Let A = {X = (x, y, z)^{T} : PX = 0 text{ and | vedclass

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(B) Given $P = \begin{bmatrix} 1 & 2 & 1 \ -2 & 3 & -4 \ 1 & 9 & -1 \end{bmatrix}$.First,we find the determinant of $P$:$|P| = 1(-3 + 36) - 2(2 + 4)

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contains exactly two elements

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(B) Given $P = \begin{bmatrix} 1 & 2 & 1 \ -2 & 3 & -4 \ 1 & 9 & -1 \end{bmatrix}$.First,we find the determinant of $P$:$|P| = 1(-3 + 36) - 2(2 + 4) + 1(-18 - 3) = 33 - 12 - 21 = 0$.Since $|P| = 0$,the system $PX = 0$ has non-trivial solutions.The equations are:$x + 2y + z = 0$  $(i)$$-2x + 3y - 4z = 0$  (ii)$x + 9y - z = 0$  (iii)From $(i)$ and (iii),adding them gives $2x + 11y = 0 \Rightarrow x = -\frac{11}{2}y$.Substituting $x$ into $(i)$: $-\frac{11}{2}y + 2y + z = 0 \Rightarrow z = \frac{7}{2}y$.Let $y = \lambda$,then $x = -\frac{11}{2}\lambda$ and $z = \frac{7}{2}\lambda$.Given $x^{2} + y^{2} + z^{2} = 1$,we substitute the values:$(-\frac{11}{2}\lambda)^{2} + \lambda^{2} + (\frac{7}{2}\lambda)^{2} = 1$$\frac{121}{4}\lambda^{2} + \lambda^{2} + \frac{49}{4}\lambda^{2} = 1$$\frac{121 + 4 + 49}{4}\lambda^{2} = 1 \Rightarrow \frac{174}{4}\lambda^{2} = 1 \Rightarrow \lambda^{2} = \frac{4}{174} = \frac{2}{87}$.Since $\lambda^{2} = \frac{2}{87}$,there are two possible values for $\lambda$ $(\lambda = \pm \sqrt{\frac{2}{87}})$.Thus,there are exactly two solutions for $(x, y, z)$.

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

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