Question

General solution to a set of equations with infinite solutions... (see comments for specific qu) :)

Answer 1

The following equations have infinitely many solutions.
−15 y − 12 z = 6
−2 x − 13 y − 11 z = 5
15 y + 12 z = −6
Give the right hand side of the vector form of the general solution, using a parameter such as s or t. (Any lowercase letter will do as a parameter, so long as it is not x, y or z.)

Answer 2

An example we have been given: for the equations
x = y + 1
y = z + 1
z = x − 2
one correct answer is
[x, y, z] = [0, −1, −2] + t [1, 1, 1]
and you would enter the right hand side of this equation in the space provided:
[0, −1, −2] + t * [1, 1, 1]
or, if you prefer:
[t, t − 1, t − 2]

**NB "in the space provided" refers to an online-submission system for answer checking**

Answer 3

\[\left[\begin{matrix}0 & -15 & -12 & 6 \\ -2 & -13& -11 & 5 \\0 & 15 & 12 & -6\end{matrix}\right]\]

will row reduce to...

Answer 4

\[\left[\begin{matrix}1 & 0 & 3/10 & 1/10 \\ 0 & 1& 4/5 & -2/5 \\0 & 0 & 0 & 0\end{matrix}\right]\]

Answer 5

so we have one free variable

Answer 6

yep, agreed :)

Answer 7

we can use z=t for the free variable
so the 2nd row gives us

y+4/5t=-2/5
so y=-4/5t-2/5

Answer 8

1st row gives us

x+3/10t=1/10

so x=-3/10t+1/10

Answer 9

so our solution is

\[\left[\begin{matrix}x \\ y \\z \end{matrix}\right]=\left[\begin{matrix}3/10 \\ -4/5 \\1 \end{matrix}\right]t+\left[\begin{matrix}1/10 \\ -2/5 \\0 \end{matrix}\right]\]

Answer 10

You are a legend! thanks so much!!

Answer 11

np

Answer 12

\[\left[\begin{matrix}x \\ y \\z \end{matrix}\right]=\left[\begin{matrix}-3/10 \\ -4/5 \\1 \end{matrix}\right]t+\left[\begin{matrix}1/10 \\ -2/5 \\0 \end{matrix}\right]\]

left of a negative sign on the 3/10 (now fixed -3/10)

Answer 13

haha yeah i was just scratching my head about that, too :P cheers!