"Let A = {X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1} where $$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$, then the set A :"

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Accepted Answer

"Let $X = \left[ {\matrix{ x \cr y \cr z \cr } } \right]$

PX = O

$$\left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

x + 2y + z = 0........(1)

-2x + 3y – 4z = 0....(2)

x + 9y - z = 0..........(3)

from (1) & (3)
$\Rightarrow$ 2x+11y =0

from (1) & (2)
$\Rightarrow$ 2x + 11y = 0

from (2) & (3)
–6x –33y = 0
$\Rightarrow$ 2x +11y = 0

putting value of x in (1), we get
–7y + 2z = 0

Now $${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$$

y²(121 + 1 + 49) = 4

y²(171) = 4

$y = \pm {2 \over {\sqrt {171} }}$

$\Rightarrow x = \pm {7 \over {\sqrt {171} }}$

$\Rightarrow z = \pm {{11} \over {\sqrt {171} }}$

$\therefore$ So, there are 2 solution set of (x, y, z) "

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

x² + y² + z² = 1} where

$$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$,

then the set A :

Solution

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Let A = {X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1} where $$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$, then the set A :

Solution

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More Matrices and Determinants questions

Drill 25 more like these. Every day. Free.

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

x² + y² + z² = 1} where

$$P = \left[ {\matrix{ 1 & 2 & 1 \cr { - 2} & 3 & { - 4} \cr 1 & 9 & { - 1} \cr } } \right]$$,

then the set A :