Augmented matrix [A|B] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 2 & 2 & 6 \\ 2 & 1 & 1 \end{matrix}|\begin{matrix} 4 \\ 7 \\ 10 \end{matrix} \right] \)
[A|B] \(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -5 \end{matrix}|\begin{matrix} 4 \\ -1 \\ 2 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -5 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 4 \\ 2 \\ -1 \end{matrix} \right] \)
Here \(\rho\) (A) = 2 [only 2 two-non zero rows]
And \(\rho\) ([A|B]) = 3 [There are 3 non-zero rows]
∴ \(\rho\) (A) ≠ \(\rho\) ([A|B])
∴ The system is inconsistent.

The matrix from of the system of equations is
\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \) ⇒ AX = B where
A =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 7 \\ 14 \end{matrix} \right] \)
Transforming augmented matrix to row-echelon form we get
[A|B] =\(\left[ \begin{matrix} 2 & -3 \\ 4 & -6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 2 & -3 \\ 0 & 0 \end{matrix}|\begin{matrix} 7 \\ 0 \end{matrix} \right] \)
Here \(\rho\) (A) = \(\rho\) [A|B] = 1
∴ \(\rho\) (A) = \(\rho\) [A|B] = 1 < the number of unknowns, the system is consistent and has one parameter family of solutions
∴ Put y = t, where t \(\in \) R
Writing the row-echelon form to equations we get
 2x - 3y = 7
∴ 2x - 3t = 7
⇒  2x = 7 + 3t
⇒ x = \(\frac { 1 }{ 2 } \)(7 + 3t)
∴ Solution set is {\(\frac { 1 }{ 2 } \)(7+3t), t} where t \(\in \) R.

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

\(\left[ \begin{matrix} 3 & -1 & 2 \\ -6 & 2 & 4 \\ -3 & 1 & 2 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }+2{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 4 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-\frac { 1 }{ 2 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \)
Note
\(\left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 8 \\ 0 & 0 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }/8 }{ \longrightarrow } \left[ \begin{matrix} 3 & -1 & 2 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix} \right] \).
This is also a row-echelon form of the given matrix.
So, a row-echelon form of a matrix is not necessarily unique.

Given adj A =\(\left[ \begin{matrix} 2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2 \end{matrix} \right] \)
We know that A = \(\pm \frac { 1 }{ \sqrt { |adjA| } } \) adj (adj A)..(1)
|adj A| = \(2\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| +4\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| +2\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \)
[Expanded along R₁]
= 2(24-0)+4(-6-14)+2(0+24)
= 2(24)+4(-20)+2(24) = 48-80+48
= 96-80 = 16
Now, adj (adj A)
=\(\left[ \begin{matrix} +\left| \begin{matrix} 12 & -7 \\ 0 & 2 \end{matrix} \right| & -\left| \begin{matrix} -3 & -7 \\ -2 & 2 \end{matrix} \right| & +\left| \begin{matrix} -3 & 12 \\ -2 & 0 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 2 \\ 0 & 2 \end{matrix} \right| & +\left| \begin{matrix} 2 & 2 \\ -2 & 2 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -2 & 0 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 2 \\ 12 & -7 \end{matrix} \right| & -\left| \begin{matrix} 2 & 2 \\ -3 & -7 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ -3 & 12 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(24-0)-(6-14)+(0+24) \\ -(-8-0)+(4+4)-(0-8) \\ +(28-24)-(-14+6)+(24-12) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} 24 & 20 & 24 \\ 8 & 8 & 8 \\ 4 & 8 & 12 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 24 & 8 & 4 \\ 20 & 8 & 8 \\ 24 & 8 & 12 \end{matrix} \right] \)
= \(4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
Substituting (2) and (3) in (1) we get,
A = \(\frac { 1 }{ \sqrt { 16 } } .4\left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)
A = \(\pm \frac { 4 }{ 4 } \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] =\pm \left[ \begin{matrix} 6 & 2 & 1 \\ 5 & 2 & 2 \\ 6 & 2 & 3 \end{matrix} \right] \)

First, we find |adj (A)| = \(\left| \begin{matrix} 7 & 7 & -7 \\ -1 & 11 & 7 \\ 11 & 5 & 7 \end{matrix} \right| \) =  7(77 - 35) - 7(-7 - 77) - 7(-5 - 121) = 1764 > 0
So, we get
A = \(\pm \frac { 1 }{ \sqrt { \left| adjA \right| } } \) adj(adj A) = \(\pm \frac { 1 }{ \sqrt { 1764 } } { \left[ \begin{matrix} +\left( 77-35 \right) & -\left( -7-77 \right) & +\left( -5-121 \right) \\ -\left( 49+35 \right) & +\left( 49+77 \right) & -\left( 35-77 \right) \\ +\left( 49+77 \right) & -\left( 49-7 \right) & +\left( 77+7 \right) \end{matrix} \right] }^{ T }\)
= \(\pm \frac { 1 }{ 42 } { \left[ \begin{matrix} 42 & 84 & -126 \\ -84 & 126 & 42 \\ 126 & -42 & 84 \end{matrix} \right] }^{ T }=\pm \left[ \begin{matrix} 1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2 \end{matrix} \right] \).

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}

Let P(x) = ax²+ bx + c
Given P(-3) = 21
[∵ P(x) ÷ x + 3, the remainder is 21]
⇒ a(-3)2 + b(-3) + c = 21
⇒ 9a - 3b + c = 21
Also, P(5) = 61
⇒ a(5)2 + b(5) + c = 61
[using remainder theorem]
⇒ 25a +5b + c = 61..........(2)
and P(1) = 9
⇒ a(1)2 + b(1) + c = 9
⇒ a + b + c = 9 ............(3)
Reducing the augment matrix to an equivalent row-echelon form using elementary row operations, we get
\(\left[ \begin{matrix} 9 & - & 1 \\ 25 & 5 & 1 \\ -1 & 1 & 1 \end{matrix}|\begin{matrix} 21 \\ 61 \\ 9 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 25 & 5 & 1 \\ 9 & -3 & 1 \end{matrix}|\begin{matrix} 9 \\ 61 \\ 21 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-9{ R }_{ 1 }\\ { R }_{ 2 }\rightarrow { R }_{ 2 }-25{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -20 & -24 \\ 0 & -12 & -8 \end{matrix}|\begin{matrix} 9 \\ -164 \\ -60 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 4 }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & -3 & -2 \end{matrix}|\begin{matrix} 9 \\ -41 \\ -15 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-\frac { 3 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & \frac { 8 }{ 5 } \end{matrix}|\begin{matrix} 9 \\ -41 \\ \frac { 48 }{ 5 } \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow 5{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & -5 & -6 \\ 0 & 0 & 8 \end{matrix}|\begin{matrix} 9 \\ -41 \\ 48 \end{matrix} \right] \)
Writing the equivalent equations from the row-echelon matrix we get,
a + b + c = 9 ............(1)
5b + 6c = 41 ................(2)
-8c = -48
⇒ c = 6
Substituting c = 6 
⇒  5b + 36 = 41
⇒ 5b = 5  
b = 1
Substituting b = 1, c = 6
a + 1 + 6 = 9
⇒ a + 7 = 9
⇒ a = 9 - 7
⇒ a = 2
∴ a = 2, b = 1, and c = 6

Transforming the augmented matrix to echelon form, we get
\(\left[ \begin{matrix} 4 & 3 & 6 \\ 1 & 5 & 7 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 25 \\ 13 \\ 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 4 & 3 & 6 \\ 2 & 9 & 1 \end{matrix}|\begin{matrix} 13 \\ 25 \\ 1 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-2{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & -17 & -22 \\ 0 & -1 & -13 \end{matrix}|\begin{matrix} 13 \\ -27 \\ -25 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -1 \right) \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -1 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 1 & 13 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 25 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 17{ R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 5 & 7 \\ 0 & 17 & 22 \\ 0 & 0 & 199 \end{matrix}|\begin{matrix} 13 \\ 27 \\ 398 \end{matrix} \right] \).
The equivalent system is written by using the echelon form:
x + 5y + 7 = 13, … (1)
17y + 22z = 27, … (2)
199z = 398..... (3)
From (3), we get z = \(\frac { 398 }{ 199 } \) = 2.
Substituting z = 2 in (2), we get y = \(\frac { 27-22\times 2 }{ 17 } =\frac { -17 }{ 17 } \) = -1
Substituting z = 2, y = -1, in (1), we get x = 13 - 5 x (-1) - 7 \(\times\) 2 = 4.
So, the solution is (x = 4, y = -1, z = 2).

First we evaluate the determinants
Δ = \(\left| \begin{matrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix} \right| \) = 6 ≠ 0, Δ₁ = \(\left| \begin{matrix} 3 & -1 & 0 \\ 17 & 3 & 4 \\ 7 & 1 & 2 \end{matrix} \right| \) = 12, Δ₂ = \(\left| \begin{matrix} 1 & 3 & 0 \\ 2 & 17 & 4 \\ 0 & 7 & 2 \end{matrix} \right| \) = -6, Δ₃ = \(\left| \begin{matrix} 1 & -1 & 3 \\ 2 & 3 & 17 \\ 0 & 1 & 7 \end{matrix} \right| \) = 24
By Cramer’s rule, we get  x₁ = \(\frac { { \Delta }_{ 1 } }{ \Delta } =\frac { 12 }{ 6 } \) = 2, x₂ = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { -6 }{ 6 } \) = -1, x₃ = \(\frac { 24 }{ 6 } \) = 4.
So, the solution is (x₁ = 2, x₂ = -1, x₃ = 4).

Δ = \(\left| \begin{matrix} 6 & -7 \\ 9 & -5 \end{matrix} \right| \) = -30 + 63 = 33
Δ₁ = \(\left| \begin{matrix} 16 & -7 \\ 35 & -5 \end{matrix} \right| \) = -80 + 245 = 165
Δ₂ = \(\left| \begin{matrix} 6 & 16 \\ 9 & 35 \end{matrix} \right| \) = 210 - 144 = 66
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 165 }{ 33 } \) = 5
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 66 }{ 33 } \) = 2
∴ Solution set is { 5, 2}

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.