Maths Problem

06-07-2006, 05:14 AM

1)Matrix question

Code:

   If |a b| *|x| = k *|x|
      |c d|   |y|        |y|
, prove that k satisfies the equation k^2 - (a+d)k+(ad-bc)=0. If the root of this quadratic equation are A and B, find the calue of A + B and AB in terms of a, b, c and d. Hence, or otherwise, prove that 
  |a b| * |  b     b   | = | b      b  | * |A 0|
  |c d|    |A-a  B-a  |   |A-a   B-a|    |0 B|

Sorry for the uglyness of the thing, if you know how to change, would you mind teaching me to do so.

**prh99** · 06-07-2006, 05:41 AM

Can I assume terms enclosed in | are in a matrix?

06-07-2006, 05:50 AM

Quote:

Originally Posted by prh99

Can I assume terms enclosed in | are in a matrix? Just curious what class is this for?

Yes, it is a matrix question.....I better add at the question to prevent misunderstanding.

KiraRin · 06-08-2006, 10:08 AM

Agh, I remeber doing this for my degree! I would have to go through my lecture notes though, and unfortunately they are at home >.<

So, no use from me then.....

H.

06-24-2006, 08:56 PM

@ReiAyanami580: Start with basics...

(matrix A) * (vector X) = (vector X)

only when (matrix A) is the identity matrix.

So (matrix A) * (vector X) = k * (vector X)

can only be true when (matrix A) is

k * (identity matrix)

Hence

a, d = k; b, c = 0

The subsitute these values into your equiation and

k^2 - (a+d)k + (ad-bc)=0

becomes

k^2 - 2k^2 + k^2 = 0; 2k^2 - 2k^2 = 0; 0 = 0

And this satifies your other equation:

(matrix A) * (matrix B) = (matrix B) * (matrix C)

because both (matrix A) and (matrix C) are equal to k * (identity matrix)

06-24-2006, 09:31 PM

Smart! Good girl!

I remember learning this stuff... some years ago... vaguely...

etherkye · 06-25-2006, 02:40 AM

i remember doing my test in this. Like almost two weeks ago XD

arch-night · 06-29-2006, 09:51 PM

:O... Holy hell.... I don't know where to start... I mean... Damn.. I'm not even remotely close to this stuff.. Especially when I'm more of a psycological person, your on your own lol.. And as for you Jinjin, crap, I've seen you reply around, and knew you were smart... But crap! I can't believe anyone can answer this lol... Cheers to you! ^^

JavaDog · 07-02-2006, 04:04 AM

Wow, lol none of that made sence to me at all.. o__o; I'm going into Geometry next year.

It looks boooorrrriiiiinnng. I'd much rather take Algebra II but that's the year after.

06-07-2006, 05:14 AM	#1 (permalink)
ReiAyanami580 Posts: n/a Points: 0 Bank: 0 Total Points: 0 Donate	Maths Problem 1)Matrix question Code: If \|a b\| \|x\| = k \|x\| \|c d\| \|y\| \|y\| , prove that k satisfies the equation k^2 - (a+d)k+(ad-bc)=0. If the root of this quadratic equation are A and B, find the calue of A + B and AB in terms of a, b, c and d. Hence, or otherwise, prove that \|a b\| * \| b b \| = \| b b \| * \|A 0\| \|c d\| \|A-a B-a \| \|A-a B-a\| \|0 B\| Sorry for the uglyness of the thing, if you know how to change, would you mind teaching me to do so. Last edited by ReiAyanami580 : 06-07-2006 at 05:50 AM.

06-07-2006, 05:41 AM	#2 (permalink)
prh99 - Join Date: Aug 2005 Posts: 1,567 Images: 3 Points: 565.00 Bank: 1,626,811.32 Total Points: 1,627,376.32 Donate	Can I assume terms enclosed in \| are in a matrix? Last edited by prh99 : 06-07-2006 at 06:13 AM.

06-08-2006, 10:08 AM	#4 (permalink)
KiraRin RevanFlame's pet Join Date: Apr 2005 Location: Sunny South Posts: 8,772 Images: 224 Points: 36,040.32 Bank: 10,993,214.29 Total Points: 11,029,254.61 Donate	Agh, I remeber doing this for my degree! I would have to go through my lecture notes though, and unfortunately they are at home >.< So, no use from me then..... H. __________________

06-24-2006, 08:56 PM	#5 (permalink)
jinjin Posts: n/a Points: 0 Bank: 0 Total Points: 0 Donate	@ReiAyanami580: Start with basics... (matrix A) * (vector X) = (vector X) only when (matrix A) is the identity matrix. So (matrix A) * (vector X) = k * (vector X) can only be true when (matrix A) is k * (identity matrix) Hence a, d = k; b, c = 0 The subsitute these values into your equiation and k^2 - (a+d)k + (ad-bc)=0 becomes k^2 - 2k^2 + k^2 = 0; 2k^2 - 2k^2 = 0; 0 = 0 And this satifies your other equation: (matrix A) * (matrix B) = (matrix B) * (matrix C) because both (matrix A) and (matrix C) are equal to k * (identity matrix) Last edited by jinjin : 06-24-2006 at 08:59 PM.

06-29-2006, 09:51 PM	#8 (permalink)
arch-night Experienced Join Date: Mar 2006 Posts: 37 Points: 505.00 Bank: 107.69 Total Points: 612.69 Donate	:O... Holy hell.... I don't know where to start... I mean... Damn.. I'm not even remotely close to this stuff.. Especially when I'm more of a psycological person, your on your own lol.. And as for you Jinjin, crap, I've seen you reply around, and knew you were smart... But crap! I can't believe anyone can answer this lol... Cheers to you! ^^ __________________ Those That Do Not Fear The Unseen Threat Are Foolish, However, Those That Stand Before That Which Is Seen, Are Heroes...

06-24-2006, 09:31 PM	#6 (permalink)
dagger906 Posts: n/a Points: 0 Bank: 0 Total Points: 0 Donate	Smart! Good girl! I remember learning this stuff... some years ago... vaguely...

06-25-2006, 02:40 AM	#7 (permalink)
etherkye Spammer Join Date: Jan 2006 Location: London Posts: 127 Points: 706.00 Bank: 110.38 Total Points: 816.38 Donate	i remember doing my test in this. Like almost two weeks ago XD

07-02-2006, 04:04 AM	#9 (permalink)
JavaDog Banned Join Date: Apr 2005 Location: Internet Posts: 105 Points: 3,356.00 Bank: 106.12 Total Points: 3,462.12 Donate	Wow, lol none of that made sence to me at all.. o__o; I'm going into Geometry next year. It looks boooorrrriiiiinnng. I'd much rather take Algebra II but that's the year after.

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