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You can pick any 6 (or generally n) consecutive numbers. Or more generally, any n numbers which have distinct values modulo n, i.e. distinct remainders when divided by n (but in practice, it's easiest to pick consecutive ones of course).

$\noindent Note that $2^6 = 64$ and $-64$ = $64 e^{i \left(2k+1\right)\pi}$ for integers $k$. So the six roots will be $2e^{i\cdot \frac{\left(2k+1\right)\pi}{6}}$, for $k=0,1,\ldots,5$.$

For (b), you may want to check out the page here: http://mathforum.org/library/drmath/view/55063.html .

Let w equal the complex number 1 + 3i. Then the first one is the point z = w + i, hence move up by 1 unit from w. The second is z + i = w, which means z = w – i, so move down by one unit from w.

Re: MATH1131 help thread Because sqrt(3)*sqrt(3) = 3, then rearrange. In general, a/sqrt(a) = sqrt(a) for any a > 0.

$\noindent There's an easier approach. The given line has direction vector $\bold{v}=\left(2,3,-1\right)$. By inspection, a normal to the plane is $\bold{n}=\left(1,-1,-1\right)$, because it gives 0 when dotted with each vector for the plane's span, so it is orthogonal to a basis for the plane's...

Re: MATH1131 help thread Conjugate pairs are ones where the angles add up to 0 (or any integer multiple of 2pi). If you convert all the arguments to principal arguments, the conjugate pairs are the ones where the angles are negatives of each other (i.e. add up to 0).

$\noindent For the first one, since they're collinear, remember that $\overrightarrow{AB}$ is a multiple of $\overrightarrow{AC}$, so this imposes some conditions on the coordinates.$ $\noindent We don't need to expand the determinant either. We can just row-reduce a tiny bit. Doing $R_2 \to...

I guess textbooks, online resources / YouTube videos, and practice are good ways to study these types of things (as well as spending time thinking about them, mentally asking Q's about the topics and trying to answer them or investigating them, etc.).

Yeah, there are other ways of proving it. It's easiest to just think about it intuitively I guess. The i-j entry of AI is the i-th row of A dotted with the j-th column of I. The j-th column of I is just the j-th standard basis vector ej for Rn (or Cn in complex case) and dotting any vector with...

You can read up about the Kronecker delta here: https://en.wikipedia.org/wiki/Kronecker_delta .

Yeah, that working is essentially correct. $\noindent And to show $AI = A = IA$, recall that $\left[I\right]_{ij}=\delta _{ij}$, where $\delta _{ij}$ is the \emph{Kronecker delta}, which is equal to $0$ if $i\neq j$ and $1$ if $i=j$ (if we think about it a bit, this is just an algebraic way of...

I don't know what those notes say, but if they call them the other way round, then yes, there is a contradiction. You can search up online for other references to left- and right- distributivity. For example, here is a page on Proof Wiki where they prove a left distributivity thing, and you can...

It's considered right distributivity (because the thing we 'distribute' over the brackets is on the right, I guess). You can see the definitions here for reference: https://en.wikipedia.org/wiki/Distributive_property#Definition .

Oh yeah you're right, sorry, forgot the Q. was right distributivity rather than left distributivity.

I'm not sure about the necessity of giving reasons for each step (I mainly gave them for your own benefit). But it would probably be good to mention why the last line at least is true (reason being that the second last line is [AB]ij + [AC]ij, which is [AB + AC]ij by definition of matrix addition).

Yeah that's correct, except make sure to put []ij around the matrices, since those sums aren't equal to matrices, they're equal to entries of matrices. So what you've shown is that the ij entry of A(B+C) is equal to the ij entry of (AB + AC), so the matrices are equal since all their entries are...

$\noindent The second line there is invalid (that'd be saying the sum of product of terms is equal to the product of the sums, which is not true, e.g. $1\times 5 + 2\times 6 \neq (1+2)\times (5+6)$). It's rather simple in fact though, here's how to do it.$ $\noindent Let $A$ have entries...

You don't need to write out an actual nxn matrix. You just need to use the definition of the i,j entry of a matrix product as a sum (essentially a dot product) and use properties of summations, denoting the i,j entry of a matrix A by say aij.

$\noindent These come down to properties of sums and the definition of the $i,j$ entry of a matrix product (since this involves a sum). For the second one, the following fact will be handy: $ $\noindent If $a_1,\ldots a_n$ are real numbers and $b_1,\ldots, b_m$ are complex numbers (and hence...