Question

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

• 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.

Explanation for correct option:

Find the elements of set $A$:

The set $A=\left\{X={\left(x,y,z\right)}^T:PX=0\text{and}{x}^2+y^2+z^2=1\right\}$

$P=\left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right]$, $X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

Determinant of $P$ is given as,

$$\begin{gathered} \left|P\right|=\left|\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right| \\ =1(-3+36)-2(2+4)+1(-18-3) \\ =+33-12-21 \\ \left|P\right|=0 \end{gathered}$$

Given,

$$\begin{gathered} PX=0 \\ \left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \\ \left[\begin{array}{c}x+2y+z\\ -2x+3y-4z\\ x+9y-z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right] \end{gathered}$$

$\begin{array}{r}x+2y+z=0\\ -2x+3y-4z=0\\ x+9y-z=0\end{array}$

$∆=0$, So the system has infinitely many solutions. [By Cramer rule]

The system of the equation reduces into two equations, put $z=\mu$

$$\begin{gathered} x+2y=-\mu \\ -2x+3y=4\mu \\ x+9y=\mu \end{gathered}$$

Solving these equations we get,

$$\begin{gathered} \Rightarrow 14y=4\mu \\ \Rightarrow y=\frac{4}{14}\mu \\ \Rightarrow y=\frac{2}{7}\mu \\ \Rightarrow x=\frac{-11}{7}\mu \left[\text{From}x+2y=-\mu \right] \end{gathered}$$

Substituting the values of $x,y,z$ in a given equation $x^2+y^2+z^2=1$

We get,

$$\begin{gathered} {\left(\frac{-11}{7}\mu \right)}^2+{\left(\mu \right)}^2+{\left(\frac{2}{7}\mu \right)}^2=1 \\ {\mu }^2\left(\frac{121}{49}+1+\frac{4}{49}\right)=1 \\ \mu =\pm \frac{1}{\sqrt{\frac{121}{49}+1+\frac{4}{49}}} \end{gathered}$$

So there are two values for $\mu$.

Therefore, the set $A$ has exactly two elements.

Hence, option (C) is the correct answer