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  • Let <img alt="A = \left \{ X = (x,y,z)^{2} : Px = 0 , and , x^{2}+y^{2}+z^{2}=1 \right \}" src="http://entrancecorner.codecogs.com/gif.latex?A%20%3D%20%5Cleft%20%5C%7B%20X%20%3D%20%28x%2Cy%2Cz%29%5E%7B2%7D%20%3A%20Px%20%3D%200%20%2C%20and%20%2C%20

Let A = \left \{ X = (x,y,z)^{2} : Px = 0 , and , x^{2}+y^{2}+z^{2}=1 \right \} where P = \begin{pmatrix} 1 & 2 &1 \\ -2& 3 &-4 \\ 1& 9 &-1 \end{pmatrix} then Set A  contains  :
Option: 1 exactly two elements  
Option: 2 is an empty set ,
Option: 3  is a singletan set ,  
Option: 4 contains more than 2 elements . 

Answers (1)

best_answer

\\\text { Given } P=\left[\begin{array}{ccc} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{array}\right],\\ |\mathrm{P}|=1(3\times (-1)-(-4)\times 9)-2((-1)\times(-2)-(-4)\times1)+1(9\times(-2)-3\times 1)\\|P|=0 \\\text {Here }|P|=0\;\; \\ \text {also given } P X=0

\Rightarrow\left[\begin{array}{ccc} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=0

\left.\begin{array}{c} x+2 y+z=0 \\ \Rightarrow-2 x+3 y-4 z=0 \\ x+9 y-z=0 \end{array}\right\} D=0, \text { so system have infinite many solutions, }

By solving these equation

\begin{aligned}{l} \text { we get } x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2} \\ \text { Also given, } x^{2}+y^{2}+z^{2}=1 \\ \Rightarrow\left(\frac{-11 \lambda}{2}\right)^{2}+\lambda^{2}+\left(\frac{7 \lambda}{2}\right)^{2}=1 \\ \Rightarrow \lambda=\pm \frac{1}{\sqrt{\frac{121}{4}+1+\frac{49}{4}}} \end{aligned}

So, there are 2 values of \lambda

Therefore,  there are 2 solution set of (x, y, z).

Posted by

Suraj Bhandari

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