Question

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

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

Answer 1

The correct option is C contains exactly two elements

Clearly $\left|P\right|=0$
$\Rightarrow PX=O$ has infinite solutions

The line of concurrence passes through $(0,0,0)$ which is centre of the sphere $x^2+y^2+z^2=1$
$\Rightarrow$ Diameter will intersect at two points
$\therefore$ Two solutions (exactly) exist