Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

For which scalars a, b, c, d do the solutions to the equation ax+by+cz=d form a subspace of R(3).

See more answers at
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly


Get your free account and access expert answers to this and thousands of other questions

In \(ax + by + cz = d\), if x and y are any number in R(1), then z is constrained to \[z=\frac{d-ax-by}{c}\] The subspace S = { any (x, y, z)} is a plane which does not pass through (0,0,0) and is therefore not a subspace.
Actually, I failed to address all the trivial cases. If a = b = c = 0, there is no solution. The empty set is not a subspace. If one or two of a,b, or c equals 0, then you can rotate your axes such that the one that is not 0 is the constant for z, in which case the theorem holds.
so your saying that if atleast a, b, or c is 0 then the others can be non zero numbers and be a subspace of R(3)?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

No, I'm saying that if d = 0 then the solution set (x, y, z) is a subspace of R(3) UNLESS all three are equal to 0.
ll three a, b, c****
and that would be the same thing if it was a, b, c to the equation ax+by=c form a subspace of R(2)... then c=0 in this case.
also, if d=0, then as long as atleast one of a, b, or c is non zero then it IS a subspace right?
actually the set u r given is a subspace if and only if d=0.
but why? and does it matter what a, b, and c are?
u have \[ \large D=\{(x,y,z):ax+by+cz=d\} \]
if d=0 then u have \[ \large D=\{(x,y,z):ax+by+cz=0\}. \] So let \(u=(x_0,y_0,z_0)\), \(v=(x_1,y_1,z_1)\) both in \(D\) and \(\alpha\in\mathbb{R}\) so (i) \(u+v=(x_0+x_1,y_0+y_1,z_0+z_1)\) so \[ \large a(x_0+x_1)+b(y_0+y_1)+c(z_0+z_1)= \] \[ \large =(ax_0+by_0+cz_0)+(ax_1+by_1+cz_1)=0+0=0 \] (ii) \(\alpha v=(\alpha x_1,\alpha y_1,\alpha z_1)\) so \[ \large a(\alpha x_1)+b(\alpha y_1)+c(\alpha z_1)=\alpha(ax_1+by_1+cz_1= \alpha0=0 \] this proves that D is a subspace of R^3.
For the converse. Let's suppose that \[ \large D=\{(x,y,z):ax+by+cz=d\} \] is a subspace of R^3. then the zero vector MUST be in D so \[ \large 0=a0+b0+c0=d \]
I do understand that for sure. but how did you know to make d=0, or how do you know it doesn't work for any non zero number?
ohhh okay.
therefore D is a subspace if and only if d=0
so that same idea applies if the question is ax+by=c, for scalars a, b, c ... subspace of R(2) right
It seems like the book is trying to trick me... why give one question stating for ax+by=c, and then the question right after that be ax+by+cz=d, when c has to be 0 for the first and d has to be 0 for the second. Its the same question isnt it, or is there more to it that I am missing.
yes it is. the book is trying to stress that for a subset of a given to be a subspace (i.e., a space on its own) the zero vector (which is unique) must also belong to the subset in question.
that's what u r supposed to get from this two exercises.
*given space*
i hope i was helpful gotta go
it most definitely was... i hope to see you around again, you have been very helpful. thank you.

Not the answer you are looking for?

Search for more explanations.

Ask your own question