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brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1Joe, if you have to go, I will understand :).

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2I need to think about this a little lol. I can see a way to solve the problem, but there might be a shorter way.

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1I just don't get how to solve these ridiculously hard questions. I would rather be doing diffEq than this!

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2First, do you notice that the vectors (1,0,1), (1,1,0), and (0,1,1) are linearly independent and form a basis?

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2Good, thats whats going to make this easy. Im going to type out the idea and post it, one sec.

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2oops oops oops, i was typing too fast. I dont mean the columns of A are linear independent, i meant the columns of that matrix next to A.

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1Oh my gosh, why did I not think of that?! I should've known to invert that sucker!

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2Here is the correction.

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1You're a much better explainer than the book is!

joemath314159
 2 years ago
Best ResponseYou've already chosen the best response.2i have to run. Im sure there is probably a shorter or more interesting solution. anyways, have a good day :)

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1Thanks you SO much, I really mean it.

imranmeah91
 2 years ago
Best ResponseYou've already chosen the best response.1I passed linear algebra because of joe

anonymoustwo44
 2 years ago
Best ResponseYou've already chosen the best response.0ok so since we have a linear transformation, we then have an induced matrix A such that T(x)=Ax now since our output is an 2x1 matrix and that the vector x we input is an 3x1 matrix, then our induced matrix A is just a 2x3 matrix cause multiplying a 2x3 matrix with a 3x1 matrix will give us an output of 2x1 matrix. So if we find this matrix A, we could now find a formula :D. ok so let our matrix A be: \[A=\left[\begin{matrix}a & b & c\\ d & e & f\end{matrix}\right]\] now we'll get:

anonymoustwo44
 2 years ago
Best ResponseYou've already chosen the best response.0I just wrote the solution :)) its hard to type lots of matrices

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1Thank you very much for your solution. But might I ask why you didn't want to just take the inverse since the 3 vectors form a basis?

anonymoustwo44
 2 years ago
Best ResponseYou've already chosen the best response.0because I won't get the induced matrix or any formula for T from that :D which means I won't get any answer from that

anonymoustwo44
 2 years ago
Best ResponseYou've already chosen the best response.0anyways goodluck with linear algebra which is used in differential equations also and has lots of implications on engineering, physics, economics, biology, chemistry, and math itself.

brinethery
 2 years ago
Best ResponseYou've already chosen the best response.1I did the inverse on both sides and came up with the same answer as what you got. I think it's easy to find the formula if one of the matrices is square. If one of them is not square, then we use the longer method, which is doing an induced matrix and solving for af. Thank you for showing me the other way so that I can use this method for nonsquare matrices.
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