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.

Given A =\(\left[ \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right] \), B =\(\left[ \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right] \) and C =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
Also,  A X B = C
Premultiply by A^-1 we get,
(A^-1 A) X B = A^-1 C
⇒ XB = A^-1C    [∵ A^-1a = 1]
Post Multiply by B^-1 we get
(X B) B^-1 = (A^-1 C) B^-1
⇒ X = (A^-1 C) B^-1
|A| = \(\left| \begin{matrix} 1 & -1 \\ 2 & 0 \end{matrix} \right| \) = 0 + 2 = 2 ≠ 0
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \)
|B| = \(\left| \begin{matrix} 3 & -2 \\ 1 & 1 \end{matrix} \right| \) = 3 + 2 = 5 ≠ 0
∴ B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] \)
A^-1C = \(\\ \frac { 1 }{ 2 } \left[ \begin{matrix} 0 & 1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 \\ 2 & 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 2 } \left[ \begin{matrix} 0+2 & 0+2 \\ -2+2 & -2+2 \end{matrix} \right] \)
=\(\frac { 1 }{ 2 } \left[ \begin{matrix} 2 & 2 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 2 } (2)\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = Z(A^-1C).B^-1
= \(\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right] \frac { 1 }{ 5 } \left[ \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 1-1 & 2+3 \\ 0+0 & 0+0 \end{matrix} \right] \)
= \(\frac { 1 }{ 5 } \left[ \begin{matrix} 0 & 5 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 5 } (5)\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] =\frac { 1 }{ 10 }\left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \)
∴ X = \(\frac { 1 }{ 10 } \left[ \begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix} \right] \).

Given A\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
Let B =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \) and 
C = \(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \)
∴ AB = C
Post multiply by B^-1 we get
A(BB^-1) = CB^-1
⇒ A = CB^-1     [∵ BB^-1 = 1]
|B| = \(\left| \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right| \)
= -10 + 3 = -7 ≠ 0
∴ B^-1 exists
B^-1 = \(\frac { 1 }{ |B| } adjB=\frac { -1 }{ 7 } \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
A = CB^-1
=\(\left[ \begin{matrix} 14 & 7 \\ 7 & 7 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(7\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left( \frac { -1 }{ 7 } \right) \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} 2 & 1 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 1 & 5 \end{matrix} \right] \)
= \(-\left[ \begin{matrix} -4+1 & -6+5 \\ -2+1 & -3+5 \end{matrix} \right] =-\left[ \begin{matrix} -3 & -1 \\ -1 & 2 \end{matrix} \right] \)
A = \(\left[ \begin{matrix} 3 & 1 \\ 1 & -2 \end{matrix} \right] \).

Given A = \(\left[ \begin{matrix} 1 & \tan { x } \\ -\tan { x } & 1 \end{matrix} \right] \)
|A| = 1 + tan² x
∴ A^-1 = \(\frac { 1 }{ |A| } \)adjA
= \(\frac { 1 }{ 1+tan^{ 2 } } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
[Interchange the elements in the leading diagonal and change the sign of elements in the off diagonal]
A^T= \(\\ \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
∴ A^TA^-1
=\(\left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & -tanx \\ tanx & 1 \end{matrix} \right] \)
=\(\frac { 1 }{ 1+tan^{ 2 }x } \left[ \begin{matrix} 1-tan^{ 2 }x & -tanx-tanx \\ tanx+tanx & -tan^{ 2 }x+1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} \frac { 1-{ tan }^{ 2 }x }{ 1+{ tan }^{ 2 }x } & \frac { -2tanx }{ 1+{ tan }^{ 2 }x } \\ \frac { 2tanx }{ 1+{ tan }^{ 2 }x } & \frac { 1-{ tan }^{ 2 }x }{ 1+tan^{ 2 }x } \end{matrix} \right] \)
A^TA^-1 =\(\left[ \begin{matrix} cos2x & -sin2x \\ sin2x & cos2x \end{matrix} \right] \)
Hence proved.

Since A² = \(\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] =\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] \).
A² + xA + yI₂ = O₂ ⇒ \(\left[ \begin{matrix} 22 & 27 \\ 18 & 31 \end{matrix} \right] +x\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] +y\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 22+4x+y & 27+3x \\ 18+2x & 31+5x+y \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \).
So, we get 22 + 4x + y = 0, 31 + 5x + y = 0, 27 + 3x = 0 and 18 + 2x = 0.
Hence x = −9 and y = 14. Then, we get  A² - 9A + 14I₂ = O₂.
Postmultiplying this equation by A^-1, we get A - 9I₂ + 14A^-1 = O₂. Hence, we get
A^-1 = \(\frac { 1 }{ 14 } \) (9I₂ - A) = \(\frac { 1 }{ 14 } \left( 9\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] -\left[ \begin{matrix} 4 & 3 \\ 2 & 5 \end{matrix} \right] \right) =\frac { 1 }{ 14 } \left[ \begin{matrix} 5 & -3 \\ -2 & 4 \end{matrix} \right] \).

We get AB = \(\left[ \begin{matrix} 0 & -3 \\ 1 & 4 \end{matrix} \right] \left[ \begin{matrix} -2 & -3 \\ 0 & -1 \end{matrix} \right] =\left[ \begin{matrix} 0+0 & 0+3 \\ -2+0 & -3-4 \end{matrix} \right] =\left[ \begin{matrix} 0 & 3 \\ -2 & -7 \end{matrix} \right] \)
(AB)^-1 = \(\frac { 1 }{ \left( 0+6 \right) } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \)...(1)
A^-1 = \(\frac { 1 }{ \left( 0+3 \right) } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] \)
B^-1 = \(\frac { 1 }{ \left( 2-0 \right) } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] =\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \)
B^-1A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right] \frac { 1 }{ 3 } \left[ \begin{matrix} 4 & 3 \\ -1 & 0 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} -7 & -3 \\ 2 & 0 \end{matrix} \right] \) ...(2).
As the matrices in (1) and (2) are same, (AB)⁻¹ = B⁻¹A⁻¹ is verified.

\(\frac { 3 }{ x } -\frac { 4 }{ y } -\frac { 2 }{ z } \) - 0, \(\frac { 1 }{ x } +\frac { 2 }{ y } +\frac { 1 }{ z } \) - 2 = 0, \(\frac { 2 }{ x } -\frac { 5 }{ y } -\frac { 4 }{ z } \) + 1 = 0
Put \(\frac { 1 }{ x } =u,\frac { 1 }{ y } =v,\frac { 1 }{ z } =w\)
We get 3u - 4v - 2w = 1, u + 2v + w = 2, 2u - 5v - 4w = -1
∴ \(\left| \begin{matrix} 3 & -4 & -2 \\ 1 & 2 & 1 \\ 2 & -5 & -4 \end{matrix} \right| =3\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(- 8 + 5) + 4'(- 4 - 2) - 2(- 5 - 4)
= 3(- 3) + 4(- 6) - 2(- 9)
= - 9 - 24 + 18 = -15
Δ₁ = \(\left| \begin{matrix} 1 & -4 & -2 \\ 2 & 2 & 1 \\ -1 & -5 & -4 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 2 & 1 \\ -5 & -4 \end{matrix} \right| +4\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -2\left| \begin{matrix} 2 & 2 \\ -1 & -5 \end{matrix} \right| \)
= 1(- 8 + 5) + 4(- 8 + 11 -2(-10 + 2)
= 1(- 3) + 4(-7) - 2(- 8)
= - 3 - 28 + 16= -15
Δ₂ = \(\left| \begin{matrix} 3 & 1 & -2 \\ 1 & 2 & 1 \\ 2 & -1 & -4 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 1 \\ -1 & -4 \end{matrix} \right| -1\left| \begin{matrix} 1 & 1 \\ 2 & -4 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| \)
= 3(- 8 + 1) - 1(- 4 - 2) - 2(- 1 - 4)
= 3(-7) - 1(- 6) - 2(- 5)
= -21 + 6 + 10 = -5
Δ₃ = \(\left| \begin{matrix} 3 & -4 & 1 \\ 1 & 2 & 2 \\ 2 & -5 & -1 \end{matrix} \right| \)
\(3\left| \begin{matrix} 2 & 2 \\ -5 & -1 \end{matrix} \right| +4\left| \begin{matrix} 1 & 2 \\ 2 & -1 \end{matrix} \right| -2\left| \begin{matrix} 1 & 2 \\ 2 & -5 \end{matrix} \right| \)
= 3(-2 + 10) + 4(-1 - 4)+ 1(-5 - 4)
= 3(8) + 4(- 5) + 1(- 9)
= 24 - 20 - 9 = - 5
∴ \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -15 }{ -15 } =1\Rightarrow \frac { 1 }{ x } =1\Rightarrow \)x = 1
v = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 3 } \Rightarrow \)y = 3
w = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 5 }{ -15 } =\frac { 1 }{ 3 } \Rightarrow \frac { 1 }{ z } =\frac { 1 }{ 3 } \Rightarrow \)z = 3
∴ Solution set is {1, 3, 3}

3x + 3y − z = 11, 2x − y + 2z = 9, 4x + 3y + 2z = 25.
Δ = \(\left| \begin{matrix} 3 & 3 & -1 \\ 2 & -1 & 2 \\ 4 & 3 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-2-6)-3(4-8)-1(6+4)·
= 3(- 8) - 3(- 4) - 1(10)
= - 24 + 12 - 10 = - 22
Δ₂ = \(\left| \begin{matrix} 11 & 3 & -1 \\ 9 & -1 & 2 \\ 25 & 3 & 2 \end{matrix} \right| \)
= \(11\left| \begin{matrix} -1 & 2 \\ 3 & 2 \end{matrix} \right| -3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -1\left| \begin{matrix} 9 & -1 \\ 25 & 3 \end{matrix} \right| \)
= 11(-2-6)-3(18-50)-1(27+25)
= 11(-8)-3(-32)-1(52)
= -88+96-52 = -44
Δ₂ = \(\left| \begin{matrix} 3 & 11 & -1 \\ 2 & 9 & 2 \\ 4 & 25 & 2 \end{matrix} \right| \)
= \(3\left| \begin{matrix} 9 & 2 \\ 25 & 2 \end{matrix} \right| -11\left| \begin{matrix} 2 & 2 \\ 4 & 2 \end{matrix} \right| -1\left| 2\begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| \)
= 3(18 - 50) -11(4 - 8) - 1(50- 36)
= 3(- 32) - 11(- 4) - 1(14)
= -96+44-14 = - 66
Δ₃ = \(\left| \begin{matrix} 3 & 3 & 11 \\ 2 & -1 & 9 \\ 4 & 3 & 25 \end{matrix} \right| \)
\(3\left| \begin{matrix} -1 & 9 \\ 3 & 25 \end{matrix} \right| -3\left| \begin{matrix} 2 & 9 \\ 4 & 25 \end{matrix} \right| -11\left| \begin{matrix} 2 & -1 \\ 4 & 3 \end{matrix} \right| \)
= 3(-25-27)-3(50-36)+ 11(6+4)
= 3(- 52) - 3(14) + 11(10)
= -156 - 42 + 110= - 88
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -44 }{ -22 } \) = 2
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -66 }{ -22 } \) = 3
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { -88 }{ -22 } \) = 4
∴ Solution set is {2, 3, 4}

x + y + 3z = 0, 4x + 3y + λz = 0, 2x + y + 2z = 0
Reducing the augmented matrix to row - echelon form we get,
[A|0] = \(\left[ \begin{matrix} 1 & 1 & 3 \\ 4 & 3 & \lambda \\ 2 & 1 & 2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & -1 & \lambda -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } }{ \underset { { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & \lambda -8 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & not\quad zero \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Case (i) when λ ≠ 8
[A|0] =\(\left[ \begin{matrix} 1 & 1 & 3 \\ 0 & -1 & -4 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 3, \(\rho \)([A|0] = 3
∴ \(\rho \)(A) = \(\rho \)([A|0]) = 3 = the number of unknowns
∴ The given system is consistent and has unique solution.
Case (ii) when λ = 8
Here \(\rho \)(A) = 2, \(\rho \)([A|0] = 2
∴ \(\rho \)(A) =\(\rho \)([A|0]) = 2 < 3 the number of unknowns,
∴ The system is consistent and has non-trivial solutions.

Assume that the system px + by + cz = 0, ax + qy + cz = 0, ax + by + rz = 0 has a non-trivial solution.
So, we have \(\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \end{matrix} \right| \) = 0, Applying R₂ ➝ R₂ - R₁ and R₃ ➝ R₃ - R₁ in the above equation,
we get \(\left| \begin{matrix} p & b & c \\ a-p & q-b & c \\ a-p & b & r-c \end{matrix} \right| \) = 0. That is,  \(\left| \begin{matrix} p & b & c \\ -\left( p-a \right) & q-b & c \\ -\left( p-a \right) & b & r-c \end{matrix} \right| \) = 0.
Since p ≠ a, q ≠ b, r ≠ c, we get (p - a)(q - b)(r - c) \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
So, we have \(\left| \begin{matrix} \frac { p }{ p-a } & \frac { b }{ q-b } & \frac { c }{ r-c } \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix} \right| \) = 0.
Expanding the determinant, we get \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 0.
That is, \(\frac { p }{ p-a } +\frac { q-\left( q-b \right) }{ q-b } +\frac { r-\left( r-c \right) }{ r-c } \) = 0
⇒ \(\frac { p }{ p-a } +\frac { b }{ q-b } +\frac { c }{ r-c } \) = 2.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \)
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 4 & -2 & 6 \\ 1 & 1 & -3 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} 8 \\ -1 \\ 21 \end{matrix} \right] \) \(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 4 & -2 & 6 \\ 15 & -3 & 9 \end{matrix}|\begin{matrix} -1 \\ 8 \\ 21 \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-15{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & -6 & 18 \\ 0 & -18 & 54 \end{matrix}|\begin{matrix} -1 \\ 12 \\ 36 \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -6 \right) , \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -18 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 1 & -3 \end{matrix}|\begin{matrix} -1 \\ -2 \\ -2 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & -3 \\ 0 & 1 & -3 \\ 0 & 0 & 0 \end{matrix}|\begin{matrix} -1 \\ -2 \\ 0 \end{matrix} \right] \).
So, ρ(A)= ρ([A | B]) = 2 < 3. From the echelon form, we get the equivalent equations
x + y − 3z = −1, y − 3z = −2 , 0 = 0.
The equivalent system has two non-trivial equations and three unknowns. So, one of the unknowns should be fixed at our choice in order to get two equations for the other two unknowns. We fix z arbitrarily as a real number t , and we get y = 3t − 2, x = −1−(3t − 2) + 3t = 1. So, the solution is (x = 1, y = 3t − 2, z = t), where t is real. The above solution set is a one-parameter family of solutions. Here, the given system is consistent and has infinitely many solutions which form a one parameter family of solutions.

Here the number of unknowns is 3.
The matrix form of the system is AX = B, where
A = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 3 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -1 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 1 \end{matrix} \end{matrix} \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 3 \\ \begin{matrix} 1 \\ \begin{matrix} 3 \\ -1 \end{matrix} \end{matrix} \end{matrix} \right] \).
The augmented matrix is [A | B] = \(\left[\begin{array}{ccc|c} 1 & 2 & -1 & 3 \\ 3 & -1 & 2 & 1 \\ 1 & -2 & 3 & 3 \\ 1 & -1 & 1 & -1 \end{array}\right]\).
Applying Gaussian elimination method on [A | B], we get
[A | B] \(\overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-3{ R }_{ 1 } \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 }, \\ { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} -7 \\ \begin{matrix} -4 \\ -3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 5 \\ \begin{matrix} 4 \\ 2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} -8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow \left( -1 \right) { R }_{ 2 }, \\ { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } \\ { R }_{ 4 }\longrightarrow \left( -1 \right) { R }_{ 4 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 4 \\ 3 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -4 \\ -2 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 0 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \)\(\overset { \begin{matrix} { R }_{ 3 }\longrightarrow 7{ R }_{ 3 }-4{ R }_{ 2 } \\ { R }_{ 4 }\longrightarrow 7{ R }_{ 4 }-3{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} -8 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} -32 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }\div \left( -8 \right) }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 4 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\longrightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 3 \\ \begin{matrix} 8 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
There are three non-zero rows in the row-echelon form of [A | B]. So, ρ([A | B]) = 3.
So, the row-echelon form of A is \(\left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 2 \\ \begin{matrix} 7 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -5 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \). There are three non-zero rows in it. So ρ(A) = 3.
Hence, ρ(A) = ρ([A | B]) = 3.
From the echelon form, we write the equivalent system of equations
x + 2y − z = 3,7y − 5z = 8, z = 4, 0 = 0.
The last equation 0=0 is meaningful. By the method of back substitution, we get
z = 4
7y − 20 = 8 ⇒ y = 4 ,
x = 3 − 8 + 4 ⇒ x = −1.
So, the solution is (x = −1, y = 4, z = 4).(Note that A is not a square matrix.)
Here the given system is consistent and the solution is unique.

Giveny = ax² + bx + c .............(1)
(-6, 8) lies on (1)
⇒ 8 = a(-6)²+b(-6)+c
⇒ 8 = 36z-6b+c ..........(2)
(-2,12) lies on (1)
⇒ -12 = a(-2)²+b(-2)+c
⇒ -12 = 4a-2b+c ...........(3)
Also (3, 8) lies on (1)
⇒ 8 = a(3)²+b(3)+c
⇒ 8 = 9a+3b+c ............(4)
Reducing the augment matrix to an equivalent row-echelon form by using elementary. row operations, we get,
\(\left[ \begin{matrix} 36 & -6 & 1 \\ 4 & -2 & 1 \\ 0 & 3 & 1 \end{matrix}|\begin{matrix} 8 \\ -12 \\ 8 \end{matrix} \right] \overset { { R }_{ 2 }\rightarrow { 9R }_{ 2 }-{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow 4{ R }_{ 3 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -12 & 8 \\ 0 & 18 & 3 \end{matrix}|\begin{matrix} 8 \\ -116 \\ 24 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }\div 4\\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div 3 }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -8 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+2{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 36 & -6 & 1 \\ 0 & -3 & 2 \\ 0 & 0 & 5 \end{matrix}|\begin{matrix} -8 \\ -29 \\ -50 \end{matrix} \right] \)
Writing the equivalent equation from the row echelon matrix, we get
36a - 6b + c = 8 ........(1)
-3b+2c = -29 ....(2)
5c = -50
⇒ c = \(\frac{-50}{5}\) = -10
Substituting c = -10 in (2) we get,
-3b+2(-10)= -29
⇒ -3b+2(-10) = -29
⇒ -3b-20 = -29
⇒ -3b = -9
⇒ b = \(\frac{-9}{-3}\) = 3
Substituting b = 3 and c = -10 in (1) we get,
36a-6(3)-10 = 8
⇒ 36a-18-10 = 8
⇒ 36a-28 = 8
⇒ 6a = 8+28 = 36
⇒ a = \(\frac{36}{36}\) = 1
∴ a = 1, b = 3, c = -10
Hence the path of the boy is
y = 1(x²)+3(x)-10
⇒ y = x²+3x-10
Since his friend is at P(7, 60),
60 = (7)²+3(7)-10
⇒ 60 = 49+21-10
⇒ 60 = 70-10 = 60
⇒ 60 = 60
Since (7, 60) satisfies his path, he can meet his friend who is at P(7, 60)

Let the price of bond invested in 6%, 8% and 9% rates be let Rs. x, Rs. y and Rs. z respectively
∴ By the given data, x + y + z = 65000 ..........(1)
\(\frac { 6\times x\times 1 }{ 100 } +\frac { 8\times y\times 1 }{ 100 } +\frac { 9\times z\times 1 }{ 100 } \) = 4800
[∵ Intrest = \(\frac { PNR }{ 100 } \)]
⇒ \(\frac { 6x }{ 100 } +\frac { 8y }{ 100 } +\frac { 9z }{ 100 } \)= 4800
⇒ 6x+8y+9z = 480000 ............(2)
Also, \(\frac { 9z }{ 100 } =600+\frac { 8y }{ 100 } \)
⇒ \(\frac { -8y }{ 100 } +\frac { 9y }{ 100 } \) = 600
⇒ -8y+9z = 60000 ............(3)
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row operation, we get
\(\left[ \begin{matrix} 1 & 1 & 1 \\ 6 & 8 & 9 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 480000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & -8 & 9 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 60000 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }+4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 1 \\ 0 & 2 & 3 \\ 0 & 0 & 21 \end{matrix}|\begin{matrix} 65000 \\ 90000 \\ 420000 \end{matrix} \right] \)
Writing the equivalent from the row echelon matrix we get,
x+y+z = 65000 ...........(1)
2y+z = 90000 ...........(2)
21z = 42000
⇒ z = \(\frac { 420000 }{ 21 } \) = 20000
Substituting z = 20,000 in (2),
2y + 3(20,000) = -90000
⇒ 2y+60,000 = 90,000
⇒ 2y = 90,000 - 60,000
= 30,000
⇒ y = \(\frac { 30,000 }{ 2 } \) = 15,000
Substitutingy = 15,000 and z = 20,000 in (1) we get,
x + 15,000 + 20,000 = 65000
⇒ x + 35,000 = 65000
⇒ x = 65,000 - 35,000
⇒ 30,000
Thus the price of 6% bond is f 30,000 the price of 8% bond is f 15,000 and the price of 9% bond is f 20,000 is Rs. 20,000.

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

Since v(3) = 64, v(6) = 133,and v(9) = 208 , we get the following system of linear equations
9a + 3b + c = 64 ,
36a + 6b + c = 133 ,
81a + 9b + c = 208 .
We solve the above system of linear equations by Gaussian elimination method.
Reducing the augmented matrix to an equivalent row-echelon form by using elementary row
operations, we get
[A | B] = \(\left[ \begin{matrix} 9 & 3 & 1 \\ 36 & 6 & 1 \\ 81 & 9 & 1 \end{matrix}|\begin{matrix} 64 \\ 133 \\ 208 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }-4{ R }_{ 1 },{ R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & -6 & -3 \\ 0 & -18 & -8 \end{matrix}|\begin{matrix} 64 \\ -123 \\ -368 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow { R }_{ 2 }\div \left( -3 \right) ,{ R }_{ 3 }\div \left( -2 \right) }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 9 & 4 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 184 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\longrightarrow 2{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 18 & 8 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 368 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }-9{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ -1 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow \left( -1 \right) { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 9 & 3 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 64 \\ 41 \\ 1 \end{matrix} \right] \).
Writing the equivalent equations from the row-echelon matrix, we get
9a + 3b + c = 64, 2b + c = 41, c = 1.
By back substitution, we get c = 1, b = \(\frac { \left( 41-c \right) }{ 2 } =\frac { \left( 41-1 \right) }{ 2 } \) = 20, a = \(\frac { 64-3b-c }{ 9 } =\frac { 64-60-1 }{ 9 } =\frac { 1 }{ 3 } \).
So, we get v(t) = \(\frac { 1 }{ 3 }\)t² + 20t + 1. Hence, v(15) = \(\frac { 1 }{ 3 }\) (225) + 20(15) + 1 = 75 + 300 + 1 = 376.

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).

Let the cost of one dosa be Rs. x
The cost of one idli be Rs. y
and the cost of one vadai be Rs. z
By the given data,
2x+ 3y + 2z = 150
2x + 2y + 4z = 200
5x + 4y + 2z = 250
∴ Δ = \(\left| \begin{matrix} 2 & 3 & 2 \\ 2 & 2 & 4 \\ 5 & 4 & 2 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 2 & 4 \\ 4 & 2 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 2 \end{matrix} \right| +2\left| \begin{matrix} 2 & 2 \\ 5 & 4 \end{matrix} \right| \)
= 2(4 - 16) - 3(4 - 20) + 2(8 - 10)
= 2(- 12) - 3(- 16) + 2(- 2)
= - 24 + 48 - 4 = 20
Δ₁ = \(\left| \begin{matrix} 150 & 3 & 2 \\ 200 & 2 & 4 \\ 250 & 4 & 2 \end{matrix} \right| \)
Taking 50 common from C3 we get,
= 100\(\left| \begin{matrix} 3 & 3 & 1 \\ 4 & 2 & 2 \\ 5 & 4 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 2 & 2 \\ 4 & 1 \end{matrix} \right| -3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 4 & 2 \\ 5 & 4 \end{matrix} \right| \right] \)
= 100[3(2 - 8) - 3(4 - 10) + 1(16 - 10)]
= 100[3(-6) - 3(- 6) + 6]
= 100[- 18 + 18 + 6] = 600
Δ₂ = \(\left| \begin{matrix} 2 & 150 & 2 \\ 2 & 200 & 4 \\ 5 & 250 & 2 \end{matrix} \right| =100\left| \begin{matrix} 2 & 3 & 1 \\ 2 & 4 & 2 \\ 5 & 5 & 1 \end{matrix} \right| \)
= \(100\left[ 3\left| \begin{matrix} 4 & 2 \\ 5 & 1 \end{matrix} \right| -3\left| \begin{matrix} 2 & 2 \\ 5 & 1 \end{matrix} \right| +1\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| \right] \)
= 100[2(4 - 10) - 3(2 - 10) + 1(10 - 20)]
= 100[2(- 6) - 3(- 8) + 1(- 10)]
= 100[- 12 + 24 - 10] = 100 [2] = 200
Δ₃ = \(\left| \begin{matrix} 2 & 3 & 150 \\ 2 & 2 & 200 \\ 5 & 4 & 250 \end{matrix} \right| =50\left| \begin{matrix} 2 & 3 & 3 \\ 2 & 2 & 4 \\ 5 & 4 & 5 \end{matrix} \right| \)
= \(50\left[ 2\left| \begin{matrix} 2 & 4 \\ 4 & 5 \end{matrix} \right| -3\left| \begin{matrix} 2 & 4 \\ 5 & 5 \end{matrix} \right| +3\left| \begin{matrix} 2 & 2 \\ 4 & 4 \end{matrix} \right| \right] \)
= 50 [2(10 - 16) - 3(10 - 20) + 3(8 - 10)]
= 50[2(- 6) - 3(- 10) +3(- 2)]
= 50 [- 12 + 30 - 6] = 50 [12] = 600
∴ x = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 200 }{ 20 } \) = 10
z = \(\frac { { \triangle }_{ 3 } }{ \triangle } =\frac { 600 }{ 20 } \) = 30
Hence, the price of one dosa be Rs. 30, one idli be Rs. 10 and the price of 1 vadai be Rs. 30.
Also the cost on dosa, six idlies and six vadai is
= 3x + 6y + 6z = 3(30) + 6(10) + 6(30)
= 90 + 60 + 180 = Rs. 330
Since the family had Rs. 350 in hand, they will be able to manage to pay the bill.

Let the pump A can fill the tank in x minutes, and the pump B can fill. the tank in y minutes In 1 minute A can fill \(\frac { 1 }{ x } \) units and in 1 minute B can fill \(\frac { 1 }{ y } \) units.
∴ \(\frac { 1 }{ x } +\frac { 1 }{ y } \) = 10
and \(\frac { 1 }{ x } -\frac { 1 }{ y } \) = 30
Put \(\frac { 1 }{ x } \) = a and \(\frac { 1 }{ y } \) = b
⇒ a + b = \(\frac { 1 }{ 10 } \) ..............(1)
and a - b = \(\frac { 1 }{ 30 } \) .............(2)
Δ = \(\left| \begin{matrix} 1 & 1 \\ 1 & -1 \end{matrix} \right| \)= -1-1= -2
Δ₁ = \(\left| \begin{matrix} \frac { 1 }{ 10 } & 1 \\ \frac { 1 }{ 30 } & -1 \end{matrix} \right| =\frac { -1 }{ 10 } -\frac { 1 }{ 30 } \)
= \(\frac { -3-1 }{ 30 } =\frac { -4 }{ 30 } =\frac { -2 }{ 15 } \)
Δ₂ = \(\left| \begin{matrix} 1 & \frac { 1 }{ 10 } \\ 1 & \frac { 1 }{ 30 } \end{matrix} \right| =\frac { 1 }{ 30 } -\frac { 1 }{ 10 } =\frac { 1-3 }{ 30 } \)
= \(\frac { -2 }{ 30 } =\frac { -1 }{ 15 } \)
∴ a = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { -2 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 15 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 15 } \Rightarrow \)= x = 15
b = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { -1 }{ \frac { 15 }{ -2 } } =\frac { 1 }{ 30 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 30 } \Rightarrow \)y = 30
Hence the pump A can fill the tank in 15 minutes and the pump B can fill the tank in 30 minutes.

Let the amount of 50% acid be x and the amount of 25% acid be y litre
By the given data, x + y = 10 ..............(1)
and \(x\left( \frac { 50 }{ 100 } \right) +y\left( \frac { 25 }{ 100 } \right) =10\left( \frac { 40 }{ 100 } \right) \)
⇒ 50x + 25y = 400 ⇒ 2x + y = 16 ...............(2)
The matrix from of the equation is \(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right] \)
\(X=\left[ \begin{matrix} x \\ y \end{matrix} \right] ,B=\left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
⇒ X = A^-1N |A| = \(\left| \begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix} \right| \) = 1 - 2 = -1
⇒ X = \(\frac { 1 }{ |A| } \)adj A.B
⇒ X= \(-1\left[ \begin{matrix} 1 & -1 \\ -2 & 1 \end{matrix} \right] \left[ \begin{matrix} 10 \\ 16 \end{matrix} \right] \)
= -\(\left[ \begin{matrix} 10-16 \\ -20+16 \end{matrix} \right] \)
⇒ X = -\(\left[ \begin{matrix} -6 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 6 \\ 4 \end{matrix} \right] \)
Thus, the amount of 50% acid is 6 litre and the amount of 25% acid is 4 litre = 10 litres of 40% acid solution.

The path y = ax² + bx + c passes through the points (10, 8), (20, 16), (40, 22). So, we get the system of equations 100a + 10b + c = 8, 400a + 20b + c = 16, 1600a + 40b + c = 22. To apply Cramer’s rule, we find
Δ = \(\left| \begin{matrix} 100 & 10 & 1 \\ 400 & 20 & 1 \\ 1600 & 40 & 1 \end{matrix} \right| =1000\left| \begin{matrix} 1 & 1 & 1 \\ 4 & 2 & 1 \\ 16 & 4 & 1 \end{matrix} \right| \) = 1000 [-2 + 12 - 6] = -6000,
Δ₁ = \(\left| \begin{matrix} 8 & 10 & 1 \\ 16 & 20 & 1 \\ 22 & 40 & 1 \end{matrix} \right| =20\left| \begin{matrix} 4 & 1 & 1 \\ 8 & 2 & 1 \\ 11 & 4 & 1 \end{matrix} \right| \) = 20[-8 + 3 + 10] = 100,
Δ₂ = \(\left| \begin{matrix} 100 & 8 & 1 \\ 400 & 16 & 1 \\ 1600 & 22 & 1 \end{matrix} \right| =200\left| \begin{matrix} 1 & 4 & 1 \\ 4 & 8 & 1 \\ 16 & 11 & 1 \end{matrix} \right| \) = 200[-3 + 48 - 84] = -7800,
Δ₃ = \(\left| \begin{matrix} 100 & 10 & 8 \\ 400 & 20 & 16 \\ 1600 & 40 & 22 \end{matrix} \right| =2000\left| \begin{matrix} 1 & 1 & 4 \\ 4 & 2 & 8 \\ 16 & 4 & 11 \end{matrix} \right| \) = 2000[-10 + 84 - 64] = 20000.
By Cramer’s rule, we get a = \(\frac { { \Delta }_{ 1 } }{ \Delta } =-\frac { 1 }{ 60 } \), b = \(\frac { { \Delta }_{ 2 } }{ \Delta } =\frac { 7800 }{ 6000 } =\frac { 78 }{ 60 } =\frac { 13 }{ 10 } \), c = \(\frac { { \Delta }_{ 3 } }{ \Delta } =\frac { 20000 }{ 6000 } =-\frac { 20 }{ 6 } =-\frac { 10 }{ 3 } \).
So, the equation of the path is y = \(\frac { 1 }{ 60 } { x }^{ 2 }+\frac { 13 }{ 10 } x-\frac { 10 }{ 3 } \).
When x = 70, we get y = 6. So, the ball went by 6 metres high over the boundary line and it is impossible for a fielder standing even just before the boundary line to jump and catch the ball. Hence the ball went for a super six and the team won the match.

Let the prices per unit for the commodities A, B and C be Rs. x, Rs. y and Rs. z.
By the given data,
2x - 4y + 5z = 15000
3x + y - 2z = 1000
-x + 3y + z = 4000
The matrix form of the system of equations is
\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 12 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
AX = B where A =\(\left[ \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \)
and B =\(\left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 2 & -4 & 5 \\ 3 & 1 & -2 \\ -1 & 3 & 1 \end{matrix} \right| \)
= \(2\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| +4\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| +5\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \)
= 2 (1 + 6) + 4 (3 - 2) + 5 (9 + 1)
= 2 (7) + 4 (1) + 5(10) = 14 + 4 + 50 = 68.
adj A = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & -2 \\ 3 & 1 \end{matrix} \right| & -\left| \begin{matrix} 3 & -2 \\ -1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 3 & 1 \\ -1 & 3 \end{matrix} \right| \\ -\left| \begin{matrix} -4 & 5 \\ 3 & 1 \end{matrix} \right| & +\left| \begin{matrix} 2 & 5 \\ -1 & 1 \end{matrix} \right| & -\left| \begin{matrix} 2 & -4 \\ -1 & 3 \end{matrix} \right| \\ +\left| \begin{matrix} -4 & 5 \\ 1 & -2 \end{matrix} \right| & -\left| \begin{matrix} 2 & 5 \\ 3 & -2 \end{matrix} \right| & +\left| \begin{matrix} 2 & -4 \\ 3 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
= \(\left[ \begin{matrix} +(1+6) & -(3-2) & +(9+1) \\ -(-4-15) & +(2+5) & -(6-4) \\ +(8-5) & -(4-15) & +(2+12) \end{matrix} \right] \)
= \(\left[ \begin{matrix} 7 & -1 & 10 \\ 19 & 7 & -2 \\ 3 & 19 & 14 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ |A| } adj=\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 68 } \left[ \begin{matrix} 7 & 19 & 3 \\ -1 & 7 & 19 \\ 10 & -2 & 14 \end{matrix} \right] \left[ \begin{matrix} 15000 \\ 1000 \\ 4000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 105000+19000+12000 \\ -15000+7000+76000 \\ 150000-2000+56000 \end{matrix} \right] \)
= \(\frac { 1 }{ 68 } \left[ \begin{matrix} 136000 \\ 68000 \\ 204000 \end{matrix} \right] =\left[ \begin{matrix} 2000 \\ 1000 \\ 3000 \end{matrix} \right] \)
∴ x = 2000, y = 1000, z = 3000
Hence the prices per unit of the commodities A, B and C are Rs. 2000, Rs. 1000 and Rs. 3000 respectively.

Let the time by one man alone be x days and one woman alone be y days
∴ By the given data,
\(\frac { 4 }{ x } +\frac { 4 }{ y } =\frac { 1 }{ 3 } \)
and \(\frac { 2 }{ x } +\frac { 5 }{ y } =\frac { 1 }{ 4 } \)
put \(\frac { 1 }{ x } \) = s and \(\frac { 1 }{ y } \) = t
∴ 4s + 4t = \(\frac { 1 }{ 3 } \)
and 2s + 5t = \(\frac { 1 }{ 4 } \)
The matrix form of the system of equation is
\(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \left[ \begin{matrix} s \\ t \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \) ⇒ AX = B where
A = \(\left[ \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} \frac { 1 }{ 3 } \\ \frac { 2 }{ 4 } \end{matrix} \right] \)
X = A^-1B
Now |A| = \(\left| \begin{matrix} 4 & 4 \\ 2 & 5 \end{matrix} \right| \) = 20 - 8 =12 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \)
∴ X = A^-1B = \(\frac { 1 }{ 12 } \left[ \begin{matrix} 5 & -4 \\ -2 & 4 \end{matrix} \right] \left[ \frac { \begin{matrix} 1 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 4 \end{matrix} } \right] \)
=\(\frac { 1 }{ 12 } \left[ \begin{matrix} \frac { 5 }{ 3 } & -1 \\ \frac { -2 }{ 3 } & +1 \end{matrix} \right] \)
= \(\frac { 1 }{ 12 } \left[ \frac { \begin{matrix} 2 \\ 3 \end{matrix} }{ \begin{matrix} 1 \\ 3 \end{matrix} } \right] =\left[ \begin{matrix} \frac { 2 }{ 3 } \times \frac { 1 }{ 12 } \\ \frac { 1 }{ 3 } \times \frac { 1 }{ 12 } \end{matrix} \right] =\left[ \begin{matrix} \frac { 1 }{ 18 } \\ \frac { 1 }{ 36 } \end{matrix} \right] \)
∴ \(\frac { 1 }{ 18 } \Rightarrow \frac { 1 }{ x } =\frac { 1 }{ 18 } \Rightarrow \)x = 18
t = \(\frac { 1 }{ 36 } \Rightarrow \frac { 1 }{ y } =\frac { 1 }{ 36 } \Rightarrow \)y = 36
one man can do 18 days
one woman can do 36 days.

Let the man's starting salary be Rs. x and his annual increment be Rs. y.
By the given data x + 3y = 19800 and x + 9y = 23,400.
The matrix form of the given system of equations is
\(\left[ \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right] \left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
⇒ AX = B where A =\(\left[ \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right] \) and B =\(\left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
⇒ X = A^-1B
|A| = \(\left| \begin{matrix} 1 & 3 \\ 1 & 9 \end{matrix} \right| \)= 9-3 = 6 ≠ 0
∴ A^-1 = \(\frac { 1 }{ |A| } adjA=\frac { 1 }{ 6 } \left[ \begin{matrix} 9 & -3 \\ -1 & 1 \end{matrix} \right] \)
∴ X = A^-1B
= \(\frac { 1 }{ 6 } \left[ \begin{matrix} 9 & -3 \\ -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 19800 \\ 23400 \end{matrix} \right] \)
= \(\frac { 1 }{ 6 } \left[ \begin{matrix} 178200 & -70200 \\ -19800 & +23400 \end{matrix} \right] =\frac { 1 }{ 6 } \left[ \begin{matrix} 108000 \\ 3600 \end{matrix} \right] \)
= \(\\ \\ \left[ \begin{matrix} 18000 \\ 600 \end{matrix} \right] \)
∴ x = 18000, y = 600
Hence the man's starting salary is Rs. 18000 and his annual increment is Rs. 600.

Given A =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \), B=\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
AB =\(\left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+3+6 & -5+2+3 & -10+1+9 \\ 7+3-10 & 7+2-3 & 14+1-15 \\ 1-3+2 & 1-2+1 & 2-1+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃
BA =\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -5+7+2 & 1+1-2 & 3-5+2 \\ -15+14+1 & 3+2-1 & 9-10+1 \\ -10+7+3 & 2+1-3 & 6-5+3 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{matrix} \right] \)= 4. I₃.
So, we get AB = BA = 4. I₃
⇒ \(\left( \frac { 1 }{ 4 } A \right) B=B\left( \frac { 1 }{ 4 } A \right) =1\)
⇒ B^-1 = \(\frac { 1 }{ 4 } \) = 1
Writing the given set of equations in matrix form we get,
\(\left[ \begin{matrix} 1 & 1 & 2 \\ 3 & 2 & 1 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(B=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] =\left[ \frac { 1 }{ 4 } A \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5 & 1 & 3 \\ 7 & 1 & -5 \\ 1 & -1 & 1 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 7 \\ 2 \end{matrix} \right] \)
= \(\frac { 1 }{ 4 } \left[ \begin{matrix} -5+7+6 \\ 7+7-10 \\ 1-7+2 \end{matrix} \right] =\frac { 1 }{ 4 } \left[ \begin{matrix} 8 \\ 4 \\ -4 \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 1 \\ -1 \end{matrix} \right] \)
∴ x = 2, y = 1, z = -1
Hence, the solution set is {2, 1, - 1}.

We find AB = \(\left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] =\left[ \begin{matrix} -4+4+8 & 4-8+4 & -4-8+12 \\ -7+1+6 & 7-2+3 & -7-2+9 \\ 5-3-2 & -5+6-1 & 5+6-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
and BA = \(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] =\left[ \begin{matrix} -4+7+5 & 4-1-3 & 4-3-1 \\ -4+14-10 & 4-2+6 & 4-6+2 \\ -8-7+15 & 8+1-9 & 8+3-3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 8 & 0 & 0 \\ 0 & 8 & 0 \\ 0 & 0 & 8 \end{matrix} \right] \) = 8I₃
So we get AB = BA = 8I₃. That is, (\(\frac { 1 }{ 8 } A\))B = B(\(\frac { 1 }{ 8 } A\)) = I₃. Hence, B^-1 = \(\frac { 1 }{ 8 } A\).
Writing the given system of equations in matrix form, we get
\(\left[ \begin{matrix} 1 & -1 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
That is B \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \).
So, \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] ={ B }^{ -1 }\left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] \)
= \(\left( \frac { 1 }{ 8 } A \right) \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -4 & 4 & 4 \\ -7 & 1 & 3 \\ 5 & -3 & -1 \end{matrix} \right] \left[ \begin{matrix} 4 \\ 9 \\ 1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} -16+36+4 \\ -28+9+3 \\ 20-27-1 \end{matrix} \right] =\frac { 1 }{ 8 } \left[ \begin{matrix} 24 \\ -16 \\ -8 \end{matrix} \right] =\left[ \begin{matrix} 3 \\ -2 \\ -1 \end{matrix} \right] \)
Hence, the solution is (x = 3, y = -2, z = -1).

Given A =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
|A| = 0-1\(\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \)
= -1(0-1) + 1(1-0) = 1 + 1 = 2
adj A =\(\left[ \begin{matrix} +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right| & -\left| \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} (0-1) & -(0-1) & +(1-0) \\ -(0-1) & +(0-1) & -(0-1) \\ +(1+0) & -(0-1) & +(0-1) \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 2 } \left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) ................(1)
Now A² =\(\left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 0+1+1 & 0+0+1 & 0+1+0 \\ 0+0+1 & 1+0+1 & 1+0+0 \\ 0+1+0 & 1+0+0 & 1+1+0 \end{matrix} \right] =\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] \)
A²- 3I =\(\left[ \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 2-3 & 1-0 & 1-0 \\ 1-0 & 2-3 & 1-0 \\ 1-0 & 1-0 & 2-3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{matrix} \right] \) .............(2)
From (1) and (2), it is proved that A^-1 = \(\frac{1}{2}\) [A² - 3I]

If A is orthogonal, then AA^T = A^TA = I₃. So, we have
AA^T = I₃ ⇒ \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] \frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & a \\ b & -2 & 6 \\ 2 & c & 3 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left[ \begin{matrix} 45+{ a }^{ 2 } & 6b+6+6a & 12-3c+3a \\ 6b+6+6a & { b }^{ 2 }+40 & 2b-2c+18 \\ 12-3c+3a & 2b-2c+18 & { c }^{ 2 }+13 \end{matrix} \right] =49\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
⇒ \(\left\{ \begin{matrix} 45+{ a }^{ 2 }=49 \\ \begin{matrix} { b }^{ 2 }+40=49 \\ \begin{matrix} { c }^{ 2 }+13=49 \\ \begin{matrix} 6b+6+6a=0 \\ \begin{matrix} 12-3c+3a=0 \\ 2b-2c+18=0 \end{matrix} \end{matrix} \end{matrix} \end{matrix} \end{matrix} \right\} \) ⇒ \(\left\{ \begin{matrix} { a }^{ 2 }=4,{ b }^{ 2 }=9,{ c }^{ 2 }=36 \\ a+b=-1,a-c=-4,b-c=-9 \end{matrix} \right\} \) ⇒ a = 2, b = −3, c = 6
So we get A = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \) and hence, we get A^-1 = A^T = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 6 & -3 & 2 \\ -3 & -2 & 6 \\ 2 & 6 & 3 \end{matrix} \right] \).

Given C₂H₆ + O₂ ⟶ H₂O + CO₂
We have to find positive integers x₁ x₂, x₃ and x₄ such that
x₁C₂H₆ + x2O₆ ⟶ x₃H₂O + x₄CO₂ .....(1)
The number of carbon atoms on the LHS of (1) should be equal to the number of carbon atoms on the RHS of (1)
∴ 2x₁ = 1x₄
⇒ 2x₁-x₄ = 0 ...(2)
Considering hydrogen atoms we get,
6x₁ = 2x₃
⇒ 6x₁- 2x₃ = 0
⇒ 3x₁- x₃ = 0....(3)
Also, considering oxygen atoms we get,
2x₂ = 1x₃ + 2x₄
⇒ 2x₂ - x₃ - 2a₄ = 0 ....(4)
Equations (2), (3) and (4) forni a homogeneous system of linear equations in 4 unknowns
∴ The augmented matrix [A|0] is
\(\left[ \begin{matrix} 2 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} -1 \\ 0 \\ 2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }\div 2 }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 3 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ 0 \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ 1 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ -2 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ 2 \end{matrix}\begin{matrix} 0 \\ -1 \\ 0 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { 3 }{ 2 } \\ \frac { -7 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 2 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -1 \end{matrix}\begin{matrix} \frac { -1 }{ 2 } \\ \frac { -7 }{ 2 } \\ \frac { 3 }{ 2 } \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow 2{ R }_{ 2 }\\ R_{ 2 }\rightarrow 2{ R }_{ 2 } }{ \underset { { R }_{ 2 }\rightarrow 2{ R }_{ } }{ \longrightarrow } } \left[ \begin{matrix} 2 \\ 0 \\ 0 \end{matrix}\begin{matrix} 0 \\ 4 \\ 0 \end{matrix}\begin{matrix} 0 \\ 0 \\ -2 \end{matrix}\begin{matrix} -1 \\ -7 \\ 3 \end{matrix}|\begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right] \)
Here \(\rho \)(A) = 1, the number of unknowns
∴ The system is consistent with one parameter family of solutions, so let x₄ = t
Writing the equations, from the row-echelon form we get
2x₁-x₄ = 0
⇒ 2x₁ = x₄
⇒ 2x = t
⇒ x₁ = \(\frac { t }{ 2 } \)
4x₂ - 7x₄ = 0
⇒ 4x₂ = 7t
⇒ x₂ = \(\frac { 7t }{ 4 } \)
-2x₃ + 3x₄ = 0
⇒ 2x₃ = 3x₄
⇒ x₃ = \(\frac { 3t }{ 4 } \)
Since x₁, x₂, x₃ and x₄ are positive integers, let us choose t = 4
∴ x₁ = \(\frac{4}{2}\) = 2

So, the balanced equation is
2C₂H₆ + 7O₂ ⟶ 6H₂O + 4CO₂

We are searching for positive integers x₁, x₂, x₃ and x₄ such that
x₁C₅H₈ + x₂O₂ ⟶ x₃CO₂ + x₄H₂O. ....(1)
The number of carbon atoms on the left-hand side of (1) should be equal to the number of carbon atoms on the right-hand side of (1) So we get a linear homogenous equation
5x₁= x₃ ⇒ 5x₁ - x₃ = 0 .........(2)
Similarly, considering hydrogen and oxygen atoms, we get respectively,
8x₁ = 2x₄ ⇒ 4x₁ - x₄ = 0 ............(3)
2x₂ = 2x₃ + x₄ ⇒ 2x₂ - 2x₄ = 0 .........(4)
Equations (2), (3), and (4) constitute a homogeneous system of linear equations in four unknowns.
The augmented matrix is [A | B] = \(\left[ \begin{matrix} 5 \\ \begin{matrix} 4 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 2 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 0 \\ -2 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -1 \\ -1 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \).
By Gaussian elimination method, we get
[A | B] \(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 5 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 0 \\ 2 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -1 \\ -2 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} 0 \\ -1 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 0 \\ 5 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -2 \\ -1 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ 0 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow 4{ R }_{ 3 }-5{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 4 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} 2 \\ 0 \end{matrix} \end{matrix}\begin{matrix} 0 \\ \begin{matrix} -2 \\ -4 \end{matrix} \end{matrix}\begin{matrix} -1 \\ \begin{matrix} -1 \\ 5 \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \right] \).
Therefore, ρ(A) = ρ([A | B]) = 3 < 4 = Number of unknowns.
The system is consistent and has infinite number of solutions.
Writing the equations using the echelon form, we get 4x₁ - x₄ = 0, 2x₂ - 2x₃ - x₄ = 0, -4x₃ + 5x₄ = 0.
So, one of the unknowns should be chosen arbitrarily as a non-zero real number.
Let us choose x₄ = t, t ≠ 0. Then, by back substitution, we get x₃ = \(\frac { 5t }{ 4 } \), x₂ = \(\frac { 7t }{ 4 } \), x₁ = \(\frac { t }{ 4 } \).
Since x₁, x₂, x₃ and x₄ are positive integers, let us choose t = 4.
Then, we get x₁ = 1, x₂ = 7, x₃ = 5 and x₄ = 4.
So, the balanced equation is C₅H₈ + 7O₂ ⟶ 5CO₂ + 4H₂O.