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}

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 equations are
2x + 3y = 7
3x + 5y = 9
Here \(\triangle =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =1\)
\(\neq 0\)
\(\therefore \) we can apply Cramer’s Rule
Now \({ \triangle }_{ x }=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =8\)  \({ \triangle }_{ y }=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =-3\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }_{ X } }{ \triangle } =\frac { 8 }{ 1 } =8\)   \(y=\frac { { \triangle }_{ y } }{ \triangle } =\frac { -3 }{ 1 } =-3\)
\(\therefore \) Solution is x = 8, y = −3

Given 
A = \(\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \) and B = \(\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
\(AB=\left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-2+5 & -2+4-1 & 3-6+1 \\ 2+6+20 & -4-12+45 & 6+18-4 \\ 3+4+15 & -6-8+3 & 9+12-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) =\left( \begin{matrix} -6 & 1 & -2 \\ 28 & -12 & 20 \\ 22 & -11 & 18 \end{matrix} \right) \)

The matrix is in echelon form and the number of non-zero rows is 2.
\(\therefore \rho (AB)=2\)
Now  \(BA=\left( \begin{matrix} 1 & -2 & 3 \\ -2 & 4 & -6 \\ 5 & 1 & -1 \end{matrix} \right) \left( \begin{matrix} 1 & 1 & -1 \\ 2 & -3 & 4 \\ 3 & -2 & 3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 1-4+9 & 1+6-6 & -1-8+9 \\ -2+8-18 & -2-12+12 & 2+16-18 \\ 5+2-3 & 5-3+2 & -5+4-3 \end{matrix} \right) \)
= \(\left( \begin{matrix} 6 & 1 & 0 \\ -12 & -2 & 0 \\ 4 & 4 & -4 \end{matrix} \right) \)

The number of non-zero rows is 2.
\(\therefore \rho (BA)=2\)

Let A = \(\left( \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix}\begin{matrix} 3 \\ -2 \\ -7 \end{matrix} \right) \)
Order of A is 3 \(\times\) 4
∴\(\rho \)(A)\(\le \)3
Consider the third order minors
\(\left| \begin{matrix} 1 & 2 & -1 \\ 2 & 4 & 1 \\ 3 & 6 & 3 \end{matrix} \right| =\) 0, \(\left| \begin{matrix} 1 & -1 & 3 \\ 2 & 1 & -2 \\ 3 & 3 & -7 \end{matrix} \right| =0\) 
\(\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & -2 \\ 3 & 6 & -7 \end{matrix} \right| =0,\) \(\left| \begin{matrix} 2 & -1 & 3 \\ 4 & 1 & -2 \\ 6 & 3 & -7 \end{matrix} \right| =0\)
Since all third order minors vanishes, \(\rho (A)\neq 3\)
Now, let us consider the second order minors,
Consider one of the second order minors \(\\ \\ \left| \begin{matrix} 2 & -1 \\ 4 & 1 \end{matrix} \right| =6\neq 0\)
There is a minor of order 2 which is not zero.
\(\therefore \rho (A)=2\)

Let A= \(\left( \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3.
∴\(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right| =0\)
Since the third order minor vanishes, therefore \(\rho (A)\neq 3\)
Consider a second order minor \(\left| \begin{matrix} 5 & 3 \\ 1 & 2 \end{matrix} \right| =7\neq 0\)
There is a minor of order 2, which is not zero.
\(\therefore \rho (A)=2\)

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.

\(\Delta =\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 2 \\ 1 & 1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1(2 -2) - 2(-4 -2) + 1(2 + 1)
= 1(0)-2(-6)+1(3)
= 12 + 3 = 15\(\neq \)0.
Since \(\Delta \neq 0\) Cramer's rule can be applied and thesystem is consistent with unique solution.
\({ \Delta }x=\left| \begin{matrix} 7 & 2 & 1 \\ 4 & -1 & 2 \\ -1 & 1 & -2 \end{matrix} \right| \)
= \(7\left| \begin{matrix} -1 & 2 \\ 1 & -2 \end{matrix} \right| -2\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 4 & -1 \\ -1 & 1 \end{matrix} \right| \)
= 7 (2 -2) -2 (-8 + 2) + 1 (4 - 1)
= 7 (0) - 2(-6) + 1(3)
= 12 + 3 = 15
\(\Delta y=\left| \begin{matrix} 1 & 7 & 1 \\ 2 & 4 & 2 \\ 1 & -1 & -2 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 4 & 2 \\ -1 & -2 \end{matrix} \right| -7\left| \begin{matrix} 2 & 2 \\ 1 & -2 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| \)
= 1 (- 8 + 2) -7(-4 -2) + 1(-2 -4)
= 1 (-6) -7 (-6) + 1 (-6)
= - 6 + 42 - 6 = 30
\(\Delta z=\left| \begin{matrix} 1 & 2 & 7 \\ 2 & -1 & 4 \\ 1 & 1 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} -1 & 4 \\ 1 & -1 \end{matrix} \right| -2\left| \begin{matrix} 2 & 4 \\ 1 & -1 \end{matrix} \right| +7\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| \)
= 1 (1 - 4) - 2(- 2 - 4) + 7(2 + 1)
= 1(-3)-2(-6)+7(3)
= - 3 + 12 + 21 = 30

\(\therefore\) Solution set is {1, 2, 2}

Transition probability matrix
(A B) T = (A B)

Where A represents the percent of people those who bought soap A and B represents the percent of people those who bought soap B.
By the given data
A = 15% = ·15
and B = 85% = ·85
Percentage after one year is
\(\left( \cdot 15\quad \cdot 85 \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) \)
= ((.15)(·65) + (·85)(-45) ·15(-35)+ ·85(-55))
= (-0975 + ·3825 ·0525 + -4675)
= (-48 ·52)
Hence, market share after one year is 48% and 52% At equilibrium,
\(\left( A\quad B \right) \left( \begin{matrix} \cdot 65 & \cdot 35 \\ \cdot 45 & \cdot 55 \end{matrix} \right) =(A\quad B)\)
(-65A + A5B ·35A +·55B) = (A B)
Equating the corresponding entries on both sides we get
\(\Rightarrow \cdot 65A+\cdot 45B=A\)
\(\Rightarrow \cdot 65A+\cdot 45(1-A)=A\)
[Since A+B = 1   B = 1-A]
\(\Rightarrow \cdot 65A+\cdot 45-\cdot 45A=A\)
\(\Rightarrow \cdot 45=A-\cdot 65A+\cdot 45A\)
\(\Rightarrow \cdot 45=A\left( \cdot 35+45 \right) \)
\(\Rightarrow \cdot 45=A(\cdot 35+45)\)
\(\Rightarrow \cdot 45=A(-8)\)
\(\Rightarrow A=\cfrac { \cdot 45 }{ \cdot 8 } =\cdot 5625=56.25\)
\(\therefore B=1-A=1-\cdot 5625=\cdot 4375\)
= 43.75%
\(\therefore\) Equilibrium is reached when A = 56.25% and B = 43.75%

Given non-homogeneous equations are
\(3x-y+\lambda z=1\)
\(2x+y+z=2\)
\(x+2y-\lambda z=-1\)
The matrix equation corresponding to the given system is

Since
\(\cfrac { 4\lambda }{ 7 } -\cfrac { 1+2\lambda }{ 3 } \)
= \(\cfrac { 12\lambda -7-14\lambda }{ 21 } =\cfrac { -7-2\lambda }{ 21 } \)
and \(\cfrac { 4 }{ 7 } -\cfrac { 4 }{ 3 } =\cfrac { 12-28 }{ 21 } \)
= \(\cfrac { -16 }{ 21 } \)
\(\therefore\) Since the system is fail to have unique solution either it can have infinitely many solution or it may be inconsistent.
This can happen only when \(\cfrac { -7-2\lambda }{ 21 } =0\)
\(\Rightarrow -7-2\lambda =0\)
\(\Rightarrow -7=2\lambda \)
\(\Rightarrow \lambda =\cfrac { -7 }{ 2 } \)

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.

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.