Need help with question: linear eqn & homogenous soln

[sigma]

Need help with this following question. Sorry if its long.

Consider the linear equation ax+by+cz = d (1) and the associated homogeneous equation ax+by+cz =0. (2)

Let (x1,y1 ,z1 ) and (x2 ,y2 ,z2 ) be two solutions to equation (1)
and let (x0 ,y0 ,z0 ) be a solution to equation (2).

a. Show that (x0 +x1,y0 +y1,z0 +z1) is a solution to equation (1).
b. Show that (x2 −x1,y2 −y1,z2 −z1) is a solution to equation (2).
c. Let k be any real number. Show that (kx0,ky0 ,kz0 ) is a solution to equation (2).

I'm not sure where to begin. I know that you have to show somehow that the solutions they give in a, b and c equal the same solutions for equation 1 or equation 2 but how do you do that? Do you set up some sorta matrix to show this? Just don't know where to begin with this question.

 

[skeeter]

for part (a) ...

\(\displaystyle ax_1 + by_1 + cz_1 = d\)
\(\displaystyle ax_0 + by_0 + cz_0 = 0\)

add the two equations together ...

\(\displaystyle a(x_0+x_1) + b(y_0+y_1) + c(z_0+z_1) = d\)

so ... what does that say about \(\displaystyle (x_0+x_1)\), \(\displaystyle (y_0+y_1)\), and \(\displaystyle (z_0+ z_1)\) ?

now ... how would you use the same general method for part (b)?

hint for part (c) ... factor out a "k"

 

[sigma]

for part (a) ...

so ... what does that say about \(\displaystyle (x_0+x_1)\), \(\displaystyle (y_0+y_1)\), and \(\displaystyle (z_0+ z_1)\) ?

That it equals a solution to equation (1)? Is that all you show? How come when you set up the 2 equations the first one is equation 1 (uses "= d") and the second is equation 2 (uses "= 0")? Aren't they only asking for equation 1?

now ... how would you use the same general method for part (b)?

Do the same thing in (a) except subtract the 2 equations? Here's what I did.

\(\displaystyle \
\L\
\begin{array}{l}
ax_2 + by_2 + cz_2 = 0 \\
ax_1 + by_1 + cz_1 = 0 \\
\\
a(x_2 - x_1 ) + b(y_2 - y_1 ) + c(z_2 - z_1 ) = 0 \\
\end{array}
\\)

Therefore this equals a solution to equation 2?

hint for part (c) ... factor out a "k"

The only thing I can think of is factoring out the k and then you have this:

\(\displaystyle \
\L\
k(x_0 , y_0 , z_0 ) = 0
\\)

Which equals a solution to equation 2? Just seems like there should be more.

 

[skeeter]

for part (a) ...

so ... what does that say about \(\displaystyle (x_0+x_1)\), \(\displaystyle (y_0+y_1)\), and \(\displaystyle (z_0+ z_1)\) ?

That it equals a solution to equation (1)? yesIs that all you show? yes, againHow come when you set up the 2 equations the first one is equation 1 (uses "= d") and the second is equation 2 (uses "= 0")? Aren't they only asking for equation 1?look at the form for equation 1 ... that is what we ended up with

now ... how would you use the same general method for part (b)?

Do the same thing in (a) except subtract the 2 equations? Here's what I did.

\(\displaystyle \
\L\
\begin{array}{l}
ax_2 + by_2 + cz_2 = 0 \\
ax_1 + by_1 + cz_1 = 0 \\
\\
a(x_2 - x_1 ) + b(y_2 - y_1 ) + c(z_2 - z_1 ) = 0 \\
\end{array}
\\)

Therefore this equals a solution to equation 2?gee... you're batting 1000

hint for part (c) ... factor out a "k"

The only thing I can think of is factoring out the k and then you have this:

\(\displaystyle \
\L\
k(x_0 , y_0 , z_0 ) = 0
\\)

Which equals a solution to equation 2? Just seems like there should be more.there is more, see below.

Let k be any real number. Show that (kx0,ky0 ,kz0 ) is a solution to equation (2).

\(\displaystyle a(kx_0) + b(ky_0) + c(kz_0) =\)
\(\displaystyle k(ax_0 + by_0 + cz_0) = k(0) = 0\)

so ... \(\displaystyle (kx_0, ky_0, kz_0)\) is a solution for equation 2.

 

[sigma]

Great. Thanks allot.