R_{3} \\ &R_{3} \rightarrow R_{3} / 8 \end{aligned}\)
\(\sim\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 1 & 2 \end{array}\right)^{2} R_{2} \rightarrow R_{2}-8 R_{1},\)
\(\sim\left(\begin{array}{lll} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\(\rho(A+B)=2\)
\(A B=\left(\begin{array}{ccc} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{array}\right)\left(\begin{array}{ccc} -1 & -2 & -1 \\ 6 & 12 & 6 \\ 5 & 10 & 5 \end{array}\right)\)
\(=\left(\begin{array}{ccc} -1+6-5 & -2+12-10 & -1+6-5 \\ -2-18+20 & -4-36+40 & -2-18+20 \\ -3-12+15 & -6-24+30 & -3-12+15 \end{array}\right)\)
\(=\left(\begin{array}{lll} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right)\)
\(\rho(A B)=0\)

(i) \(A=\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 2 & -1 & 3 & 4 \\ 5 & -1 & 7 & 11 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 0 & -3 & 1 & -2 \\ 0 & -6 & 2 & -4 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-2 R_{1} \\ &R_{3} \rightarrow R_{3}-5 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 1 & 1 & 3 \\ 0 & -3 & 1 & -2 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-2 R_{2}\)
\(\rho(A)=2\)
(ii) \(\left(\begin{array}{llll} 4 & 2 & 1 & 3 \\ 6 & 3 & 4 & 7 \\ 2 & 1 & 0 & 1 \end{array}\right)\)
\( \left(\begin{array}{cccc} 1 & 2 & 4 & 3 \\ 4 & 3 & 6 & 7 \\ 0 & 1 & 2 & 1 \end{array}\right) C_{1} \leftrightarrow C_{3} \\ \left(\begin{array}{cccc} 1 & 2 & 4 & 3 \\ 0 & -5 & -10 & -5 \\ 0 & 1 & 2 & 1 \end{array}\right) R_{2} \rightarrow R_{2}-4 R_{1} \)
\( \sim\left(\begin{array}{llll} 1 & 2 & 4 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & 1 \end{array}\right) R_{2} \rightarrow-R_{2} / 5 \\ \sim\left(\begin{array}{llll} 1 & 2 & 4 & 3 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2} \)
\(\rho(A)=2\)
(iii) \(\left(\begin{array}{cccc} 3 & 1 & -5 & -1 \\ 1 & -2 & 1 & -5 \\ 1 & 5 & -7 & 2 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -2 & 1 & -5 \\ 3 & 1 & -5 & -1 \\ 1 & 5 & -7 & 2 \end{array}\right)^{-} R_{1} \leftrightarrow R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & -2 & 1 & -5 \\ 0 & 7 & -8 & 14 \\ 0 & 7 & -8 & 7 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & -2 & 1 & -5 \\ 0 & 7 & -8 & 14 \\ 0 & 0 & 0 & -7 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\(\rho(A)=3\)
(iv) \(\left(\begin{array}{cccc} 3 & 1 & 2 & 0 \\ 1 & 0 & -1 & 0 \\ 2 & 1 & 3 & 0 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 0 & -1 & 0 \\ 3 & 1 & 2 & 0 \\ 2 & 1 & 3 & 0 \end{array}\right) R_{1} \leftrightarrow R_{2}\)
\(\sim\left(\begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & 5 & 0 \\ 0 & 1 & 5 & 0 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-3 R_{1} \\ &R_{3} \rightarrow R_{3}-2 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & 5 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-R_{2}\)
\(\rho(A)=2\)
(v) \(\left(\begin{array}{cccc} 0 & 1 & 2 & 1 \\ 2 & -3 & 0 & -1 \\ 1 & -1 & -1 & 0 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 2 & -3 & 0 & -1 \\ 0 & 1 & 2 & 1 \end{array}\right) R_{1} \leftrightarrow R_{3}\)
\(\sim\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 0 & -1 & 2 & -1 \\ 0 & 1 & 2 & 1 \end{array}\right) R_{2} \rightarrow R_{2}-2 R_{1}\)
\(\sim\left(\begin{array}{cccc} 1 & -1 & -1 & 0 \\ 0 & -1 & 1 & -1 \\ 0 & 0 & 4 & 0 \end{array}\right) R_{3} \rightarrow R_{3}+R_{2}\)
\(\rho(A)=3\)
(vi) \(\left(\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 2 & 4 & 1 & -2 \\ 3 & 6 & 3 & -7 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 6 & -16 \end{array}\right) \begin{aligned} &R_{2} \rightarrow R_{2}-2 R_{1} \\ &R_{3} \rightarrow R_{3}-3 R_{1} \end{aligned}\)
\(\sim\left(\begin{array}{cccc} 1 & 2 & -1 & 3 \\ 0 & 0 & 3 & -8 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-2 R_{2}\)
\(\rho(A)=2\)
(vii) \(\left(\begin{array}{cccc} 1 & -2 & 3 & 4 \\ -2 & 4 & -1 & -3 \\ -1 & 2 & 7 & 6 \end{array}\right)\)
\(\sim\left(\begin{array}{cccc} 1 & -2 & 3 & 4 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 10 & 10 \end{array}\right) R_{2} \rightarrow R_{2}+2 R_{1}\)
\(\sim\left(\begin{array}{cccc} 1 & -2 & 3 & 4 \\ 0 & 0 & 5 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right) R_{3} \rightarrow R_{3}-2 R_{2}\)
\(p(A)=2\)

Transition probability matrix

Shares after one week
\(\left( \cdot 6\cdot 4 \right) \left( \begin{matrix} \cdot 7 & \cdot 3 \\ \cdot 6 & \cdot 8 \end{matrix} \right) \)
= (-6\(\times\)·7+-4x·2 ·6\(\times\)·3+·4\(\times\)·8)
= z:(-42+·08 ·18+·32)= (·50 ·50)
\(\Rightarrow\) A = 50% and B = 50%
Shares after two weeks \(\left( \cdot 5\cdot 5 \right) \left( \begin{matrix} \cdot 7 & \cdot 3 \\ \cdot 2 & \cdot 8 \end{matrix} \right) \)
= (-5\(\times\)·7+·5\(\times\).2 ·5\(\times\)·3 +·5\(\times\)·8)
= (-35+·10 ·15 + 40) = (-45 ·55)
A = 45% and B = 55%

Given non-homogeneous equations are
2x + 3y = 5  6X + 5y = 11
\(\Delta =\left| \begin{matrix} 2 & 3 \\ 6 & 5 \end{matrix} \right| =10-18=-8\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution
\(\Delta x=\left| \begin{matrix} 5 & 3 \\ 11 & 5 \end{matrix} \right| =25-33=-8\)
\(\Delta y=\left| \begin{matrix} 2 & 5 \\ 6 & 11 \end{matrix} \right| =22-30=-8\)
\(\therefore x=\cfrac { \Delta x }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -8 }{ -8 } =1\)
\(\therefore \) Solution set is {1, 1}

GivenAX B
\(\left( \begin{matrix} 2 & 4 \\ 4 & 3 \end{matrix} \right) ,\left( \begin{matrix} n \\ 1 \end{matrix} \right) \left( \begin{matrix} 8 \\ 11 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} 2n+4 \\ 4n+3 \end{matrix} \right) =\left( \begin{matrix} 8 \\ 11 \end{matrix} \right) \)
Equating the corresponding entries on both sides, we get
2n +4 = 8
2n = 8-4
2n=4
\(n=\cfrac { 4 }{ 2 } \)
2 = 2

\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right] \)
\(\Rightarrow \left( \begin{matrix} x0+0 \\ 0+0+z \\ 0+y+0 \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right) \)
\(\Rightarrow x=2\quad z=-1\quad y=3\)
\(\therefore\) Solution set is {2,3, -1}

The non-homogeneous equations are
2x - 3y - 1 = 0, 5x + 2y - 12 = 0
\(\Delta =\left| \begin{matrix} 2 & -3 \\ 5 & 2 \end{matrix} \right| =4+15=19\neq 0\)
Since \(\Delta \neq 0\) Cramer's rule can be applied and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 1 & -3 \\ 12 & 2 \end{matrix} \right| =2+36=38\)
\(\Delta y=\left| \begin{matrix} 2 & 1 \\ 5 & 12 \end{matrix} \right| =24-5=19\)
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 38 }{ 19 } =1\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { 19 }{ 19 } =1\)
\(\therefore \) Solution set is {2, 1}

Given non-homogeneous equations are
x- 3y + 4z = 3, 2x - 5y + 7z = 6, 3x - 8y + 11z = 1

Clearly \(\rho (A)=2\) and \(\rho (A,B)=3\)
\(\rho (A,B)\neq \rho (A)\)
Hence, the given system is inconsistent and has no solution.

Given non-homogeneous equations are
x + 2y = 3,y - z = 2,x + Y + z = 1

Obviously,\(\rho (A)=2\) and \(\rho (A,B)\)
Hence \(\rho (A)=2\quad \rho\) (A, B) = 2
\(\therefore\) The system is consistent and has infinite number of solutions.

The non-homogeneous equation are
2x - y + z = 7, 3x + y - 5z = 13, x + y + z = 5

Clearly \(\rho (A)=3\) and \(\rho (A,B)\) = 3 = Number of unknowns 
\(\therefore\) The given system is consistent and has unique solution.

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
\(\therefore \rho (A)\le min(3,3)\)
\(\Rightarrow \rho (A)\le 3\)

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