Given \(\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right) \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x+0+x \\ 0+y+0 \\ 0+0+z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow \left( \begin{matrix} x \\ y \\ z \end{matrix} \right) =\left( \begin{matrix} 1 \\ -1 \\ 0 \end{matrix} \right) \)
\(\Rightarrow x=1,y=-1,z=0\)

The matrix A is singular, if
\(\left| \begin{matrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{matrix} \right| =0\)
\(1\left| \begin{matrix} 2 & 1 \\ 2 & -3 \end{matrix} \right| +2\left| \begin{matrix} 1 & 1 \\ x & -3 \end{matrix} \right| +3\left| \begin{matrix} 1 & 2 \\ x & 2 \end{matrix} \right| =0\)
\(\Rightarrow\) (-8) -6 - 2x + 6 - 6x = 0
\(\Rightarrow\) -8-2x-6x = 0
\(\Rightarrow\) -8-8x = 0
\(\Rightarrow\) -8 = 8x
\(\Rightarrow\) \(x=\cfrac { -8 }{ 8 } =-1\)

Given non-homogeneous equations are x + y + z = 6, x + 2y + 3z = 14, x + 4y + 7z = 30

Here \(\rho (A)=\rho (A,B) = 2\) 
\(\therefore\) The given system is consistent

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

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.

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
\(\therefore \rho (A)=3\)

Let ‘x’ be the cost of a pencil
Let ‘y’ be the cost of an eraser
\(\therefore \) By given data, we get the following equations
11x + 3y = 64
8x + 3y = 49
\(\triangle =\left| \begin{matrix} 11 & 3 \\ 8 & 3 \end{matrix} \right| =9\neq 0,\) It has unique solution.
\({ \triangle }_{ x }\left| \begin{matrix} 64 & 3 \\ 49 & 3 \end{matrix} \right| =45\)
\({ \triangle }_{ y }\left| \begin{matrix} 11 & 64 \\ 8 & 49 \end{matrix} \right| =27\)
\(\therefore \) By Cramer’s rule
\(x={ \frac { \triangle x }{ \triangle } =\frac { 45 }{ 9 } =5 }\)
\(y={ \frac { \triangle y }{ \triangle } =\frac { 27 }{ 9 } =3 }\)
\(\therefore \) The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Let the price of one share of A be x
Let the price of one share of B be y
\(\therefore \) By given data, we get the following equations
10x + 5y = 125
9x + 12y = 150
\(\triangle =\left| \begin{matrix} 10 & 5 \\ 9 & 12 \end{matrix} \right| =75\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 125 & 5 \\ 150 & 12 \end{matrix} \right| =750\)
\({ \triangle }_{ y }=\left| \begin{matrix} 10 & 125 \\ 9 & 150 \end{matrix} \right| =375\)
\(\therefore \) Cramer’s rule 
\(x=\frac { \triangle x }{ \triangle } =\frac { 750 }{ 75 } =10\)
\( y=\frac { \triangle y }{ \triangle } =\frac { 375 }{ 75 } =5\)
The price of the share A is Rs10 and the price of the share B is Rs. 5.

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.
\(\therefore \rho (A)=2\)

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 ‘x’ be the cost of a Business Mathematics book
Let ‘y’ be the cost of a Accountancy book.
Let ‘z’ be the cost of a Commerce book.
\(\therefore \) 3x + 2y + z = 840
2x + y + z = 570
x + y + 2z = 630
Here \({ \triangle }=\left| \begin{matrix} 3 & 2 & 1 \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} \right| =-2\neq 0\)
\({ \triangle }_{ x }=\left| \begin{matrix} 840 & 2 & 1 \\ 570 & 1 & 1 \\ 630 & 1 & 2 \end{matrix}\begin{matrix} 1 \\ 1 \\ 2 \end{matrix} \right| =-240 \)
\({ \triangle }_{ y }=\left| \begin{matrix} 3 & 840 & 1 \\ 2 & 570 & 1 \\ 1 & 630 & 2 \end{matrix} \right| =-300 \)
\({ \triangle }_{ z }=\left| \begin{matrix} 3 & 2 & 840 \\ 2 & 1 & 570 \\ 1 & 1 & 630 \end{matrix} \right| =-360\)
\(\therefore \) By Cramer’s rule
\(x=\frac { { \triangle }x }{ { \triangle } } =\frac { -240 }{ -2 } =120 \)
\(y=\frac { { \triangle }y }{ { \triangle } } =\frac { 300 }{ -2 } =150 \)
\(z=\frac { { \triangle }z }{ { \triangle } } =\frac { 360 }{ -2 } =180\) 
\(\therefore \) The cost of a Business Mathematics book is Rs. 120,
the cost of a Accountancy book is Rs. 150 and
the cost of a Commerce book is Rs. 180.

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 \(\cfrac { -7-2\lambda }{ 21 } =0\)
\(\Rightarrow -7-2\lambda =0\)
\(\Rightarrow -7=2\lambda \)
\(\Rightarrow \lambda =\cfrac { -7 }{ 2 } \)

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 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.

non-singular matrix

\({ R }_{ 1 }\leftrightarrow { C }_{ 1 }\)

\({ P }_{ jk }\le 0\)

3

Trivial solution