Question

Let $A=\left\{X=(x,y,z{)}^T:PX=0\text{and}{x}^2+y^2+z^2=1\right\}$ where $P=\left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right]$ then the set $A$ :

• A. is a singleton
• B. contains exactly two elements
• C. contains more than two elements
• D. is an empty set

Answer 1

Answer: (B)

Given $P=$ $\left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& 1\end{array}\right]$ Here $|P|=0\&$ also
given $PX=0$
$\Rightarrow \left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& 1\end{array}\right]\left[\begin{array}{c}X\\ Y\\ Z\end{array}\right]=0$

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 $(x,y,z)$