Let A = \{ X = (x, y, z)^T : PX = 0 x² + y² + z² = 1 \} where P = {bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 {bmatrix} andthen the set A: [2 Sep 2020 Shift 2]

Correct Answer is : contains exactly two elements

contains more than two elements

Correct Answer is : contains exactly two elements

Test Index

Matrices and Determinants Part 1

Section: Mathematics

Question : 2 of 100

Marks: +1, -0

Let

A = \{ X = (x, y, z)^T : PX = 0

x^{2} + y^{2} + z^{2} = 1 \}

P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}

Solution:

Given

P = \begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}

, Here

|P| = 0 \&

also given PX = 0

⇒

\begin{bmatrix} 1 & 2 & 1 \\ -2 & 3 & -4 \\ 1 & 9 & -1 \end{bmatrix}

\begin{bmatrix} x \\ y \\ z \end{bmatrix} = 0

⇒

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

D = 0, so system have infinite many solutions,

By solving these equation

we get

x = \frac{-11 \lambda}{2}; y = \lambda; z = \frac{7 \lambda}{2}

Also given,

x^{2} + y^{2} + z^{2} = 1

\Rightarrow \left( \frac{-11 \lambda}{2} \right)^{2} + (\lambda)^{2} + \left( \frac{7 \lambda}{2} \right)^{2} = 1

\Rightarrow \lambda = \pm \frac{1}{\sqrt{ \frac{121}{4} + 1 + \frac{49}{4} }}

so, there are 2 values of

\lambda

\therefore

so, there are 2 solution set of

(x, y, z)

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