To be honest this is a bit beyond my grasp of matrix algebra. I can have a go at finding the answer, probably with Google's help, if you want but I can't give you the answer straight off.
In mathematics, an elementary matrix is a matrix which differs from the identity matrix by one single elementary row operation. The elementary matrices generate the general linear group of invertible matrices. Left multiplication (pre-multiplication) by an elementary matrix represents elementary row operations, while right multiplication (post-multiplication) represents elementary column operations.
Elementary row operations are used in Gaussian elimination to reduce a matrix to row echelon form. They are also used in Gauss-Jordan elimination to further reduce the matrix to reduced row echelon...
If E is an elementary matrix, as described below, to apply the elementary row operation to a matrix A, one multiplies A by the elementary matrix on the left, Eā A. The elementary matrix for any row operation is obtained by executing the operation on the identity matrix.
@Abcd there are only three elementary row operations.
Take the scaling one as an example because it's simple.
$$ \left( \begin{matrix} X & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right) \left( \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}\right) = \left( \begin{matrix} Xa & Xb & Xc \\ d & e & f \\ g & h & i \end{matrix}\right) $$
So to multiply the first row by $X$ perform this operation on I to get:
So the point of theorem 1 is not to give you a faster/better/whatever way to do an elementary transformation, but to provide a tool for working with more complicated calculations.
@Abcd OK, suppose we want to do an elementary row transformation on the product AB - I don't know what the notation for this is but let's use $X(AB)$ to mean the elementary transformation $X$ applied to $AB$
Any lens is a combination of a focussing device (the lens) and an aperture through which the light passes. For a regular circular lens the aperture is just the area covered by the lens.
The image created by this is a combination of a perfect image combined with the Fourier transform of the aperture.
For a regular circular lens the image is a perfect image convolved with an Airy disk, which is the Fourier transform of a circular aperture.
Under most circumstances the size of the Airy disk is far smaller than the size of the image so at worst all this does is cause a very slight blurring of the image.
In your example you have a semicircular aperture, so the image will be convolved with whatever the Fourier transform of a semicircle is. Offhand I don't know what it is.
@NehalSamee the image remains the same, but it is convolved with a different Fourier transform. However under most circumstances this will have little effect on the image.
Two bodies of same mass m rotate around each other (like binary stars) at a distance r from each other. Show that, each has a velocity of $v=1/2*\sqrt{Gm}{r}$
Hi! I have very little calculus experience and the following came across in one of my exercises: area of circular sector (without using circle formula)... More like mathematical-physics, but can anyone check if my reasoning is good: $$A=\int \mathrm{d}A\implies A=2\int_0^{\alpha/2}R^2\mathrm{d}\theta/2=\\=R^2\int_0^{\alpha/2}\mathrm{d}\theta=R^2\alpha/2$$ Where $\alpha$ is in radians. It does check off for the full circle and its fractions so I guess it's fine, but is it rigorous and correct?
Where $\alpha$ is the central angle of the circular sector, $R$ is its radius and $\mathrm{d}\theta$ is an infinitesimal change in the angle. Method used: cutting sector into infinitesimally small triangular pieces
I am particularly doubtful about my very first equality, it seems... Ugh IDK, strange ($A=\int \mathrm{d}A$)
I guess so, you are likely right. But I thought it's quite strange because it basically proves the circle area formula... in 2 lines? Anyway, I guess it's just a consequence of the power of calculus.
Thanks!
I was also more cautious than usual because although it gives the same answer as the Wikipedia entry, that one uses a more sophisticated integral