The matrix form of the system is
\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
AX = B where
A=\(\left[ \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
|A| =\(\left| \begin{matrix} 3 & 1 & 9 \\ 3 & 2 & 12 \\ 2 & 1 & 7 \end{matrix} \right| =3\left| \begin{matrix} 2 & 12 \\ 1 & 7 \end{matrix} \right| -1\left| \begin{matrix} 3 & 12 \\ 2 & 7 \end{matrix} \right| +9\left| \begin{matrix} 3 & 2 \\ 2 & 1 \end{matrix} \right| \)
= 3 (14 - 12) - 1 (21 -24) + 9 (3 -4)
= 3 (2) -1 (-3) + 9 (-1)
= 6 + 3 - 9 = 9 - 9 = 0
Since |A| = 0, the homogeneous system of equations have non-trivial solutions also.

Let A =\(\left[ \begin{matrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{matrix} \right] \)
Now |A| = \(\left| \begin{matrix} 6 & -5 \\ -2 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & -5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} -15 & 6 \\ 5 & -2 \end{matrix} \right| \)
= 3 (12 - 10) + 1 (-30 + 25) + 1 (30 - 30)
= 3(2) + 1 (-5) + 0 = 6 - 5 = 1 ≠ 0
∴ Rank of A is 3.

Agumented matrix
[A|B] \(\left[ \begin{matrix} 2 & 5 \\ 6 & 15 \end{matrix}|\begin{matrix} 7 \\ 13 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & 5 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ -8 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 and \(\rho\)([A|B]) = 3
∴ \(\rho\) (a) ≠ \(\rho\) ([AIB])
Hence the system is inconsistent.

Given A (adj A) =\(\left[ \begin{matrix} 10 & 0 \\ 0 & 10 \end{matrix} \right] =10\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)...(1)
We know A (adj A) = (adj A) A = |A|. I₂ ...(2)
Comparing (1) and (2), we get |A| = 10

\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} -1 & 3 \\ 4 & -7 \\ 3 & -4 \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 2
∴ \(\rho \)(A) ≤ min (3, 2) = 2
The highest order of minor of A is 2
It is \(\left| \begin{matrix} -1 & 3 \\ 4 & -7 \end{matrix} \right| \)= 7-12 = 5 ≠ 0
∴ \(\rho \)(A) = 2

\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 2 & -4 \\ -1 & 2 \end{matrix} \right] \)
A is a matrix of order 2 \(\times\) 2
∴ \(\rho \)(A) ≤ min(2,2) = 2
The highest order of minor of A is 2
it is \(\left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| \)= 4 - 4 = 0
So, \(\rho \)(A)<2
Next consider the minor of order 1 |2| = 2 ≠ 0
∴ \(\rho \)(A) = 1

Let A = \(\left[ \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right] \). Then A is a matrix of order 3 × 3 and ρ(A) ≤ 3
The third order minor |A| = \(\left| \begin{matrix} 2 & 0 & -7 \\ 0 & 3 & 1 \\ 0 & 0 & 1 \end{matrix} \right| \) = (2)(3)(1) = 6 ≠ 0. So, ρ(A) = 3.
Note that there are three non-zero rows.

Given adj (A) =\(\left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
We know that A^-1 = ±\(\frac { 1 }{ \sqrt { |adjA| } } \) (adj A) ...............(1)
|adj A| = 0 + 2\(\left| \begin{matrix} 6 & -6 \\ -3 & 6 \end{matrix} \right| \) + 0
[Expanded along R₁]
= 2(36-18) = 2(18) = 36
∴ A^-1 = \(\pm \frac { 1 }{ \sqrt { 36 } } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \)
= \(\pm \frac { 1 }{ 6 } \left[ \begin{matrix} 0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6 \end{matrix} \right] \).

AB =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] =\left[ \begin{matrix} 8+3 & 10+4 \\ 20+9 & 25+12 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
|AB| =\(\left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right]\)
= 407 - 406 = 1 ≠ 0
(AB)^-1 = \(\frac { 1 }{ |AB| } \) adj(AB)
= \(\frac { 1 }{ 1 } \left[ \begin{matrix} 11 & 14 \\ 29 & 37 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \) ....(1)
|A| =\(\left[ \begin{matrix} 2 & 1 \\ 5 & 3 \end{matrix} \right] \) = 6 - 5 =1
|B| =\(\left[ \begin{matrix} 4 & 5 \\ 3 & 4 \end{matrix} \right] \) = 16 - 15 = 1
B^-1 = \(\frac { 1 }{ |B| } adjB=\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \)
A^-1 = \(\frac { 1 }{ |A| } adjA=\left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
∴ B^-1A^-1 =\(\left[ \begin{matrix} 4 & -5 \\ -3 & 4 \end{matrix} \right] \left[ \begin{matrix} 3 & -1 \\ -5 & 2 \end{matrix} \right] \)
=\(\\ \left[ \begin{matrix} 12+25 & -4-10 \\ -9-20 & 3+8 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 37 & -14 \\ -29 & 11 \end{matrix} \right] \)....(2)
From (1) and (2), (AB)^-1 = B^-1A^-1

Δ = \(\left| \begin{matrix} 2 & 3 \\ 1 & 6 \end{matrix} \right| \) = 12 - 3 = 9 ≠ 0
Δ₁ = \(\left| \begin{matrix} 10 & 3 \\ 4 & 6 \end{matrix} \right| \) = 60 - 12 = 48
Δ₂ = \(\left| \begin{matrix} 2 & 10 \\ 1 & 4 \end{matrix} \right| \) = 8 - 10 = -2
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 48 }{ 9 } =\frac { 16 }{ 3 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -2 }{ 9 } \)
∴ Solution set is \(\left\{ \frac { 16 }{ 3 } ,\frac { -2 }{ 9 } \right\} \).

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

2x-y = 8, 3x+2y+2 = -2
The matrix form of the system is
\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ AX = B where A =\(\\ \left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)
B =\(\left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
⇒ X = A^-1N
Now, |A| =\(\left[ \begin{matrix} 2 & -1 \\ 3 & 2 \end{matrix} \right] \)= 4 + 3 = 7
∴ A^-1= \(\frac { 1 }{ |A| } \)adj A
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 1 \\ -3 & 2 \end{matrix} \right] \left[ \begin{matrix} 8 \\ -2 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 16-2 \\ -24-4 \end{matrix} \right] \)
= \(\frac { 1 }{ 7 } \left[ \begin{matrix} 14 \\ -28 \end{matrix} \right] =\left[ \begin{matrix} \frac { 14 }{ 7 } \\ \frac { -28 }{ 7 } \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -4 \end{matrix} \right] \)
∴ x = 2, y = -4
Hence, the solution set is {2, -4}

\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 8 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 2 \\ 1 \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 3 \\ 2 \end{matrix} \end{matrix} \right] \)
A is a matrix of order 3 \(\times\) 4
∴ \(\rho \)(A) ≤ min(3, 4) = 3 
The highest order of minor of A is 3
It is \(\left| \begin{matrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 8 & 1 & 0 \end{matrix} \right| \) = 0+0-8(4-4) = 0
[Expanded along C₁]
Also, \(\left| \begin{matrix} 0 & 2 & 1 \\ 0 & 4 & 3 \\ 8 & 0 & 2 \end{matrix} \right| =0+0-8\left| \begin{matrix} 2 & 1 \\ 4 & 3 \end{matrix} \right| \)
[Expanded along C₁]
= -8(6-4) = -8(2) = -16 ≠ 0
∴ \(\rho \)(A) = 3

\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right] \)
Expending along R₁,
|A| = \(5\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| +1\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \)
= 5 (25 - 1)-1 (5 - 1)+ 1 (1 - 5)
= 5 (24) - 1(4) + 1(- 4)
= 120 - 4 - 4 = 120 - 8 = 112 ≠ 0
Since A is non singular, A^-1 exit
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 5 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 5 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 5 & 1 \end{matrix} \right| & -\left| \begin{matrix} 5 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 5 & 1 \\ 1 & 5 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(25-1)-(5-1)+(1-5) \\ -(5-1)+(25-1)-(5-1) \\ +(1-5)+(5-1)+(25-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & -4 & -4 \\ -4 & 24 & -24 \\ -4 & -4 & 24 \end{matrix} \right] \)
Taking 4 common from every entry we get,
adj A = \(4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 112 } .4\left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \)
= \(\frac { 1 }{ 28 } \left[ \begin{matrix} 6 & -1 & -1 \\ -1 & 6 & -1 \\ -1 & -1 & 6 \end{matrix} \right] \).

\(\left[ \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right] \)
Let A =\(\left( \begin{matrix} 2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2 \end{matrix} \right) \)
adj A =\(\left( \begin{matrix} +\left| \begin{matrix} 4 & 1 \\ 7 & 2 \end{matrix} \right| & -\left| \begin{matrix} 3 & 1 \\ 3 & 2 \end{matrix} \right| & +\left| \begin{matrix} 3 & 4 \\ 3 & 7 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & 1 \\ 7 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & 7 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & 1 \\ 4 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 1 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 3 & 4 \end{matrix} \right| \end{matrix} \right) \)
=\(\left[ \begin{matrix} +(8-7)-(6-3)+(21-12) \\ -(6-7)+(4-3)-(14-9) \\ +(3-4)-(2-3)+(8-9) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 1 & -3 & 9 \\ 1 & 1 & -5 \\ -1 & 1 & -1 \end{matrix} \right] ^{ T }\)
adj A =\(\left[ \begin{matrix} 1 & 1 & -1 \\ -3 & 1 & 1 \\ 9 & -5 & -1 \end{matrix} \right] \)

Let x represent the number of question with correct answer and y represent the number of questions with wrong answers.
By the given data, x + y 100 ............... (1)
x - \(\frac { 1 }{ 4 } \)y = 80
Multiplying by 4 we get we get
4x - y = 320...............(2)
From (1) and (2)
Δ = \(\\ \left| \begin{matrix} 1 & 1 \\ 4 & -1 \end{matrix} \right| \)= -1 - 4 = -5
Δ₁ = \(\left| \begin{matrix} 100 & 1 \\ 320 & -1 \end{matrix} \right| \) = -100 - 320 = -420
Δ₂ = \(\left| \begin{matrix} 1 & 100 \\ 4 & 320 \end{matrix} \right| \) = 320 - 400 = -80
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -720 }{ -5 } \) = +84
and y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -80 }{ -5 } \) = 16
Hence, the number of questions with correct answer is 84 and wrong question is 16.

5x − 2y + 16 = 0, x + 3y − 7 = 0
Given Δ = \(\left| \begin{matrix} 5 & -2 \\ 1 & 3 \end{matrix} \right| \) = 15+2 = 17
Δ₁ = \(\left| \begin{matrix} -16 & -2 \\ 7 & 3 \end{matrix} \right| \) = -48+14 = -34
Δ₂ = \(\left| \begin{matrix} 5 & -16 \\ 1 & 7 \end{matrix} \right| \) = 35+16 = 51
∴ x = \(\frac { \triangle _{ 1 } }{ \triangle } =\frac { -34 }{ 7 } \) = -2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 51 }{ 17 } \)
∴ Solution set is {-2, 3}

The augmented matrix [A|B] is \(\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 1 & \lambda \\ 1 & -7 & 8 \end{matrix}|\begin{matrix} 12 \\ \mu \\ 17 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 3 & 1 & \lambda \\ 2 & -3 & 5 \end{matrix}|\begin{matrix} 17 \\ \mu \\ 12 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2R_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 22 & \lambda -51 \\ 0 & 11 & -11 \end{matrix}|\begin{matrix} 17 \\ \mu -51 \\ -22 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 3 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 11 }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 0 & \lambda -2 \\ 0 & 1 & -1 \end{matrix}|\begin{matrix} 17 \\ \mu -7 \\ -2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & \lambda -2 \end{matrix}|\begin{matrix} 17 \\ -2 \\ \mu -7 \end{matrix} \right] \)
Case (i) : when λ ≠ 2,
\(\rho\) ([A|B]) = 3 and \(\rho\)(A) = 3
∴ \(\rho\)([AIB])= \(\rho\)(A) = 3 = the number of unknowns
∴ The system has unique solution
Case (ii) : when λ = 2, μ =7
\(\left[ \begin{matrix} 1 & -7 & 8 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 17 \\ -2 \\ 0 \end{matrix} \right] \)
Here \(\rho\)(A) = 2 and \(\rho\)([A|B]) = 2
∴ \(\rho\)(A) = \(\rho\)([A|B]) = 2 < number of unknowns
Thus the system is consistent with infinitely many solutions.
Case (iii) : When λ= 2 and μ ≠ 7
\(\rho\) (A) = 2 and \(\rho\) ([A|B]) = 3
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
Thus, the given system of equations is inconsistent.

Let the required numbers be x, y and z
By the given data,
x + y + z = 20  ....(1)
2z + x = 23 ⇒ x + 2z = 23...(2)
y + z + 3x = 46 ⇒  3x + y + z = 46..(3)
Δ = \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 2 \\ 3 & 1 & 1 \end{matrix} \right| =1\left| \begin{matrix} 0 & 2 \\ 0 & 1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 2 \\ 1 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -2 + 5 + 1 =4
Δ₁ = \(\left| \begin{matrix} 20 & 1 & 1 \\ 23 & 0 & 2 \\ 46 & 1 & 1 \end{matrix} \right| =20\left| \begin{matrix} 0 & 2 \\ 1 & 1 \end{matrix} \right| -1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| +1\left| \begin{matrix} 23 & 0 \\ 46 & 1 \end{matrix} \right| \)
= -40 + 69 + 23 = 52
Δ₂ = \(\left| \begin{matrix} 1 & 20 & 1 \\ 1 & 23 & 2 \\ 3 & 46 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 23 & 2 \\ 46 & 1 \end{matrix} \right| -20\left| \begin{matrix} 1 & 2 \\ 3 & 1 \end{matrix} \right| +1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| \)
= -69 + 100 - 23 = 8
Δ₃ = \(\left| \begin{matrix} 1 & 1 & 20 \\ 1 & 2 & 23 \\ 3 & 1 & 46 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 0 & 23 \\ 1 & 46 \end{matrix} \right| -1\left| \begin{matrix} 1 & 23 \\ 3 & 46 \end{matrix} \right| +20\left| \begin{matrix} 1 & 0 \\ 3 & 1 \end{matrix} \right| \)
= -23 + 23 + 20 = 20
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 52 }{ 4 } \) = 13
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 8 }{ 4 } \) = 2 and z =\(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 20 }{ 4 } \) = 5
Hence the required numbers are 13, 2 and 5.

Put \(\frac { 1 }{ x } \) = a, \(\frac { 1 }{ y } \) = b, \(\frac { 1 }{ z } \) = c
∴ 2a + 3b + 10c = 4  ....(1)
4a- 6b - 5c = 1 .....(2)
6a + 9b -20c = 2 ...(3)
Δ = \(\left| \begin{matrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{matrix} \right| \)
= \(2\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (120 - 45) -3 (-80 - 30) + 10 (36 + 36)
= 150 + 330 + 720 = 1200
Δ₁ = \(\left| \begin{matrix} 4 & 3 & 10 \\ 1 & -6 & 5 \\ 2 & 9 & -20 \end{matrix} \right| \)
= \(4\left| \begin{matrix} -6 & 5 \\ 9 & -20 \end{matrix} \right| -3\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| +10\left| \begin{matrix} 1 & -6 \\ 2 & 9 \end{matrix} \right| \)
= 4 (120 - 45) -3 (-20 - 10) + 10 (9 + 12)
= 300 + 90 + 120 = 600
Δ₂ = \(\left| \begin{matrix} 2 & 4 & 10 \\ 4 & 1 & 5 \\ 6 & 2 & -20 \end{matrix} \right| =2\left| \begin{matrix} 1 & 5 \\ 2 & -20 \end{matrix} \right| -4\left| \begin{matrix} 4 & 5 \\ 6 & -20 \end{matrix} \right| +10\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| \)
= 2 (-2 - 10) - 4 (-80 - 30) + 10 (8 - 6)
= -60 + 440 + 20 = 400
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 4 \\ 4 & -6 & 1 \\ 6 & 9 & 2 \end{matrix} \right| 2\left| \begin{matrix} -6 & 1 \\ 9 & 2 \end{matrix} \right| -3\left| \begin{matrix} 4 & 1 \\ 6 & 2 \end{matrix} \right| +10\left| \begin{matrix} 4 & -6 \\ 6 & 9 \end{matrix} \right| \)
= 2 (-12 - 9) -3 (8 - 6) + 4 (36 + 36)
= - 42 - 6 + 288 = 240
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 1200 } =\frac { 1 }{ 2 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 2 } \) ⇒ x = 2
∴ b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 400 }{ 1200 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \) ⇒ y = 1 
∴ c = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 240 }{ 1200 } =\frac { 1 }{ 5 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 5 } \) ⇒ z = 5
∴ Solution set is {2, 3, 5}

2x + 3y − z = 0, x − y − 2z = 0, 3x + y + 3z = 0
Reducing the augmented matrix to row - echelon form we get
[A|0]=\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 2 & 3 & -1 \\ 3 & 1 & 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 4 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & -2 \\ 0 & 5 & 3 \\ 0 & 0 & \frac { 33 }{ 5 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|0] = 3
So, \(\rho \)(A) = \(\rho \)(A|0]) = 3 = Number of unknowns Hence, the system is consistent with unique solutions.
Thus, the system has trivial solution only.
x = 0, y = 0, z = 0

x+y+z-2 = 0, 6x-4y+5z-31= 0, 5x+2y+2z = 13
The matrix form of the system is
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ 13 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 1 & 1 \\ 6 & -4 & 5 \\ 5 & 2 & 2 \end{matrix} \right| =1\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| -1\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| +1\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \)
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} -4 & 5 \\ 2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 6 & 5 \\ 5 & 2 \end{matrix} \right| & +\left| \begin{matrix} 6 & -4 \\ 5 & 2 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 5 & 2 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ -4 & 5 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 6 & 5 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 6 & -4 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(-8-10) & -(12-25) & +(12+20) \\ -(2-2) & +(2-5) & -(2-5) \\ +(5+4) & -(5-6) & +(-4-6) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 13 & 32 \\ 0 & -3 & 3 \\ 9 & 1 & -10 \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -18 & 0 & 9 \\ 13 & -3 & 1 \\ 32 & 3 & -10 \end{matrix} \right] \left[ \begin{matrix} 2 \\ 31 \\ 13 \end{matrix} \right] \)
= \(\frac { 1 }{ 27 } \left[ \begin{matrix} -36 & +0 & +117 \\ 26 & -93 & +13 \\ 64 & +93 & -130 \end{matrix} \right] =\frac { 1 }{ 27 } \left[ \begin{matrix} 81 \\ -54 \\ 27 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ 1 \end{matrix} \right] \)
∴ x = 3, y = -2, z = 1
∴ Solution set is {3, -2, 1}

2x + 3y - z = 9, x + y + z = 9, 3x - y - z = -1
The matrix form of the system is
\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right] \)
X =\(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
⇒ X = A^-1N
|A| = \(\left| \begin{matrix} 2 & 3 & -1 \\ 1 & 1 & 1 \\ 3 & -1 & -1 \end{matrix} \right| =2\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| -3\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \)
= 2(-1+1)-3(-1-3)-1(-1-3)
= 0-3(-4) = 12+4 = 16
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 1 \\ -1 & -1 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| & +\left| \begin{matrix} 1 & 1 \\ 3 & -1 \end{matrix} \right| \\ -\left| \begin{matrix} 3 & -1 \\ -1 & -1 \end{matrix} \right| & +\left| \begin{matrix} 2 & -1 \\ 3 & -1 \end{matrix} \right| & -\left| \begin{matrix} 2 & 3 \\ 3 & -1 \end{matrix} \right| \\ +\left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(-1+1) & -(-1-3) & +(-1-3) \\ -(-3-1) & +(2+3) & -(-2-9) \\ +(3+1) & -(2+1) & +(2-3) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 0 & 4 & -4 \\ 4 & 1 & 11 \\ 4 & -3 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adjA\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0 & 4 & 4 \\ 4 & 1 & -3 \\ -4 & 11 & -1 \end{matrix} \right] \left[ \begin{matrix} 9 \\ 9 \\ -1 \end{matrix} \right] \)
= \(\frac { 1 }{ 16 } \left[ \begin{matrix} 0+36-4 \\ 36+9+3 \\ -36+99+1 \end{matrix} \right] =\frac { 1 }{ 16 } \left[ \begin{matrix} 32 \\ 48 \\ 64 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \\ 4 \end{matrix} \right] \)
∴ x = 2, y = 3, z = 4
∴ Solution set is {2, 3, 4}

Here the number of equations is equal to the number of unknowns.
Transforming into echelon form (Gaussian elimination method), the augmented matrix becomes
\(\left[ \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & -2 & -5 \\ 0 & -4 & -9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow R_{ 2 }\div \left( -1 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div 7{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 4 & 9 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)\(\overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) }{ { \longrightarrow } } \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 5 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ([A | O]) = 3 = Number of unknowns.
Hence, the system has a unique solution. Since x = 0, y = 0, z = 0 is always a solution of the homogeneous system, the only solution is the trivial solution x = 0, y = 0, z = 0.
Note
In the above example, we find that
|A| = \(\left| \begin{matrix} 1 & 2 & 3 \\ 3 & 4 & 4 \\ 7 & 10 & 12 \end{matrix} \right| \) = 1(48 - 40) - 2(36 - 28) + 3(30 - 28) = 8 - 16 + 6 = -2 ≠ 0.

2x+3y = 9, 7x+3y-5z = 8, 2x+3y+⋋z = μ
The matrix form of the system is AX = B where
A =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ,B=\left[ \begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \)
Applying elementary row operations augmented matrix [A|B] we get
[A|B] =\(\left[ \begin{matrix} 2 & 3 & 5 \\ 7 & 3 & -5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 9 \\ 8 \\ \mu \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 2 & 3 & 5 \\ 2 & 3 & \lambda \end{matrix}|\begin{matrix} 8 \\ 9 \\ \mu \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-\frac { 2 }{ 7 } { R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & \frac { 15 }{ 7 } & \frac { 45 }{ 7 } \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ \frac { 45 }{ 7 } \\ 4-9 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\times 7 }{ \longrightarrow } \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & \lambda -5 \end{matrix}|\begin{matrix} -8 \\ 47 \\ \mu -9 \end{matrix} \right] \)
Case (i): when λ = 5
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ -4 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 3
So, \(\rho \)(A) ≠ \(\rho \)[A|B]
Hence the system is inconsistent and has no solution
Case (ii) : When λ ≠ 5, μ ≠ 9
[A|B] =\(\\ \left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} -8 \\ 47 \\ not\quad zero \end{matrix} \right] \)
Here \(\rho \)(A) = 3 and \(\rho \)[A|B] = 3
∴ \(\rho \)(A) = \(\rho \)[A|B] = 3 = number of unknowns
Hence, the system is consistent with solution
Case (iii) : When λ = 5, μ = 9
[A|B] =\(\left[ \begin{matrix} 7 & 3 & -5 \\ 0 & 15 & 45 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -8 \\ 47 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 2 and \(\rho \)[A|B] = 2
∴ \(\rho \)(A) = \(\rho \)[A|B] = 2
∴ The system is consistent and has infinite number of solutions.

2x - y + z = 2, 6x - 3y + 3z = 6, 4x - 2y + 2z = 4
The matrix form of the given system is AX = B where
A =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] B=\left[ \begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \)
Applying elementary row operations on the augment matrix [A|B], we get,
[A|B] =\(\left[ \begin{matrix} 2 & -1 & 1 \\ 6 & -3 & 3 \\ 4 & -2 & 2 \end{matrix}|\begin{matrix} 2 \\ 6 \\ 4 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 2 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1 [∵ only one non zero row]
and \(\rho \)[A|B] = 1 [∵ only one-zero row]
∴ \(\rho \)(A) =\(\rho \)(A|B] = 1< 3 the given system is consistent and has two parameter family of solutions.
So, z = t and y = s where, t \(\in \)R
Writing the equivalent equations from the rowechelon matrix, we get
2x-y+z = 2 .............(1)
y = s
z = t
Substituting (2) and (3) In (1) we get
2x-s+t = 2
⇒ 2x-s+t = 2
⇒ x = \(\frac{1}{2}\)[s-t+2]
∴ Solution set is {\(\frac{1}{2}\)(s-t+2),s,t} here s, t \(\in \) R.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix[A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & -2 & 2 \\ 3 & -3 & 3 \end{matrix}|\begin{matrix} -9 \\ -18 \\ -27 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-3{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -9 \\ 0 \\ 0 \end{matrix} \right] \).
So, ρ(A) = ρ ([A | B]) = 1 < 3.
From the echelon form, we get the equivalent equations x - y + z = -9, 0 = 0, 0 = 0.
The equivalent system has one non-trivial equation and three unknowns.
Taking y = s, z = t arbitrarily, we get x - s + t = -9; x = -9 + s - t.
So, the solution is (x = -9 + s - t, y = s, z = t), where s and t are parameters.
The above solution set is a two-parameter family of solutions.
Here, the given system of equations is consistent and has infinitely many solutions which form a two parameter family of solutions.