Question

Can anyone help

Answer 1

Sure whats your question

Answer 2

So I have two linear equations


\[E_{1} = x-y-z = 3 \]

\[E_{2} = x-10y+10z = 0\]

Answer 3

E1-E2

= (x-y-z =3) - (x-10y-10z = 0)

= 9y-11z = 3

Answer 4

@Michele_Laino can you help

Answer 5

I have the answer, it says there is linear dependency here but my whole issue is how ?

Answer 6

usually when there is linear dependency the two equations are usually the same equation: for example

Ax+by+Cz

3Ax+3by+3Cz

Answer 7

@pooja195 @parthkohli

Answer 8

@Hero

Answer 9

Yes...


Ax+by+Cz

3Ax+3by+3Cz

are essentially the same expression; the 2nd is the 1st, times 3.


Better yet:


Ax+by+Cz =4

3Ax+3by+3Cz = 12

...is a set of linear equations; the 2nd is simply 3 times the first.
Thus, the graphs are identical, and you have infinitely many solutions.

Answer 10

Look up "linear dependency" and use what you find to determine whether or not these 2 equations are "linearly independent."

Answer 11

@mathmale

This is what I did

\[10E_{1} = 10(x-y-z = 3) = 10x-10y-10z = 30\]

\[10E_{1}-E_{2} = (10x-10y-10z)-(x-10y+10z) = 9x-20z = 30\]

\[E_{3} = 9x-20z = 30 \]

Answer 12

\[-1*(x-y-z = 3) = (-x+y+z = -3) \]

\[(x-10y+10z) + (-x+y+z) = -9y+11z = -3\]

Answer 13

@Redcan

Answer 14

I'm not sure I understand the question? You have to vectors, so the only way they can be linearly dependent on each other is if they are multiples of each other i.e. \[E_1 = c E_2\] for some \[c\in\mathbb{R}\]

Answer 15

This is the thing, you see the answer to the question says that they are linear dependent on each-other.

Answer 16

But I'm having trouble explaining it.

Answer 17

assume it's true, they are lin. dep. then
\[cE_1=E_2\] for some c. So
\[c(x-y-z-3)+(x-10y+10z)=0\]
\[(1+c)x-(c+10)y+(10-c)z-3c=0\] which yields four equation that must all
hold simultaneously, but c=-1 from the first and from the last -3c=0, a contradiction! Therefore \[cE_1\ne E_2.\]

Answer 18

It's worth noting, if you write them in matrix form, and then row reduce. if you don't get a row of zeros, there aren't lin. dep.