Let A be a diagonal matrix and of the form \(\large A=\begin{bmatrix}a&0&0\\0&b&0\\0&0&c\end{bmatrix}\)

\(\large\therefore |A|=\begin{vmatrix}a&0&0\\0&b&0\\0&0&c\end{vmatrix}=abc\neq0\)

 \(\large\therefore A^{-1}\) exist and \(\large A^{-1}=\Large{1\over abc}\large\begin{bmatrix}bc&0&0\\0&ac&0\\0&0&ab\end{bmatrix}^t=\begin{bmatrix}1\over a&0&0\\0&1\over b&0\\0&0&1\over c\end{bmatrix}\)

which is also diagonal matrix.

If we take \(\large (N,.)\) then we can say that it's inverse element does not esist.

Here identity element is \(\large e=1\)

The inverse of \(\large 2\in N\) is \(\Large {1\over2}\) which is not an element of \(\large N\)

So inverse does not exist.

So \(\large (N,.)\) is not a group.

Given a * b = 1 + ab

Now b * a = 1 + ba = 1 + ab

So b * a = a * b

∴ is commutative

(a * b) * c = (1 + ab) * c = 1 + (1 + ab) c = 1 + c + abc

and a * (b * c) = a * (1 + bc) = 1 + a (1 + bc) = 1 + a + abc

We found (a * b) * c ≠ a * (b * c)

∴ Operation * is not associate

\(\large\alpha\) and \(\large\beta\) are the roots of \(\large x^2-ax+b^2=0\). Then 

  \(\large \alpha+\beta=a , \space \alpha\beta=b^2\)

  \(\large\therefore (\alpha+\beta)^2=a^2\) 

\(\large\Rightarrow \alpha^2+\beta^2+2\alpha\beta=a^2\\\large\Rightarrow \alpha^2+\beta^2+2b^2=a^2\\\large\Rightarrow \alpha^2+\beta^2=a^2-2b^2\)

Here \(\large \vec a=i+2j+2k\) and let \(\large \vec b=xi+yj+zk\)

\(\large\therefore |\vec a|=\sqrt{1^2+2^2+2^2}=3\)       (1)

Given \(\large |\vec b|=5\)       (2)

angle between \(\large \vec a\) and \(\large \vec b\) is \(\Large{\pi\over 6}\)

 \(\large\therefore \Large{\vec a.\vec b\over |\vec a||\vec b|}\large=\cos\Large{\pi\over 6}\)

\(\Large \Rightarrow \Large {x+2y+2z\over 3\times 5}={\sqrt3\over2}\)     (using (1) and (2))

\(\\\large \Rightarrow x+2y+2z=\Large{15\sqrt3\over2}\)    (3)

\(\large\therefore |\vec a\times\vec b|^2=|\vec a|^2\times|\vec b|^2-(a.b)^2\) 

                  \(\large=3^2\times 5^2-(x+2y+2z)^2\)    (using (1) and (2))

                 \(\large=225-\Large{225\times 3\over4}={225\over4}\)     (using (3))

\(\large\therefore |\vec a\times\vec b|=\Large{15\over2}\)

Therefore the area of the triangle formed by these two vectors is \(\Large={1\over 2}\large|\vec a\times \vec b|=\Large{1\over2}{15\over 2}={15\over4}\)

Let \(\large{ y=tan^{-1}\Large ({x\over\sqrt{1-x^2}}) }\) and \(\large{ z=sin^{-1}(3x-4x^3) }\)

Then we have to find \(\Large { {\mathrm d}y\over{\mathrm d}z }\) 

Let \(\large { x=sin\theta }\).

Then \(\large { y= tan^{-1}\Large({ sin\theta\over \sqrt{1-sin^2\theta}}) }\)

            \(\large { =tan^{-1}\Large{sin\theta\over cos\theta}=\large tan^{-1}tan\theta=\theta }\).

\(\Large{ \therefore {{\mathrm d}y\over{\mathrm d}\theta}=\large 1 }\)

and \(\large { z=sin^{-1}(3sin\theta-4sin^3\theta)=sin^{-1}sin3\theta=3\theta }\).

\(\Large{ {{\mathrm d}z\over {\mathrm d}\theta}=\large 3 }\)

Therefore

\(\Large { {{\mathrm d}y\over{\mathrm d}z}={{\mathrm d}y/{\mathrm d\theta}\over{\mathrm d}z /{\mathrm d \theta}}={1\over3} }\).                           

Equation of the curve is \(\large y=x^2-\Large{1\over x^2}\)     (1)

Therefore \(\Large {\mathrm{d}y\over \mathrm{d}x}\large=2x+\Large{2\over x^3}\)

So, the slope of the tangent of the curve at \(\large (-1,0)\) is \(\large m_1=2(-1)+\Large{2\over(-1)^3}\large=-4\)

If \(\large m_2\) be the slope of the normal to the curve (1), then

        \(\large m_1.m_2=-1\Rightarrow (-4)m_2=-1\)

i.e., \(\large m_2=\Large{1\over4}\)

Let \(\large Z=\Large{2+3i\sin\theta\over1-2i\sin\theta}\)

           \(\large=\Large{(2+3i\sin\theta)(1+2i\sin\theta)\over(1-2i\sin\theta)(1+2i\sin\theta)}\\\large=\Large{(2-6\sin^2\theta)+7i\sin\theta\over1+4\sin^2\theta}\\\large=\Large{2-6\sin^2\theta\over 1+4\sin^2\theta}\large+i\Large{7\sin\theta\over1+4\sin^2\theta}\)

Now \(\large Z\) will be purelly imaginary if

     \(\Large{2-6\sin^2\theta\over 1+4\sin^2\theta}\large=0\)

\(\large\Rightarrow 2-6\sin^2\theta=0\\\large\Rightarrow \sin^2\theta =\Large{1\over3}\\\large\Rightarrow \sin\theta =\pm\Large{1\over\sqrt3}\\\large\Rightarrow\theta=\pm\sin^{-1}\Large{1\over\sqrt3}\)