Let \({ cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) =\theta \Rightarrow cos\theta =\frac { 1 }{ 2 } \)
\(\Rightarrow sin\theta =\sqrt { 1-{ cos }^{ 2 }\theta } =\sqrt { 1-\frac { 1 }{ 4 } } =\sqrt { \frac { 3 }{ 4 } } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore sin\left( { cos }^{ -1 }\left( \frac { 1 }{ 2 } \right) \right) =\frac { \sqrt { 3 } }{ 2 } \)

Given A³ = I ⇒ IA³I = |I|
⇒ |A.A.A| = 1
⇒ |A|. |A|·|A| = 1
⇒ |A|³ = 1
∴ |A| ≠ 0
Hence, A is non-singular.

\(\left[ \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right] \)
Let A = \(\left( \begin{matrix} -3 & 4 \\ 6 & 2 \end{matrix} \right) \)
adj A = \(\left( \begin{matrix} 2 & -4 \\ -6 & -3 \end{matrix} \right) \)
[Interchange the elements in the leading diagonal and change the sign of the elements in off diagonal]

We have z = (2+3i)(1−i) = (2+3)+(3−2)i = 5+i
\(\Rightarrow\) \({ z }^{ -1 }=\frac { 1 }{ z } =\frac { 1 }{ 5+i } \)
Multiplying the numerator and denominator by the conjugate of the denominator, we get
\({ z }^{ -1 }=\frac { \left( 5-i \right) }{ \left( 5+i \right) \left( 5-i \right) } =\frac { 5-i }{ { 5 }^{ 2 }+{ I }^{ 2 } } =\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)
\(\Rightarrow\)\({ z }^{ -1 }=\frac { 5 }{ 26 } -i\frac { 1 }{ 26 } \)

LHS =\({ tan }^{ -1 }\left( \frac { 1-x }{ 1+x } \right) -{ tan }^{ -1 }\left( \frac { 1-y }{ 1+y } \right) \)
= tan^-1(1) - tan^-1 (x) - (tan^-1(1) - tan^-1(y)
\(\left[ \because { tan }^{ -1 }(\frac { x-y }{ 1+xy } )={ tan }^{ -1 }x-{ tan }^{ -1 }y \right] \)
= tan^-1(1) - tan^-1 (x) - tan^-1(1) + tan^-1(y)
= tan^-1(y) - tan^-1(x)
= \({ tan }^{ -1 }\left( \frac { y-x }{ 1+xy } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { 1+\left( yx \right) ^{ 2 }+\left( y-x \right) ^{ 2 } } } \right) \)
= \({ sin }^{ -1 }\left( \frac { y-x }{ \sqrt { (1+{ x }^{ 2 })(1+{ x }^{ 2 }) } } \right) \)
RHS

4^x- 3(2^{x + 2}) + 2⁵ = 0
\(\Rightarrow \left( { 2 }^{ 2 } \right) ^{ x }-3({ 2 }^{ x })({ 2 }^{ 2 })+{ 2 }^{ 5 }=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-3\left( { 2 }^{ x } \right) ({ 2 }^{ 2 })=0\)
\(\left( { 2 }^{ x } \right) ^{ 2 }-12\left( { 2 }^{ x } \right) +32=0\)
Put \({ 2 }^{ x }=y\)
\({ y }^{ 2 }-12y+32=0\)
(y - 8)(y - 4) = 0
y = 8, 4 
Case (i) when \(y=8,{ 2 }^{ x }=8\Rightarrow { 2 }^{ x }={ 2 }^{ 3 }\Rightarrow x=3\)
Case (ii) when \(y=4,{ 2 }^{ x }=4\Rightarrow { 2 }^{ x }={ 2 }^{ 2 }x=\pm 2\)
∴ The roots are 2, 3

6x⁴- 35x³+ 62x²- 35x + 6 = 0
This equation is type I even degree reciprocal equation.
Hence, it can be rewritten as
\(6\left( { x }^{ 2 }+\frac { 1 }{ x } \right) -35(x+\frac { 1 }{ x } )+62=0 ...(1)\)
putting \(x+\frac { 1 }{ x } =y\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2={ y }^{ 2 }\)
\(\Rightarrow { x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ={ y }^{ 2 }-2\)
∴ (1) becomes as,
\(\Rightarrow 6({ y }^{ 2 }-2)-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-12-35y+62=0\)
\(\Rightarrow { 6y }^{ 2 }-35y+50=0\)
\(\Rightarrow (3y-10)(2y-5)=0\)
\(\Rightarrow y=\frac { 10 }{ 3 } ,\frac { 5 }{ 2 } \)
Case (i) when \(y=\frac { 10 }{ 3 } ,x+\frac { 1 }{ x } =\frac { 10 }{ 3 } \)


\(\Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 10 }{ 3 } \)
\(\Rightarrow { 3x }^{ 2 }-10x+3=10x\)
\(\Rightarrow { 3x }^{ 2 }-10x+3=0\)
\(\Rightarrow (x-3)(3x-1)=0\)
\(\Rightarrow x=3,\frac { 1 }{ 3 } \)
Case (ii) when \(y=\frac { 5 }{ 2 } ,x+\frac { 1 }{ x } =\frac { 5 }{ 2 } \Rightarrow \frac { { x }^{ 2 }+1 }{ x } =\frac { 5 }{ 2 } \)
\(\Rightarrow { 2x }^{ 2 }+2=5x\Rightarrow { 2x }^{ 2 }-5x+2=0\)
\(\Rightarrow (x-2)(2x-1)=0\)
Hence the roots are \(2,\frac { 1 }{ 2 } ,3,\frac { 1 }{ 3 } \)

Applying Gauss-Jordan method, we get
[A | I₂] = \(\left[ \begin{matrix} 0 & 5 \\ -1 & 6 \end{matrix}|\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \overset { { R }_{ 1 }\leftrightarrow { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} -1 & 6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow \left( -1 \right) { R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 5 \end{matrix}|\begin{matrix} 0 & -1 \\ 1 & 0 \end{matrix} \right] \overset { { R }_{ 2 }\longrightarrow \frac { 1 }{ 5 } { R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & -6 \\ 0 & 1 \end{matrix}|\begin{matrix} 0 & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \)\(\overset { { R }_{ 1 }\longrightarrow { R }_{ 1 }+6{ R }_{ 2 } }{ \longrightarrow } \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}|\begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] \).
So, we get A^-1 = \(\left[ \begin{matrix} \left( 6/5 \right) & -1 \\ \left( 1/5 \right) & 0 \end{matrix} \right] =\frac { 1 }{ 5 } \left[ \begin{matrix} 6 & -5 \\ 1 & 0 \end{matrix} \right] \).

Given F (\(\alpha\)) = \(\left[ \begin{matrix} \cos { \alpha } & 0 & \sin { \alpha } \\ 0 & 1 & 0 \\ -\sin { \alpha } & 0 & \cos { \alpha } \end{matrix} \right] \)
Expanding along R₁ we get,
|F(\(\alpha\))| = cos \(\alpha\) \(\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| -0+sin\alpha \left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \)
= cos \(\alpha\) (cos - 0) + sin \(\alpha\) (0 + sin \(\alpha\))
= cos² + sin² \(\alpha\) = 1 ≠ 0
Since F (\(\alpha\)) is a non-singular matrix, [F(\(\alpha\))]^-1 exists
Now, adj (F(\(\alpha\))) = \(\left[ \begin{matrix} +\left| \begin{matrix} 1 & 0 \\ 0 & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} 0 & 0 \\ sin\alpha & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} 0 & 1 \\ -sin\alpha & 0 \end{matrix} \right| \\ -\left| \begin{matrix} 0 & sin\alpha \\ 0 & cos\alpha \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & sin\alpha \\ -sin\alpha & cos\alpha \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & 0 \\ sin\alpha & 0 \end{matrix} \right| \\ +\left| \begin{matrix} 0 & sin\alpha \\ 1 & 0 \end{matrix} \right| & -\left| \begin{matrix} cos\alpha & sin\alpha \\ 0 & 0 \end{matrix} \right| & +\left| \begin{matrix} cos\alpha & 0 \\ 0 & 1 \end{matrix} \right| \end{matrix} \right] ^{ T }\)
=\(\left[ \begin{matrix} +(cos\alpha -0) & -(0) & +(0+sin\alpha ) \\ -(0) & +(cos^{ 2 }\alpha +sin^{ 2 }\alpha & -(0) \\ +(0-sin\alpha ) & -(0) & +(cos-0) \end{matrix} \right] ^{ T }\)
\(\left[ \begin{matrix} cos\alpha & 0 & +sin\alpha \\ 0 & 1 & 0 \\ -sin\alpha & 0 & cos\alpha \end{matrix} \right] =\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
∴ F(\(\alpha\))^-1 = \(\frac { 1 }{ |F(\alpha )| } \) adj (F(\(\alpha\)))
[F(\(\alpha\))]^-1 = \(\frac { 1 }{ 1 } \left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ +sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(1)
Now, F(-\(\alpha\))=\(\left[ \begin{matrix} cos(-\alpha ) & 0 & sin(-\alpha ) \\ 0 & 1 & 0 \\ -s9n(-\alpha ) & 0 & cos(-\alpha ) \end{matrix} \right] \)
=\(\left[ \begin{matrix} cos\alpha & 0 & -sin\alpha \\ 0 & 1 & 0 \\ sin\alpha & 0 & cos\alpha \end{matrix} \right] \) ..............(2)
[∵ cos \(\alpha\) is an even function, cos (-\(\alpha\)) = cos \(\alpha\) and sin \(\alpha\) is an odd function, sin (-\(\alpha\)) = -sin\(\alpha\)]
From (1) and (2)
[F(\(\alpha\))]^-1 = F (-\(\alpha\))

AB = |(3+2i) - (0+5i)| = |3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
BC = |(0+5i) - (-3+2i)| = |3+3i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
CD = |(-3+2i) - (0-i) = |-3+i|
= \(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
DA = |(0-i) - (3+2i)| = |-3-3i|
=\(\sqrt { 9+9 } =\sqrt { 18 } =3\sqrt { 2 } \)
∴ AB = BC = CD = DA
Also AC = |(3+2i) - (-3+2i)|
= |6| = \(\sqrt { 36 } \) = 6
∴ AC = BD
Hence ABCD is a square

Let A =\(\left[ \begin{matrix} 4 \\ -2 \\ 1 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 0 \\ -1 \\ 8 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
A\(\overset { { R }_{ 1 }\leftrightarrow { R }_{ 3 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ -2 \\ 4 \end{matrix}\begin{matrix} 4 \\ 3 \\ 4 \end{matrix}\begin{matrix} 8 \\ -1 \\ 0 \end{matrix}\begin{matrix} 3 \\ 5 \\ 7 \end{matrix} \right] \)
\(\overset { { R }_{ 2 }\rightarrow { R }_{ 2 }+2{ R }_{ 1 }\\ { R }_{ 3 }\rightarrow { R }_{ 3 }-4{ R }_{ 1 } }{ \longrightarrow } \left[ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix}\begin{matrix} 4 \\ 11 \\ -12 \end{matrix}\begin{matrix} 8 \\ 15 \\ -32 \end{matrix}\begin{matrix} 7 \\ 19 \\ -25 \end{matrix} \right] \)
A is in row - echelon form and it has 3 non-zero rows.
∴ \(\rho\) (A) = 3

(1+i)¹⁸
Let 1+ i = \(r(cos\theta +isin\theta )\). Then , we get
\(r=\sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 2 } ;\alpha ={ tan }^{ -1 }\left( \frac { 1 }{ 1 } \right) =\frac { \pi }{ 4 } \)
\(\theta =\alpha =\frac { \pi }{ 4 } \) (\(\because\) 1+i lies in the first Quadrant)
Therefore 1+ i = \(\sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
Raising the power 18 on both sides
\(\left( 1+i \right) ^{ 18 }=\left[ \sqrt { 2 } \left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \right] ^{ 18 }=\sqrt { 12 } ^{ 18 }\left( cos\frac { \pi }{ 4 } +isin\frac { \pi }{ 4 } \right) \)
By de Moivre’s theorem
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \({ 2 }^{ 9 }\left( cos\left( 4\pi +\frac { \pi }{ 2 } \right) +isin\left( 4\pi +\frac { \pi }{ 2 } \right) \right) ={ 2 }^{ 9 }\left( cos\frac { \pi }{ 2 } +isin\frac { \pi }{ 2 } \right) \)
\(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }\left( cos\frac { 18\pi }{ 4 } +isin\frac { 18\pi }{ 4 } \right) \)
= \(\left( 1+i \right) ^{ 18 }={ 2 }^{ 9 }(i)=512i\)

Given A = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \)
A^T = \(\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)...............(1)
We know that (\(\lambda\)A) -1 = \(\frac { 1 }{ \lambda } \)A-1
A-1 =\(\left\{ \frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \right\} ^{ -1 }=\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] ^{ T }\)
where \(\lambda\) = \(\frac{1}{9}\)
A^-1= 9B^-1 where B =\(\left[ \begin{matrix} -8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4 \end{matrix} \right] \) ............(2)
Now, |B|=\(-8\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| -1\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| +4\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \)
= -8(16+56)-1(16-7)+4(-32-4)
= -8(72)-1(9)+4(-36) = -576-9-144
= -729
adj B =\(\left[ \begin{matrix} +\left| \begin{matrix} 4 & 7 \\ -8 & 4 \end{matrix} \right| & -\left| \begin{matrix} 4 & 7 \\ 1 & 4 \end{matrix} \right| & +\left| \begin{matrix} 4 & 4 \\ 1 & -8 \end{matrix} \right| \\ -\left| \begin{matrix} 1 & 4 \\ -8 & 4 \end{matrix} \right| & +\left| \begin{matrix} -8 & 4 \\ 1 & 4 \end{matrix} \right| & -\left| \begin{matrix} -8 & 1 \\ 1 & -8 \end{matrix} \right| \\ +\left| \begin{matrix} 1 & 4 \\ 4 & 7 \end{matrix} \right| & -\left| \begin{matrix} -8 & 4 \\ 4 & 7 \end{matrix} \right| & +\left| \begin{matrix} -8 & 1 \\ 4 & 1 \end{matrix} \right| \end{matrix} \right] \)
=\(\left[ \begin{matrix} +(16+56)-(16-7)+(-32-3) \\ -(4+32)+(-32-4)+(64-1) \\ +(7-16)-(-56-16)+(-3-4) \end{matrix} \right] \)
=\(\left[ \begin{matrix} 72 & -9 & -36 \\ -36 & -36 & -63 \\ -9 & 72 & -36 \end{matrix} \right] ^{ T }=\left[ \begin{matrix} 72 & -36 & -9 \\ -9 & -36 & 72 \\ -36 & -63 & -36 \end{matrix} \right] \)
= \(9\left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
∴ B^-1=\(\frac { 1 }{ |B| } adjB=\frac { -9 }{ 729 } \left[ \begin{matrix} 8 & -4 & -1 \\ -1 & -4 & +8 \\ -4 & -7 & -4 \end{matrix} \right] \)
= \(\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \)
Substituting this in (2) we get,
A^-1 = \(9.\frac { 1 }{ 81 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] =\frac { 1 }{ 9 } \left[ \begin{matrix} -8 & 4 & 1 \\ 1 & 4 & -8 \\ 4 & 7 & 4 \end{matrix} \right] \) ...............(3)
From (1) and (3)we get,
A^T = A^-1