Question

qn below

Answer 1

wat

Answer 2

wait I am copying it

Answer 3

DO IT NOW

Answer 4

show some patience

Answer 5

ene That's a really hard question. What class is this for?

Answer 6

add all 3.

x+y+z + a(x+y+z) = 0

=> x + y + z + ax + ay + az = 0

x(a+1) + y(a+1) + z(a+1) = 0

(a+1)(x+y+z) = 0

so i think if u consider a+1 to be zero then a would be -1 or something? but i dunno; just a guess :)

Answer 7

12th grade

Answer 8

12th grade; i've just finished 10th :P

Answer 9

ene So I'm taking that next year. WOW. I might not survive.

Answer 10

Really, all real numbers would work as long as 0 is one of the numbers.

Answer 11

lol afterall you The hero @TheHero

Answer 12

i think a = -1.

Answer 13

why 0 should be one of them?

Answer 14

hahaha xD

Answer 15

@AravindG bcoz (a+1)(x+y+z) = 0; one of them HAS TO BE 0.

In that case am telling if a+1 = 0 then a = -1.

Answer 16

Thank you

Answer 17

The options are :

a){1,0,-1}
b)R-{-1}
c)R-{1}
d){1,-1}

Answer 18

I suppose then it will be b

Answer 19

Nope i think it can be A .

Answer 20

If a=-1 it will become an identity

Answer 21

here's how -

(a+1)(x+y+z) = 0

if A = 0 then it is x+y+z = 0.

and u plug A = 0 in all those 3 eqns, u get x = 0 y = 0 z = 0 which again on adding give u x+y+z = 0. correct?

Answer 22

I see so x+y+z=0 implies x=y=z=0 ?

Answer 23

are you familiar with cramers method?

Answer 24

@abb0t nope. :P

Answer 25

Nope :/

Answer 26

@agent0smith what do you think?

Answer 27

I agree with @luckythebest that a=0 has to be a solution - it gives a unique solution

x + ay = 0 gives x = 0
y + az = 0 gives y = 0
z + ax = 0 gives z =0

Answer 28

so which option will be right?

Answer 29

I think -1 works as you said, as it gives x=y=z (though that seems to imply infinite solutions not a unique solution?)

a=1 gives
x=-y
y=-z
z=-x

I'm not sure what this notation means, though... the negative?
b)R-{-1}
c)R-{1}

I still think it's (a), though.

Answer 30

It means we are removing -1 from set of real numbers

Answer 31

R-{-1} ^^

Answer 32

ahhh okay. That changes things then, i thought it meant only -1.

Answer 33

:)

Answer 34

0 gives a unique solution for sure. -1 gives an identity, not unique solutions, so it would seem all real numbers except -1, which makes sense from

(a+1)(x+y+z) = 0

when x+y+z=0. And i think x=y=z=0 (has to equal zero) from
x + ay = 0
y + az = 0
z + ax = 0

Answer 35

x + ay = 0
y + az = 0
z + ax = 0

using substitution gives
z = -ax
y = a^2 x
x = -a^3 x
or
x(1+a^3) = 0

which is true if x=0 for all a. And we know if a=-1 then it gives an identity.

Answer 36

lol @ rman "DO IT NOW"

Answer 37

lol I laughed at that too.