Let `A={X=(x, y, z)^(T): PX=0" and "x^(2)+y^(2)+z^(2)=1}`, where `P=[(1,2,1),(-2,3,-4),(1,9,-1)]`, then the set A:

Options

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

Answer

To solve the problem, we need to find the set \( A \) defined by the conditions \( PX = 0 \) and \( x^2 + y^2 + z^2 = 1 \), where \( P \) is given as:

\[ P = \begin{pmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{pmatrix} \]

Let \( X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \).

### Step 1: Set up the equations from \( PX = 0 \)

We can express the equation \( PX = 0 \) as follows:

\[ \begin{pmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]

This results in the following system of linear equations:

1. \( x + 2y + z = 0 \) (Equation 1)
2. \( -2x + 3y - 4z = 0 \) (Equation 2)
3. \( x + 9y - z = 0 \) (Equation 3)

### Step 2: Solve the system of equations

We can solve these equations to express \( x, y, z \) in terms of a parameter.

From Equation 1, we can express \( z \) in terms of \( x \) and \( y \):

\[ z = -x - 2y \]

Substituting \( z \) into Equation 2:

\[ -2x + 3y - 4(-x - 2y) = 0 \]
\[ -2x + 3y + 4x + 8y = 0 \]
\[ 2x + 11y = 0 \implies x = -\frac{11}{2}y \]

Now substituting \( x \) back into the expression for \( z \):

\[ z = -\left(-\frac{11}{2}y\right) - 2y = \frac{11}{2}y - 2y = \frac{11}{2}y - \frac{4}{2}y = \frac{7}{2}y \]

Now we have:

\[ x = -\frac{11}{2}y, \quad z = \frac{7}{2}y \]

### Step 3: Substitute into the sphere equation

Now we substitute \( x \), \( y \), and \( z \) into the equation \( x^2 + y^2 + z^2 = 1 \):

\[ \left(-\frac{11}{2}y\right)^2 + y^2 + \left(\frac{7}{2}y\right)^2 = 1 \]

Calculating each term:

\[ \frac{121}{4}y^2 + y^2 + \frac{49}{4}y^2 = 1 \]

Combining terms:

\[ \left(\frac{121}{4} + 1 + \frac{49}{4}\right)y^2 = 1 \]
\[ \left(\frac{121 + 4 + 49}{4}\right)y^2 = 1 \]
\[ \left(\frac{174}{4}\right)y^2 = 1 \implies \frac{87}{2}y^2 = 1 \implies y^2 = \frac{2}{87} \]

### Step 4: Find \( y, x, z \)

Now we can find \( y \):

\[ y = \pm \sqrt{\frac{2}{87}} \]

Then substituting back to find \( x \) and \( z \):

\[ x = -\frac{11}{2}y = -\frac{11}{2}\left(\pm \sqrt{\frac{2}{87}}\right) = \mp \frac{11\sqrt{2}}{2\sqrt{87}} \]

\[ z = \frac{7}{2}y = \frac{7}{2}\left(\pm \sqrt{\frac{2}{87}}\right) = \pm \frac{7\sqrt{2}}{2\sqrt{87}} \]

### Step 5: Conclusion

Thus, the set \( A \) consists of the points \( (x, y, z) \) that satisfy the equations derived above, leading to infinitely many solutions parameterized by \( y \).