Every point on the line \(\frac { x-1 }{ 2 } =\frac { y-2 }{ 3 } =\frac { z-3 }{ 4 } \) = s (say) is of the form (2s + 1, 3s + 2, 4s + 3) and every point on the line \(\frac { x-4 }{ 5 } =\frac { y-1 }{ 2 } =z=t\) (say) is of the form (5t + 4, 2t + 1, t).
So, at the point of intersection, for some values of s and t, we have
(2s + 1, 3s + 2, 4s + 3) = (5t + 4, 2t + 1, t)
Therefore, 2s - 5t = 3, 3s - 2t = -1 and 4s - t = -3. Solving the first two equations we get t = -1, s = -1. 
These values of s and t satisfy the third equation. Therefore, the given lines intersect. 
Substituting, these values of t or s in the respective points, the point of intersection is (-1,- 1, -1)

Given A =\(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \)
A² = \(\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] =\left[ \begin{matrix} 25-3 & 15-6 \\ -5+2 & -3+4 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22 & 9 \\ -2 & 1 \end{matrix} \right] \)
∴ A²- 3A - 7I₂
=\(\left[ \begin{matrix} 22 & 9 \\ -3 & 1 \end{matrix} \right] -3\left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -7\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \)
=\(\left[ \begin{matrix} 22-15-7 & 9-9+0 \\ -3+3+0 & 1+6-7 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix} \right] \)= O₂
Hence proved.
∴ A²-3A-7I₂ = O₂
Postmultiplying by A^-1 we get,
A²-A^-1-3AA^-1-7I₂A^-1 = 0.A^-1
⇒ A(AA^-1)-3(AA^-1)-7(A^-1) = 0
[∵ I₂A^-1 = A^-1 and | (0)A^-1= 0]
⇒ AI-3I-7A-1 = 0  [∵ AA^-1= 1]
⇒ AI-3I = 7A^-1
⇒ A^-1 = \(\frac { 1 }{ 7 } \)[A - 3I] [∴ AI = A]
⇒ A^-1 = \(\frac { 1 }{ 7 } =\left[ \left[ \begin{matrix} 5 & 3 \\ -1 & -2 \end{matrix} \right] -3\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \right] \)
⇒ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 5-3 & 3-0 \\ -1-0 & -2-3 \end{matrix} \right] =\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \)
∴ A^-1 = \(\frac { 1 }{ 7 } \left[ \begin{matrix} 2 & 3 \\ -1 & -5 \end{matrix} \right] \).

Let cosec⁻¹(−1) = y. Then, cosec y = −1
Since the range of principal value branch of y = cosec^-1x is\(\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0} and
cosec \((-\frac{\pi}{2})=-1,\) 
We have y = \(-\frac{\pi}{2}\).
Note that \(-\frac{\pi}{2}\in\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right] \)\{0}.
Thus, the principal value of cosec⁻¹(−1) is −\(\frac{\pi}{2}\)

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}

Here the number of equations is 4 and the number of unknowns is 3. Reducing the augmented matrix to echelon-form, we get
[A | O] = \(\left[ \begin{matrix} 1 \\ \begin{matrix} 2 \\ \begin{matrix} 3 \\ 6 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -3 \\ \begin{matrix} -7 \\ -9 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 1 \\ \begin{matrix} 10 \\ 10 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 } \\ { R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }-6{ R }_{ 1 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} -5 \\ \begin{matrix} -10 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} 5 \\ \begin{matrix} 16 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 2 }\rightarrow { R }_{ 2 }\div \left( -5 \right) \\ { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -2 \right) \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 5 \\ -15 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -8 \\ 22 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
\(\overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }-5{ R }_{ 2 }, \\ { R }_{ 4 }\rightarrow { R }_{ 4 }+15{ R }_{ 2 } \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} -3 \\ 7 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { \begin{matrix} { R }_{ 3 }\rightarrow { R }_{ 3 }\div \left( -3 \right) \\ { R }_{ 4 }\rightarrow { R }_{ 4 }\div 7 \end{matrix} }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 1 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \overset { { R }_{ 4 }\rightarrow { R }_{ 4 }-{ R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} 1 \\ \begin{matrix} 1 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix}\begin{matrix} -2 \\ \begin{matrix} -1 \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{matrix} \end{matrix}|\begin{matrix} 0 \\ \begin{matrix} 0 \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{matrix} \end{matrix} \right] \)
So, ρ(A) = ρ([A | O]) = 3 =  number of unknowns
Hence the system has trivial solution only.

Let the equation of the parabola be
(x - h)² = -4a(y - k).
Here the vertex is (0.5, 4)
Equation of the parabola is (x - 0.5)²
= -4a(y-4) ...(1)
O(0, 0) is a point on the parabola
(0 - 0.5)² = -4a (0 - 4)
⇒ \({ \left( \frac { -1 }{ 2 } \right) }^{ 2 }=-4a(-4)\)
⇒ \(\frac { 1 }{ 4 } =16a\Rightarrow a=\frac { 1 }{ 64 } \)
∴ (1) becomes as (x - 0.5)² = \(-4\times \frac { 1 }{ 64 } (y-4)\)
Also D(0.75, y₁) is a point on the parabola
∴ (0.75 - 0.5)² = \(\frac { -1 }{ 16 } ({ y }_{ 1 }-4)\)
⇒ \({ \left( \frac { 1 }{ 4 } \right) }^{ 2 }=\frac { -1 }{ 6 } ({ y }_{ 1 }-4)\)
\(\Rightarrow \frac{1}{\not 16}=\frac{1}{\not16}\left(y_{1}-4\right)\)
⇒ 1 = -y₁ + 4
⇒  y₁ = -1 + 4 = 3m
Height of the water at a horizontal distance of 0.75m is 3m

z³ = -27 = (-1 \(\times\) 3)³ = -1 \(\times\) 33
z = \((-1)^{ \frac { 1 }{ 3 } }\times 3^{ 3\times \frac { 1 }{ 3 } }=(-1)^{ \frac { 1 }{ 3 } }\)\(\times\) 3
∴ z = 3\(\left[ cos\pi +isin\pi \right] ^{ \frac { 1 }{ 3 } }\)
[∵ cos π = -1 and sin π = 0]
= 3\(\left[ cos\frac { 1 }{ 3 } (2k\pi +\pi )isin\frac { 1 }{ 3 } (2k\pi +\pi ) \right] \)
k = 0, 1, 2
When k = 0,
z = 3\(\left[ cos\frac { 1 }{ 3 } (\pi )isin\frac { 1 }{ 3 } (\pi ) \right] =3cos\frac { \pi }{ 3 } \)
When k = 1
z = 3\(\left[ cos\frac { 1 }{ 3 } (3\pi )isin\frac { 1 }{ 3 } (3\pi ) \right] \)
= 3[cos π + i sin π] = 3(-1+0) 
When k = 2
z = 3\(\left[ cos\frac { 1 }{ 3 } (5\pi )isin\frac { 1 }{ 3 } (5\pi ) \right] =3\left[ cos5\frac { \pi }{ 3 } \right] \)
Hence, the roots are 3 cis\(\frac { \pi }{ 3 } \), -3, 3 c is 5\(\frac { \pi }{ 3 } \)

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

This equation is Type I even degree reciprocal equation. Hence it can be rewritten as
x²\(\left[ \left( { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } \right) -10 \left( x+\frac { 1 }{ x } \right) +26 \right] \)= 0 Since x \(\neq\) 0, we get \(\left(x^{2}+\frac{1}{x^{2}}\right)-10\left(x+\frac{1}{x}\right)+26=0\)
Let y = \(\left( x+\frac { 1 }{ x } \right) \)
[(y²-2)-10y+26] = 0 ⇒ (y²-10y+24) = 0 ⇒ (y-6)(y-4) = 0 ⇒ y = 6 or y = 4
Case (i)
y = 6 ⇒ x +\(\frac{1}{x}\) = 6 ⇒ x = 3+2\(\sqrt{2}\), x = 3 - 2\(\sqrt{2}\)
Case (ii)
y = 4 ⇒ x = 2+\(\sqrt{3}\), x = 2-\(\sqrt{3}\).
Hence, the roots are \(3 \pm 2 \sqrt{2}, 2 \pm \sqrt{3}\)

By the definition of a conic  \(\frac{SP}{PM}\)= e or SP² = e²PM²
Then, (x−2)² + (y−3)² = \(\frac { 1 }{ 4 } \) (x-7)² 
3x²+ 4y²−2x − 24y + 3 = 0
\({ 3\left( x-\frac { 1 }{ 3 } \right) }^{ 2 }+4(y-3)^{ 2 }=3\left( \frac { 1 }{ 9 } \right) +4\times 9-3=\frac { 100 }{ 3 } \)
\(\frac { { \left( x-\frac { 1 }{ 3 } \right) }^{ 2 } }{ \frac { 100 }{ 9 } } +\frac { (y-3{ ) }^{ 2 } }{ \frac { 100 }{ 12 } } \) = 1 which is in the standard form.
Therefore, the length of major axis = 2a = 2\(\sqrt { \frac { 100 }{ 9 } = } \frac { 20 }{ 3 } \) and
the length of minor axis = 2b = 2\(\sqrt { \frac { 100 }{ 12 } = } \frac { 10 }{ \sqrt { 3 } } \).