Let A = { X = ( x, y, z )T : PX = 0 and x2 + y2 + z2 = 1}

Question:

Let $A=\left\{X=(x, y, z)^{T}: P X=0\right.$ and

$\left.x^{2}+y^{2}+z^{2}=1\right\}$ where $P=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1\end{array}\right]$

then the set $\mathrm{A}$ :

  1. is a singleton

  2. contains exactly two elements

  3. contains more than two elements

  4. is an empty set


Correct Option: , 2

Solution:

Given $\mathrm{P}=\left[\begin{array}{ccc}1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & 1\end{array}\right]$, Here $|\mathrm{P}|=0$ \& also

given $\mathrm{PX}=0$

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

$\left.\begin{array}{c}x+2 y+z=0 \\ -2 x+3 y-4 z=0 \\ x+9 y-z=0\end{array}\right\}$ - $\mathrm{D}=0$, so system have

infinite many solutions, By solving these equation

we $\operatorname{get} x=\frac{-11 \lambda}{2} ; y=\lambda ; z=\frac{7 \lambda}{2}$

Also given, $x^{2}+y^{2}+z^{2}=1$

$\Rightarrow\left(\frac{-11 \lambda}{2}\right)^{2}+(\lambda)^{2}+\left(\frac{7 \lambda}{2}\right)^{2}=1$

$\Rightarrow \lambda=\pm \frac{1}{\sqrt{\frac{121}{4}+1+\frac{49}{4}}}$

so, there are 2 values of $\lambda$.

$\therefore$ so, there are 2 solution set of $(x, y, z)$.

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