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 $\mathrm{A}$ :
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.