\(\frac { 3 }{ x } -\frac { 4 }{ y } -\frac { 2 }{ z } \) - 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}

\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Let A =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix} \right] \)
Applying Gauss - Jordan method, we get
[A|I₃] =\(\left[ \begin{matrix} 1 & -1 & 0 \\ 1 & 0 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 6 & -2 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-6{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 1 & 0 \\ 0 & 1 & -1 \\ 0 & 4 & -3 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -6 & 0 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -1 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} 0 & 1 & 0 \\ -1 & -4 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\\ \overset { { R }_{ 1 }\rightarrow { R }_{ 1 }+R_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & -1 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -1 & 1 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}|\begin{matrix} -2 & -3 & 1 \\ -3 & - & 1 \\ -2 & -4 & 1 \end{matrix} \right] \)
So, We get A^-1=\(\left[ \begin{matrix} -2 & -3 & 1 \\ -3 & -3 & 1 \\ -2 & -4 & 1 \end{matrix} \right] \).

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 = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix} \right] \), X = \(\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] \), B = \(\left[ \begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \).
Applying elementary row operations on the augmented matrix [A | B], we get
[A | B] = \(\left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 1 & \lambda \\ 1 & 3 & -5 \end{matrix}|\begin{matrix} 7 \\ \mu \\ 5 \end{matrix} \right] \overset { { R }_{ 2 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 1 & 3 & -5 \\ 1 & 1 & \lambda \end{matrix}|\begin{matrix} 7 \\ 5 \\ \mu \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\longrightarrow { R }_{ 2 }-{ R }_{ 1 }, \\ { R }_{ 3 }\longrightarrow { R }_{ 3 }-{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & -1 & \lambda -1 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -7 \end{matrix} \right] \overset { { R }_{ 3 }\longrightarrow { R }_{ 3 }+{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 2 & 1 \\ 0 & 1 & -6 \\ 0 & 0 & \lambda -7 \end{matrix}|\begin{matrix} 7 \\ -2 \\ \mu -9 \end{matrix} \right] \).
(i) If λ =7 and μ \(\neq\) 9, then ρ(A) = 2 and ρ([A | B]) = 3. So ρ(A) ≠ ρ([A | B]) Hence the given system is inconsistent and has no solution.
(ii) If λ ≠ 7 and μ is any real number, then ρ(A) = 3 and ρ([A | B]) = 3.
So, ρ(A) = ρ([A | B]) = 3 = Number of unknown. Hence the given system is consistent and has a unique solution.
(iii) If λ = 7 and μ = 9, then ρ(A) = 2 and ρ([A | B]) = 2.
So, ρ(A) = ρ([A | B]) = 2 < Number of unknown. Hence the given system is consistent and has infinite number of solutions.

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.