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 The set of all values of   $\lambda$  for  which system of linear  equations ,$2X_{1}-2X_{2}+X_{3}=\lambda X_{1}$, $2X_{1}-3X_{2}+2X_{3}=\lambda X_{2}$   and  $-X_{1}+2X_{2}=\lambda X_{3}$   has   a non-trival solution.

Answer:

Explanation:

 Given system of linear equations

     $2X_{1}-2X_{2}+X_{3}=\lambda X_{1},$

$\Rightarrow$             $(2-\lambda)X_{1}-2X_{2}+X_{3}=0        .......(i)$

     $2X_{1}-3X_{2}+2X_{3}=\lambda X_{2}$

  $\Rightarrow$     $2X_{1}-(3+\lambda)X_{2}+2X_{3}=0        .......(ii)$

                  $-X_{1}+2X_{2}=\lambda X_{3}$

   $\Rightarrow$  $-X_{1}+2X_{2}-\lambda X_{3}=0$                         .......(iii)

 Since, the system has non-trival solution.

      $\therefore$           $\begin{bmatrix}2-\lambda & -2&1 \\2 & -(3+\lambda)& 2  \\ -1&2&-\lambda \end{bmatrix}=0$

  $\Rightarrow  (2-\lambda)(3\lambda+\lambda^{2}-4)+2(-2\lambda+2)+1(4-3-\lambda)=0$

  $\Rightarrow  (2-\lambda)(\lambda^{2}+3\lambda-4)+4(1-\lambda)+(1-\lambda)=0$

$\Rightarrow  (2-\lambda)(\lambda+4) (\lambda-1)+5(1-\lambda)=0$

$\Rightarrow  (\lambda-1)[(2-\lambda)(\lambda+4)-5]=0$

$\Rightarrow  (\lambda-1)(\lambda^{2}+2\lambda-3)=0$

 $\Rightarrow  (\lambda-1)[(\lambda-1)(\lambda+3)]=0$

$\Rightarrow  (\lambda-1)^{2}(\lambda+3)=0$

  $\Rightarrow  \lambda=1,1,-3

 If  $A = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}$ is a matrix satisfying the equation AA^T =9l, where I  is 3x3 identity matrix , then the ordered pair (a,b)  is equal to

Answer:

Explanation:

Given   ,  $A = \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}$

              $A^{T} = \begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$

$A.A^{T} =  \begin{bmatrix}1 & 2 & 2\\2 & 1 &-2 \\ a&2&b\end{bmatrix}\begin{bmatrix}1 & 2 & a\\2 & 1 &2 \\ 2&-2&b\end{bmatrix}$

$=  \begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}$

 It is given that A.A^T  =9I

  =   $ \begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}$

  = 9  $ \begin{bmatrix}1 & 0 & 0\\0 & 1 &0 \\ 0&0&1\end{bmatrix}$

   $\Rightarrow$   $\begin{bmatrix}9 & 0 & a+4+2b\\0 & 9 &2a+2-2b \\ a+4+2b&2a+2-2b&a^{2}+4+b^{2}\end{bmatrix}= \begin{bmatrix}9 & 0 & 0\\0 & 9 &0 \\ 0&0&9\end{bmatrix}$

On comparing we get, a+4+2b=0

  $\Rightarrow$         a+2b=-4      ......(i)

   2a+2-2b=0

 $\Rightarrow$     a-b=-1

 and     $a^{2}+4+b^{2}=9$           .........(iii)

   On solving Eqs. (i) and (ii) , we get,

                    a=-2, b=-1

This satisfies Eq.(iii)

   Hence, (a,b)= (-2,-1)

Let  $\alpha$ and  $\beta$  be thr roots of equation   $x^{2}-6x-2=0$ . If   $a_{n}=\alpha^{n}-\beta^{n}$ , for   $n\geq 1$  , then the value  of   $\frac{a_{10}-2a_{8}}{2a_{9}}$  is equal to

Answer:

Explanation:

 Given,    $\alpha$ and  $\beta$  are the roots of the equation  $x^{2}-6x-2=0$

 $\therefore$    $a_{n}=\alpha^{n}-\beta^{n}$       $n\geq 1$ 

   $\therefore$         $a_{10}=\alpha^{10}-\beta^{10}$

                              $a_{8}=\alpha^{8}-\beta^{8}$

                              $a_{9}=\alpha^{9}-\beta^{9}$

Now, consider

       $\frac{a_{10}-2a_{8}}{2a_{9}}=\frac{\alpha^{10}-\beta^{10}-2(\alpha^{8}-\beta^{8})}{2(\alpha^{9}-\beta^{9})}$

  $=\frac{\alpha^{8}(\alpha^{2}-2)-\beta^{8}(\beta^{2}-2))}{2(\alpha^{9}-\beta^{9})}$

      $=\frac{\alpha^{8}.6\alpha-\beta^{8}6\beta}{2(\alpha^{9}-\beta^{9})}$

$=\frac{6\alpha^{9}.-6\beta^{9}}{2(\alpha^{9}-\beta^{9})}=\frac{6}{2}=3$

{   $ \therefore $   $\alpha $  and  $\beta$  are the roots of the equation:

     $x^{2}-6x-2=0$ 

      or                 $\alpha^{2}$ =6 $\alpha$ +2

     $\Rightarrow$     $\alpha^{2}-2=6\alpha$                              

     and      $\beta^{2}=6\beta-2$

     $\Rightarrow$    $\beta^{2}-2=6\beta$  }

 Alter :

   Since ,  $\alpha$ and  $\beta$  be thr roots of equation   $x^{2}-6x-2=0$ 

   or    $x^{2}=6x+2$

  $\therefore$     $\alpha^{2}$ =6 $\alpha$ +2

   $\Rightarrow$     $\alpha^{10}=6\alpha^{9}+2\alpha^{8}$  ......(i)

Similarly,    $\beta^{10}=6\beta^{9}+2\beta^{8}$.................(ii)

On subtracting  Eq. (ii) from Eq.(i) , we get

     $\alpha^{10}-\beta^{10}=6(\alpha^{9}-\beta^{9})+2(\alpha^{8}-\beta^{8})$  ( $\Rightarrow a_{10}=6a_{9}+2a_{8}  (\because a_{n}=\alpha^{n}-\beta^{n})$ )

 $\Rightarrow a_{10}-2a_{8}=6a_{9}\Rightarrow \frac{a_{10}-2a_{8}}{2a_{9}}=3$

A complex number z is said to be unimodular, if |z| =1, suppose  z₁ and z₂ are complex numbers such that $\frac{z_{1}-2z_{2}}{z-z_{1}z_{2}}$  is unimodular and z₂ is not unimodular.   Then , the point z₁ lies on a

Answer:

Explanation:

Central Idea:

if z is unimodular, then |z|=1, Also , use property of modulus i.e,   $z\overline{z}=|z|^{2}$

Given,   $z_{2}$  is not unimodular i.e, |$z_{2}$|≠ 1

 and   $\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}$ is unimodular.

$\Rightarrow$    $\mid\frac{z_{1}-2z_{2}}{2-z_{1}\overline{z_{2}}}\mid$  =1

$\Rightarrow$      $\mid z_{1}-2z_{2}\mid ^{2}=\mid2-z_{1}\overline{z_{2}}\mid ^{2}$

$\Rightarrow$           $(z_{1}-2z_{2})(\overline{z_{1}}-2\overline{z_{2}})=(2-z_{1}\overline{z_{2}})(2-\overline{z_{1}}z_{2})$                                             $(\because  z\overline{z}= |z|^{2})$

$\Rightarrow$     $|z_{1}|^{2}+4|z_{2}|^{2}-2\overline{z_{1}}z_{2}-2z_{1}\overline{z_{2}}$

$\Rightarrow$     $(|z_{2}|^{2}-1)(|z_{2}|^{2}-4)=0$

  $\because$      $|z_{2}|\neq 1$

   $\because$   $|z_{1}|= 2$

 Let   $z_{1}=x+iy\Rightarrow x^{2}+y^{2}=(2)^{2}$

 $\therefore$    point z₁ lies on a circle of radius  2.

Lat A and B be the two sets containing four and two elements respectively. Then, the number  of subsets of the  set A X B , each having at least three elements are,

Answer:

Explanation:

Given, n(A)=4,   n(B)=2

  $\Rightarrow$     n( A X B)= 8

   Total number of subsets of set

                       (A X B) =   $2^{8}$

Number of subsets of set A X B  hacing no element   (i.e, Φ)=1

 Number  od subsets of set A XB  having one element=   $^{8}C_{1}$

Number of subsets of set AX B  having two elements = $^{8}C_{2}$

$\therefore$       Number of subsets having atleast three elements

                             =   $2^{8}-(1+^{8}C_{1}+^{8}C_{2})$

                            =   $2^{8}-1-8-28$

                               =  $2^{8}-37$

                                  =256-37

                                 =219

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