Solve the following
Question:

Let $\mathrm{A}=\left\{\mathrm{X}=(x, y, z)^{\mathrm{T}}: \mathrm{PX}=0\right.$ 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. (1) is a singleton

2. (2) is an empty set

3. (3) contains more than two elements

4. (4) contains exactly two elements

Correct Option: , 4

Solution:

$\because|P|=1(-3+36)-2(2+4)+1(-18-3)=0$

Given that $P X=0$

$\therefore$ System of equations

$x+2 y+z=0 ; 2 x-3 y+4 z=0$

and $x+9 y-z=0$ has infinitely many solution.

Let $z=k \in \mathbf{R}$ and solve above equations, we get

$x=-\frac{11 k}{7}, y=\frac{2 k}{7}, z=k$

But given that $x^{2}+y^{2}+z^{2}=1$

$\therefore k=\pm \frac{7}{\sqrt{174}}$

$\therefore$ Two solutions only.