Let A = X = (x, y, z)T: PX = 0 and x2 + y2 + z2 = 1 where [P = [ pmatrix 1 & 2 & 1 - 2 & 3…

Question

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

x2 + y2 + z2 = 1} where

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

then the set A :

Accepted Answer

Let $$X = \left[ {\begin{pmatrix} x  \
   y  \
   z  \

\end{pmatrix} } \right]$$

PX = O

$$\left[ {\begin{pmatrix} 1 & 2 & 1  \
   { - 2} & 3 & { - 4}  \
   1 & 9 & { - 1}  \

\end{pmatrix} } \right]\left[ {\begin{pmatrix} x  \
   y  \
   z  \

\end{pmatrix} } \right] = \left[ {\begin{pmatrix} 0  \
   0  \
   0  \

\end{pmatrix} } \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)

Solutions · 1