\({ cot }^{ -1 }(1)+{ sin }^{ -1 }\left( -\frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }(-\sqrt { 2 } )\)
Let \(\Rightarrow cot\ x=1\Rightarrow tan\ x=1\Rightarrow tan\ x=tan\frac { \pi }{ 4 } \)
\(\Rightarrow x=\frac { \pi }{ 4 } \)
\({ sin }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } \right) =y\Rightarrow sin\ y=\frac { -\sqrt { 3 } }{ 2 } =-sin\frac { \pi }{ 3 } \)
\(\Rightarrow sin\ y=sin\left( { \frac { -\pi }{ 3 } } \right) \)
\(\Rightarrow y=\frac { -\pi }{ 3 } \)
\({ sec }^{ -1 }\left( -\sqrt { 2 } \right) =z\)
\(\Rightarrow sec\ z=-\sqrt { 2 } \Rightarrow cosz=\frac { -1 }{ \sqrt { 2 } } \)
\(\Rightarrow cos\ z=-cos\frac { \pi }{ 4 } \)
\(\Rightarrow cos\ z=cos\left( \pi -\frac { \pi }{ 4 } \right) \)
\(\Rightarrow cos\ z=cos\left( \frac { 3\pi }{ 4 } \right) \)
\(\Rightarrow z=\frac { 3\pi }{ 4 } \)
\(\therefore { cot }^{ -1 }(1)+{ sin }^{ -1 }\left( \frac { -\sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }\left( \frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }\left( -\sqrt { 2 } \right) \)
= \(\frac { \pi }{ 4 } -\frac { \pi }{ 3 } -\frac { 3\pi }{ 4 } \)

= \(-\frac { 5\pi }{ 6 } \)
\(\therefore\) \({ cot }^{ -1 }(1)+{ sin }^{ -1 }\left( -\frac { \sqrt { 3 } }{ 2 } \right) -{ sec }^{ -1 }(-\sqrt { 2 } )\) =\(-\frac { 5\pi }{ 6 } \)

\(\frac { 3 }{ x } \) + 2y = 12, \(\frac { 2 }{ x } \) + 3y = 13
Let \(\frac { 1 }{ x } \)
∴ z+2y = 12, 2z+3z = 13
∴ Δ = \(\left| \begin{matrix} 3 & 2 \\ 2 & 3 \end{matrix} \right| \)= 9 - 4 = 5
Δ₁ = \(\left| \begin{matrix} 12 & 2 \\ 13 & 3 \end{matrix} \right| \)= 36 - 26 = 10
Δ₂ = \(\left| \begin{matrix} 3 & 12 \\ 2 & 13 \end{matrix} \right| \)= 39 - 26 = 10
∴ z = \(\frac { { \triangle }_{ 1 } }{ \triangle } =\frac { 10 }{ 5 } =2\Rightarrow \frac { 1 }{ x } =2\Rightarrow x=\frac { 1 }{ 2 } \)
y = \(\frac { { \triangle }_{ 2 } }{ \triangle } =\frac { 15 }{ 5 } \) = 3
∴ Solution set {\(\frac{1}{2}\), 3}

2x+5y = -2, x+2y = -3
The matrix form of the system is
\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) \left( \begin{matrix} x \\ y \end{matrix} \right) =\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
⇒ AX = B where
A =\(\left( \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right) ,B=\left( \begin{matrix} -2 \\ -3 \end{matrix} \right) \)
X =\(\left( \begin{matrix} x \\ y \end{matrix} \right) \)
⇒ = A^-1B
|A| = \(\left| \begin{matrix} 2 & 5 \\ 1 & 2 \end{matrix} \right| \)= 4 - 5 = -1 ≠ 0
∴ A^-1 =\(\frac { 1 }{ |A| } adjA=\frac { 1 }{ -1 } \left[ \begin{matrix} 2 & -5 \\ -1 & 2 \end{matrix} \right] \)
=\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \)
∴ X = A^-1B =\(\left[ \begin{matrix} -2 & 5 \\ 1 & -2 \end{matrix} \right] \left[ \begin{matrix} -2 \\ -3 \end{matrix} \right] \)
= \(\left[ \begin{matrix} 4-15 \\ -2+6 \end{matrix} \right] =\left[ \begin{matrix} -11 \\ 4 \end{matrix} \right] \)
∴ Solution set is x = -11, y = 4

sin²x - 5 sinx + 4 = 0
put y = sin x
⇒ y²-5y+4 = 0
⇒ (y-4)(y-1) = 0
⇒ y = 4, 1
Case(i) 
When y = 4, sin x = 4 and no solution for sin x = 4 since the range the sin function is [-1, 1]
Case (ii)
When y = 1, sin x = 1
⇒ sin x = sin \(\frac{\pi}{2}\) [\(\because sin \frac {\pi}{2}=1\)]
\(x=n \pi+(-1)^{n} \frac{\pi}{2} \forall n \in z\).

It is enough to prove that the sides of the triangle are equal.
Let z₁ = 1, \({ z }_{ 2 }=\frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \) and \({ z }_{ 3 }=\frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \)
The length of the sides of the triangles are
\(\left| { z }_{ 1 }-{ z }_{ 2 } \right| =\left| 1-\left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\left| \cfrac { 3 }{ 2 } -\cfrac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\frac { 2\sqrt { 3 } }{ 2 } =\sqrt { 3 } \)
\(\left\lfloor { z }_{ 2 }-{ z }_{ 3 } \right\rfloor =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -\left( \frac { -1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) \right| =\sqrt { \left( \sqrt { 3 } \right) ^{ 2 } } =\sqrt { 3 } \)
\(\left| { z }_{ 3 }-{ z }_{ 1 } \right| =\left| \left( \frac { -1 }{ 2 } +i\frac { \sqrt { 3 } }{ 2 } \right) -1 \right| =\left| \frac { -3 }{ 2 } -\frac { \sqrt { 3 } }{ 2 } i \right| =\sqrt { \frac { 9 }{ 4 } +\frac { 3 }{ 4 } } =\sqrt { 3 } \)
Since the sides are equal, the given points form an equilateral triangle

Given ∝, β and ૪ are the roots of ax³ + bx² + cx + d = 0
\(\therefore \alpha +\beta +\gamma =\frac { -b }{ a } \)
\(\alpha \beta +\beta \gamma +\gamma \alpha =\frac { c }{ a } \)
\(\alpha \beta \gamma =\frac { -d }{ a } \)
Now, \(\sum { \frac { \alpha }{ \beta \gamma } } =\frac { \alpha }{ \beta \gamma } +\frac { \beta }{ \gamma \alpha } +\frac { \gamma }{ \alpha \beta } \)
\(=\frac { { \alpha }^{ 2 }+{ \beta }^{ 2 }+{ \gamma }^{ 2 } }{ \alpha \beta \gamma } \)
\(=\frac { { (\alpha +\beta +\gamma ) }^{ 2 }-2(\alpha \beta +\beta \gamma +\gamma \alpha ) }{ \alpha \beta \gamma } \)
\(=\frac { { \left( -\frac { b }{ a } \right) }^{ 2 }-2\left( \frac { c }{ a } \right) }{ -\frac { d }{ a } } \)
\(=\frac { \frac { { b }^{ 2 } }{ { a }^{ 2 } } -\frac { 2c }{ a } }{ -\frac { d }{ a } } \Rightarrow \frac { { b }^{ 2 }-2ac }{ { a }^{ 2 } } \times \frac { -a }{ d } \)
\(\therefore \frac { \sum { \alpha } }{ \beta \gamma } =-\frac { \left( { b }^{ 2 }-2ac \right) }{- ad } =\frac { 2ac-{ b }^{ 2 } }{ ad } \)

The cubic equation is
x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ = 0
\( x^{3}-x^{2}\left(2+\frac{1}{2}+1\right)+x\left(1+\frac{1}{2}+2\right)-(2)\left(\frac{1}{2}\right)(1)=0\)
\(x^{3}-x^{2}\left(\frac{4+1+2}{2}\right)+x\left(\frac{2+1+4}{2}\right)-1=0 \)
\(x^{3}-x^{2}\left(\frac{7}{2}\right)+x\left(\frac{7}{2}\right)-1=0\)
2x³ – 7x² + 7x – 2 = 0

 \(y=-sin\left( { \frac { 1 }{ 3 } x } \right) \)
 The amplitude sin x is 1
\(\Rightarrow \) amplitude of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is also 1.
The period of \(-sin\left( \frac { 1 }{ 3 } x \right) \) is \(\frac { 1 }{ 3 } x=2\pi \Rightarrow x=6\pi \)

ang \(z=\frac { -2 }{ 1+i\sqrt { 3 } } \)
= arg(-2)-arg \(\left( 1+i\sqrt { 3 } \right) \) \(\left( \because arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg{ z }_{ 1 }=g_{ 2 } \right) \)
= \(\left( \pi -{ tan }^{ -1 }\left( \frac { 0 }{ 2 } \right) \right) -tan^{ -1 }\left( \frac { \sqrt { 3 } }{ 1 } \right) \)
= \(\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 } \)
This implies that one of the values of arg z is \(\frac { 2\pi }{ 3 } \)
Since \(\frac { 2\pi }{ 3 } \) lies between \(-\pi \), the principal argument Argz is \(\frac { 2\pi }{ 3 } \)

We compute \(\left| 6-8i \right| =\sqrt { { 6 }^{ 2 }+\left( -8 \right) ^{ 2 } } =10\)
and applying the formula for square root, we get
\(\sqrt { 6-8i } =\pm \left( \sqrt { \frac { 10+6 }{ 2 } } -i\sqrt { \frac { 10-6 }{ 2 } } \right) \) (\(\therefore\) b is negative\( \frac{b}{|b|}=-1 \))
= \(\pm \left( \sqrt { 8 } +i\sqrt { 2 } \right) \)
= \(\pm \left( 2\sqrt { 2 } -i\sqrt { 2 } \right) \)

 Let \(f(x)={ tan }^{ -1 }\sqrt { 9-{ x }^{ 2 } } \)
\(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\) but \(\sqrt { 9-{ x }^{ 2 } } \varepsilon R\)
\(\therefore \ 9-{ x }^{ 2 }\ge 0\)
\(\Rightarrow { x }^{ 2 }-9\ge 0\)
\(\Rightarrow (x+3)(x-3)\le 0\)
\(\Rightarrow \) Domain is [-3, 3]

Let A = \(\left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] \). Then, A^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }^{ T }=\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
So, we get
AA^T = \({ \left[ \begin{matrix} \cos { \theta } & -\sin { \theta } \\ \sin { \theta } & \cos { \theta } \end{matrix} \right] }\left[ \begin{matrix} \cos { \theta } & \sin { \theta } \\ -\sin { \theta } & \cos { \theta } \end{matrix} \right] \)
= \(\left[ \begin{matrix} \cos ^{ 2 }{ \theta +\sin ^{ 2 }{ \theta } } & \cos { \theta \sin { \theta } } -\sin { \theta \cos { \theta } } \\ \sin { \theta \cos { \theta -\cos { \theta \sin { \theta } } } } & \sin ^{ 2 }{ \theta } +\cos ^{ 2 }{ \theta } \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right] \) = I₂
Similarly, we get A^TA = I₂. Hence AA^T = A^TA = I₂ ⇒ A is orthogonal.

Let there be N radioactive nuclei in a sample at any time t and let N_o be the initial number of radioactive nuclei.
Then \(\frac{dN}{dt}\infty N\)
\(\Rightarrow \frac { dN }{ dt } =-\lambda N\)
Where \(\lambda>0\) is a constant
\(\Rightarrow \frac { dN }{ N } =-\lambda dt\)
\(\int { \frac { dN }{ N } } =-\lambda dt\)
\(\int { \frac { dN }{ N } =-\lambda \int { dt } } \)
\(\Rightarrow log\ N=-\lambda t+C\ ...(1)\)
\(T\quad t=0,\ we\ have\ N={ N }_{ 0 }\)
\(\therefore log{ N }_{ 0 }=0+C\)
\(\Rightarrow C=log{ N }_{ 0 }\)
\(\therefore\)(1) becomes, log N = -\(\lambda t+log{ N }_{ 0 }\)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } =-\lambda t\quad ...(2)\)
It is given that 10% of the original number of nuclei have undergone disintegration in a period of 100 years.
Whent= 100 \(N={ N }_{ 0 }-\frac { 10 }{ 100 } \times { N }_{ 0 }=\frac { { 9N }_{ 0 } }{ 10 } \)
Substituting in (2) we get
\(log\quad \frac { 9 }{ 10 } =-100\lambda \)
\(\Rightarrow \lambda =-\frac { 1 }{ 100 } log\frac { 9 }{ 10 } \)
Substituting in (2) we get,
\(log\frac { N }{ { N }_{ 0 } } =\left( \frac { 1 }{ 100 } log\frac { 9 }{ 10 } \right) t\)
when t = 1000,
\(log\frac { N }{ { N }_{ 0 } } =\frac { 1 }{ 100 } log\left( \frac { 9 }{ 10 } \right) \times 1000\)
\(=10log\left( \frac { 9 }{ 10 } \right) \)
\(\Rightarrow log\frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } ={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\)
\(\Rightarrow \frac { N }{ { N }_{ 0 } } \times 100={ \left( \frac { 9 }{ 10 } \right) }^{ 10 }\times 100=\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \)
Hence, \(\frac { { 9 }^{ 10 } }{ { 10 }^{ 8 } } \%\) of radioactive nuclei will remain after 1000 years,

(x²+y²)dy = xy dx
\(\frac { dy }{ dx } =\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } ...(1)\)
\(\therefore put=vx\Rightarrow \frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
\(\therefore\)(1) becomes,
\(v+x\frac { dv }{ dx } =\frac { xvx }{ { x }^{ 2 }+{ v }^{ 2 }{ x }^{ 2 } } \)
\(=\frac { { x }^{ 2 }v }{ { x }^{ 2 }(1+{ v }^{ 2 }) } =\frac { v }{ 1+{ v }^{ 2 } } \)
\(x\frac { dv }{ dx } =\frac { v }{ 1+{ v }^{ 2 } } v=\frac { v-v-{ v }^{ 3 } }{ 1+{ v }^{ 2 } } =\frac { -{ v }^{ 3 } }{ 1+{ v }^{ 2 } } \)
Separating the variables we get,
\(\frac { 1+{ v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \frac { 1 }{ { v }^{ 3 } } +\frac { { v }^{ 2 } }{ { v }^{ 3 } } dv=\frac { -dx }{ x } \)
\(\Rightarrow \int { { v }^{ -3 }dv } +\int { \frac { dv }{ v } =-\int { \frac { dx }{ x } } } \)
\(\Rightarrow \frac { { v }^{ -2 } }{ -2 } +log\ v=-logx+logc\)
\(\Rightarrow -\frac { 1 }{ 2{ v }^{ 2 } } +log\ v=-log\ x+log\quad c\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } -log\ v=logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =logv+logx-logc\)
\(\Rightarrow \frac { 1 }{ 2{ v }^{ 2 } } =log\left( \frac { vx }{ c } \right) \)
\(\Rightarrow \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =log\left( \frac { y }{ c } \right) \Rightarrow { e }^{ \frac { { x }^{ 2 } }{ { e }^{ 2{ y }^{ 2 } } } }=\frac { y }{ c } \)
\(\Rightarrow y={ ce }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } } ...(2)\)
Given y(1) = 1
\(1={ ce }^{ \frac { 1 }{ 2 } }\Rightarrow 1=c\sqrt { e } \)
\(\Rightarrow c=\frac { 1 }{ \sqrt { e } } \)
\(\therefore\)(2) becomes,
\(y=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } }\)
\(Also\ y({ x }_{ 0 })=e\Rightarrow e=\frac { 1 }{ \sqrt { e } } { e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow e\sqrt { e } ={ e }^{ \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =log\quad e\sqrt { e } =log{ e }^{ \frac { 3 }{ 2 } }\)
\(\Rightarrow \frac { { x }_{ 0 }^{ 2 } }{ 2{ e }^{ 2 } } =\frac { 3 }{ 2 } { log }_{ e }^{ e }=\frac { 3 }{ 2 } (1)\)
\(\left[ \because { log }_{ e }^{ e }=1 \right] \)
\(\Rightarrow { x }_{ 0 }^{ 2 }=\frac { 3 }{ 2 } (2{ e }^{ 2 })={ 3e }^{ 2 }\)
\(\Rightarrow { x }_{ 0 }=\pm \sqrt { 3{ e }^{ 2 } } =\pm \sqrt { 3 } .e\)
\(\therefore { x }_{ 0 }=\pm \sqrt { 3 } .e\)

Let \({ z }^{ 3 }+8i=0\)
\(\Rightarrow\) z³ = -8i
= \(8(-i)=8\left( cos\left( -\frac { \pi }{ 2 } +2k\pi \right) isin\left( -\frac { \pi }{ 2 } +2k\pi \right) \right) \),k\(\in Z\)
\(z=\sqrt [ 3 ]{ 8 } \left( cos\left( \frac { -\pi +4k\pi }{ 6 } \right) +isin\left( \frac { -\pi +4k\pi }{ 6 } \right) \right) \)
Taking k = 0, 1, 2 we get,
k = 0, \(z=2\left( cos\left( -\frac { \pi }{ 6 } \right) +isin\left( -\frac { \pi }{ 6 } \right) \right) =2\left( -\frac { 1 }{ 2 } -i\frac { \sqrt { 3 } }{ 2 } \right) =2\left( \frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) \)
k = 1, \(z=2\left( cos\left( \frac { \pi }{ 2 } \right) +isin\left( \frac { \pi }{ 2 } \right) \right) =2=\left( 0+i \right) =0+2i=2i\)
k = 2,\(z=2\left( xcos\left( \frac { 7\pi }{ 6 } \right) +isim\left( \frac { 7\pi }{ 6 } \right) \right) =2\left( cos\left( \pi +\frac { \pi }{ 6 } \right) \right) +isin\left( \pi +\frac { \pi }{ 6 } \right) \)
= \(2\left( -cos\left( \frac { \pi }{ 6 } \right) -isin\left( \frac { \pi }{ 6 } \right) \right) =2\left( -\frac { \sqrt { 3 } }{ 2 } -i\frac { 1 }{ 2 } \right) =-\sqrt { 3 } -i\)
The values of z are \(\sqrt { 3 } -i,2i\) and \(-\sqrt { 3 } -i\)

Let cos⁻¹x = \(\alpha\) and cos-1 y = \(\beta\).
Then, x = cos\(\alpha\) and y cos =\(\beta\)
cos^-1x + x + cos^-1 y + cos^-1x = \(\pi\) gives \(\alpha\)+\(\beta\) = \(\pi\) -cos^-1z.
Now, cos(\(\alpha\)+\(\beta\)) = cos\(\alpha\)cos\(\beta\)-sin\(\alpha\)sin\(\beta\) = xy-\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
-cos(cos^-1 z) = xy\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
so, \(-z=xy-\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \), Which gives -xy - z = -\(\sqrt { 1-{ x }^{ 2 } } \sqrt { 1-{ y }^{ 2 } } \)
Squaring on both sides and simplifying, we get x² + y² + z² + 2xyz = 1.

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}.

Let f(x) x⁶-3x⁵-5x⁴+22x³-39x²-39x+135
Given (1+2i) is a root \(\Rightarrow\)(-2i) is also a root
Also \(\sqrt3\) is a root \(\Rightarrow\)-\(\sqrt3\) is also a root.
Hence, the factors of f(x) are [x - (1 + 2i)]
[x-(1-2i)] [x\(\sqrt3\)] [x+\(\sqrt3\)]
[(x-1)-2i] [(x-1)+2i] [x-\(\sqrt3\)][x+\(\sqrt3\)]
((x-1)²+2²)(x²-3) = (x²-2x+1+4)(x²-3)
\(\Rightarrow\) factor of f(x) is (x²-2x+5)(x²-3)
\(\Rightarrow\)x⁴-3x²-2x³+6x+5x²-15
\(\Rightarrow\)(x⁴-3x²-2x³+6x-15) is a factor of f(x) 
To find the other factor, let us divide f(x) by
x⁴ - 2x³ + 2x² + 6x - 15

The other factor is x² - x - 9
\(\Rightarrow x=\frac { 1\pm \sqrt { { (-1) }^{ 2 }-4(1)(-9) } }{ 2 } \left[ \because x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \right] \)
\(\Rightarrow x=\frac { 1\pm \sqrt { 37 } }{ 2 } \)
Hence the roots are
1 - 2i, 1 + 2i, \(\sqrt { 3 }, -\sqrt { 3 }, \frac { 1+\sqrt { 37 } }{ 2 } ,\frac { 1-\sqrt { 37 } }{ 2 } \).