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Q.
Let $A =\left\{ X =( x , y , z )^{ T }: PX =0\, \text{and} \,x^2 + y^2 + z^2 = 1 \right\}$ where $P= \begin{bmatrix}1&2&1\\ -2&3&-4\\ 1&9&-1\end{bmatrix}$
then the set $A $ :
Given $P=$ $\begin{bmatrix}1&2&1\\ -2&3&-4\\ 1&9&1\end{bmatrix}$ Here $| P |=0 \&$ also
given $PX = 0$
$\Rightarrow \begin{bmatrix}1&2&1\\ -2&3&-4\\ 1&9&1\end{bmatrix}\begin{bmatrix}X\\ Y\\ Z\end{bmatrix}=0$
infinite many solutions,
By solving these equation
we 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 $( x , y , z )$
