Let \(A=\left( \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right) \)
Order of A is 2 \(\times\) 2
\(\therefore \rho (A)\le 2\)  [Since minimum of (2, 2) is 2]
Consider the second order minor
\(\left| \begin{matrix} 5 & 6 \\ 7 & 8 \end{matrix} \right| =40-42\)
= \(-2\neq 0\)
There is a minor of order 2, which is not zero
\(\therefore \rho (A)=2\)

Let A =\(\begin{pmatrix} 1 & 5 \\ 3 & 9 \end{pmatrix}\)
Order of A is 2 \(\times\) 2
∴ \(\rho \) (A) \(\le \) 2
Consider the second order minor
\(\begin{vmatrix} 1 & 5 \\ 3 & 9 \end{vmatrix}=-6\neq 0\)
There is a minor of order 2, which is not zero.
∴ \(\rho \) (A) \(\le \) 2

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

Given A =\(\left( \begin{matrix} -2 & 1 & 3 \\ 0 & 1 & 1 \\ 1 & 3 & 4 \end{matrix}\begin{matrix} 4 \\ 2 \\ 7 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ -2 & 1 & 3 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 1 }\rightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 7 & 11 \end{matrix}\begin{matrix} 7 \\ 2 \\ 18 \end{matrix} \right) { R }_{ 2 }\rightarrow { { R_{ 2 }+2{ R }_{ 1 } } }\)
\(\sim \left( \begin{matrix} 1 & 3 & 4 \\ 0 & 1 & 1 \\ 0 & 0 & 4 \end{matrix}\begin{matrix} 7 \\ 2 \\ 4 \end{matrix} \right) { R }_{ 3 }\rightarrow { R_{ 3 }-{ 7R }_{ 2 } }\)
The last equivalent matrix is in echelon form and there are 3 non - zero rows.
\(\therefore \rho (A)=3\)

Transition probability matrix
T = \(_{ B }^{ A }\left( \begin{matrix} \overset { A }{ 0.9 } & \overset { B }{ 0.1 } \\ 0.3 & 0.7 \end{matrix} \right) \)
At equilibrium, (A B) T = (AB) where A + B = 1
(A B) \(\left( \begin{matrix} 0.9 & 0.1 \\ 0.3 & 0.7 \end{matrix} \right) \) = (A B)
0.9A + 0.3B = A
0.9A + 0.3(1−A) = A
0.9A−0.3A + 0.3 = A
0.6A + 0.3 = A
0.4A = 0.3
A = \(\frac { 0.3 }{ 0.4 } =\frac { 3 }{ 4 } \)
B = 1-\(\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)
Hence the market share of brand A is 75% and the market share of brand B is 25%

\(\Delta =\left| \begin{matrix} 2 & 3 \\ 3 & 5 \end{matrix} \right| =10-9=1\neq 0\)
Since \(\Delta \neq 0\) 
we can apply Cramer's rule and the system is consistent with unique solution.
\(\Delta x=\left| \begin{matrix} 7 & 3 \\ 9 & 5 \end{matrix} \right| =7(5)-9(3)\)
= 35 - 27 = 8
\(\Delta y=\left| \begin{matrix} 2 & 7 \\ 3 & 9 \end{matrix} \right| =2(9)-3(7)\)
= 18 - 21 = -3
 \(\therefore\) \(x=\cfrac { \Delta x }{ \Delta } =\cfrac { 8 }{ 1 } =8\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
\(\therefore\) Solution set is (8, -3)

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

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 3 & -2 \\ 6 & -4 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \end{matrix} \right) \)
AX = B

\(\therefore \)\(\rho ([A,B])=2\), \(\rho (A)=1\)
\(\rho (A)\neq \rho \left( [A,B] \right) \)
\(\therefore \) The given system is inconsistent and has no solution.

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 the amount invested in the rate of 2%, 3% and 6% be Rs. x, Rs. y and Rs. z respectively
By the given data,
x+ y+z = 8500
\(\cfrac { 2x }{ 100 } +\cfrac { 3y }{ 100 } +\cfrac { 6z }{ 100 } =380\)
\(\Rightarrow \cfrac { 2x+3y+6z }{ 100 } =380\)
\(\because Interest=\cfrac { PNR }{ 100 } =\cfrac { x\times 1\times 2 }{ 100 } =\cfrac { 2x }{ 100 } \)
\(\Rightarrow 2x+3y+6z=38000\)
Also,z = x+y
x+y-z  =0
\(\Delta =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| \)
\(1\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1(-3 - 6) - 1(-2 -6) + 1 (2 - 3)
= 1(-9) - 1 (-8) + 1(-1)
= -9 + 8 - 1 = 2 \(\neq \) 0 
Since \(\Delta \neq 0\),Cramer's rule can be applied and the system is consistent with unique solution
\({ \Delta x }=\left| \begin{matrix} 8500 & 1 & 1 \\ 38000 & 3 & 6 \\ 0 & 1 & -1 \end{matrix} \right| \)
= \(8500\left| \begin{matrix} 3 & 6 \\ 1 & -1 \end{matrix} \right| -1\left| \begin{matrix} 38000 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 38000 & 3 \\ 0 & 1 \end{matrix} \right| \)
= 8500 (- 3 - 6) - 1(-38000 -0) + 1(38000 - 0)
= 8500(-9) - 1(-38000) + 1(38000)
= - 76500 + 38000 + 38000
= -500
\(\Delta y=\left| \begin{matrix} 1 & 8500 & 1 \\ 2 & 38000 & 6 \\ 1 & 0 & -1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -8500\left| \begin{matrix} 2 & 6 \\ 1 & -1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| \)
= 1 (-38000 - 0) - 8500 (-2 -6) + 1(0 - 38000)
= - 38000 - 8500 (-8) - 38000
= - 38000 + 68000 - 38000
= - 8000
\(\Delta z=\left| \begin{matrix} 1 & 1 & 8500 \\ 2 & 3 & 38000 \\ 1 & 1 & 0 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 38000 & 6 \\ 0 & -1 \end{matrix} \right| -1\left| \begin{matrix} 2 & 38000 \\ 1 & 0 \end{matrix} \right| +8500\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= 1 (0 - 38000) - 1(0 -38000) +85000 (2 - 3)
= - 38000 + 38000 + 8500 (-1)
= - 8500


Hence, the amount invested in the three accounts are Rs. 250, Rs. 4000 and Rs. 4250 respectively.

Let the weight assigned to the three varieties be Rs. x, Rs. y and Rs. z respectively By the given data,
x + 2y + 3z = 11
2x + 4y + 5z = 21
3x + 5y + 6z = 27
\(\Delta =\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{matrix} \right| =1\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(24 - 25) -2(12 - 15) + 3(10 - 12)
= 1(-1) -2 (-3) + 3(-2)
= -1+6 - 6 = -1\(\neq \) 0.
Since \(\Delta \neq 0\) the system is consistent with unique solution and Cramer's rule can be applied.
\(\Delta x=\left| \begin{matrix} 11 & 2 & 3 \\ 21 & 4 & 5 \\ 27 & 5 & 6 \end{matrix} \right| \)
\(=11\left| \begin{matrix} 4 & 5 \\ 5 & 6 \end{matrix} \right| -2\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| +3\left| \begin{matrix} 21 & 4 \\ 27 & 5 \end{matrix} \right| \)
= 11(24 - 25) - 2(126 - 135) + 3(105 - 108)
= 11(-1) - 2(-9) + 3 (-3)
= 11+18-9
= -2
\(\Delta y=\left| \begin{matrix} 1 & 11 & 3 \\ 2 & 21 & 5 \\ 3 & 27 & 6 \end{matrix} \right| \)
= \(\left| \begin{matrix} 21 & 5 \\ 27 & 6 \end{matrix} \right| -11\left| \begin{matrix} 2 & 5 \\ 3 & 6 \end{matrix} \right| +3\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| \)
= 1(126 - 135) - 11(12 -15) + 3(54 - 63)
= - 9 - 11(-3) + 3(-9)
= - 9 + 33 - 27
= 3
\(\Delta z=\left| \begin{matrix} 1 & 2 & 11 \\ 2 & 4 & 21 \\ 3 & 5 & 27 \end{matrix} \right| =1\left| \begin{matrix} 4 & 21 \\ 5 & 27 \end{matrix} \right| -2\left| \begin{matrix} 2 & 21 \\ 3 & 27 \end{matrix} \right| +11\left| \begin{matrix} 2 & 4 \\ 3 & 5 \end{matrix} \right| \)
= 1(108 - 105) - 2(54 - 63) + 11(10 - 12)
= 1(3) - 2(-9) + 11(-2)
= 3 + 18 - 22
= -1
\(x=\cfrac { \Delta x }{ \Delta } =\cfrac { -2 }{ -1 } =2\)
\(y=\cfrac { \Delta y }{ \Delta } =\cfrac { -3 }{ 1 } =3\)
and \(z=\cfrac { \Delta z }{ \Delta } =\cfrac { -1 }{ -1 } =1\)
Hence, the weights assigned to the three varieties are 2, 3 and 1 respectively

Here \(\triangle =\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =13\neq 0\)
\(\therefore \) We can apply Cramer’s Rule and the system is consistent and it has unique solution.
\({ \triangle }_{ x }=\left| \begin{matrix} 1 & 1 & 1 \\ 2 & -1 & 3 \\ 3 & 2 & -1 \end{matrix} \right| =-13\)
\( { \triangle }_{ y }=\left| \begin{matrix} 1 & 4 & 1 \\ 2 & 1 & 3 \\ 3 & 1 & -1 \end{matrix} \right| =39\)
\( { \triangle }_{ z }=\left| \begin{matrix} 1 & 1 & 4 \\ 2 & -1 & 1 \\ 3 & 2 & 1 \end{matrix} \right| =26\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -13 }{ 13 } =-1\)
\( y=\frac { { \triangle }y }{ { \triangle } } =\frac { 39 }{ 13 } =3\)
\( z=\frac { { \triangle }z }{ { \triangle } } =\frac { 26 }{ 13 } =2\)
\(\therefore \) The solution is (x, y, z) = (−1, 3, 2)

Given non-homogeneous equations are
5x+ 3y + 7z = 4
3x + 26y + 2z = 9
7x + 2y + 10z = 5
The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 5 & 3 & 7 \\ 3 & 26 & 2 \\ 7 & 2 & 10 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 4 \\ 9 \\ 5 \end{matrix} \right) \)

\(\therefore\) The system is consistent with infinitely many solutions let us rewrite the above echelon form into matrix form
\(\left( \begin{matrix} 1 & \frac { 26 }{ 3 } & \frac { 2 }{ 3 } \\ 0 & \frac { -11 }{ 3 } & \frac { 1 }{ 3 } \\ 0 & 0 & 0 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 3 \\ -1 \\ 0 \end{matrix} \right) \)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
\(x+\cfrac { 26 }{ 3 } y+\cfrac { 2 }{ 3 } z=3\)
let z = k where k\(\in\) R
\((2)\Rightarrow \cfrac { -11 }{ 3 } y+\cfrac { k }{ 3 } =-1\)

\(\Rightarrow \ -11y=-3-k\) 
11y = 3 + k
\(\Rightarrow \quad y=\cfrac { 1 }{ 11 } \left( 3+k \right) \)
Substituting \(y=\cfrac { 1 }{ 11 } \left( 3+k \right) \) and z = k in (1) we get,
\(x+\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) +\cfrac { 2 }{ 3 } k=3\)
\(=\cfrac { 26 }{ 3 } \left( \cfrac { 3+k }{ 11 } \right) -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k }{ 33 } -\cfrac { 2k }{ 3 } +3\)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 78-26k-22k+99 }{ 33 } \)
\(\cfrac { 21-48k }{ 33 } =\cfrac { 3(7-16k) }{ 33 } \)
= \(\cfrac { 1 }{ 11 } (7-6k)\)
 \(\therefore\) Solution set is \(\left\{ \cfrac { 1 }{ 11 } \left( 7-16k \right), \cfrac { 1 }{ 11 } (3+k),k \right\} \)K \(\in\) R
Hence, for different values of k, we get infinitely many solutions.

The matrix equation corresponding to the given system is
\(\left( \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 2 & a \end{matrix} \right) \left( \begin{matrix} X \\ Y \\ Z \end{matrix} \right) =\left( \begin{matrix} 6 \\ 10 \\ b \end{matrix} \right) \)
AX = B

Case (i) For no solution:
The system possesses no solution only when \(\rho (A)\neq ([A,B])\) which is possible only when a−3 = 0 and b −10 \(\neq \) 0.
Hence for a = 3, b \(\neq \) 10, the system possesses no solution.
Case (ii) For a unique solution:
The system possesses a unique solution only when \(\rho (A)= ([A,B])\)=number of unknowns.
i.e when \(\rho (A)=\rho ([A,B])\) = 3
Which is possible only when a−3 \(\neq \) 0 and b may be any real number as we can observe .
Hence for a \(\neq \) and b \(\in \) R, the system possesses a unique solution.
Case (iii) For an infinite number of solutions:
The system possesses an infinite number of solutions only when
\(\rho (A)=\rho ([A,B])\) i,e when \(\rho (A)=\rho ([A,B])\) = 2 < 3 ( number of unknowns) which is possible only when a − 3 = 0, b−10 = 0
Hence for a = 3, b = 10, the system possesses infinite number of solutions.