[hihogyu]

can't believe i'm already stuck now.

it's edexcel Heinemann textbook Review Exercise Q3.


The point P(x,y,z) is mapped on the point Q(X,Y,Z) by the relation
(just gonna omit the bit brakets.. dunno how to draw)

|X| |x| |-1 2 0|
|Y| = M|y|, where M = |2 0 -2|
|Z| |z| |0 -2 1|

Show that Q lies on the plane with equation 2X + Y + 2Z = 0.
Show also that M^3 = λM and give the value of λ.


and.. the P6 exam is tomorrow

[Gregball_87]

so youve got (Xi + Yj + Zk) = M(xi + yj + zk)

i: X = -x + 2y
j: Y = 2x - 2z
k: Z = -2y + z

2X + Y + 2Z = 2(-x + 2y) + (2x - 2z) + 2(-2y + z), which all cancells to give 0 as wanted, theres probs a better way but... its just cos uv got a vector equal to another vector, so u just have to compare the coefficients

[hihogyu]

Originally Posted by Gregball_87

so youve got (Xi + Yj + Zk) = M(xi + yj + zk)

i: X = -x + 2y
j: Y = 2x - 2z
k: Z = -2y + z

2X + Y + 2Z = 2(-x + 2y) + (2x - 2z) + 2(-2y + z), which all cancells to give 0 as wanted, theres probs a better way but... its just cos uv got a vector equal to another vector, so u just have to compare the coefficients

wow, thanks!
don't really think there is a better way, no? but this is gd enough for me

[JohnSPals]

Originally Posted by hihogyu

can't believe i'm already stuck now.

Show also that M^3 = λM and give the value of λ.

and.. the P6 exam is tomorrow

Haha. I remember doing this bit. Just get it shoved in a graphical calculator if it comes up on an exam!

Lazy bum, I am.