In all the following questions you are expected to have an "elegant" solution, not a brute force one. No if statements or loops. Unless where specifically noted, no MATLABĀ® functions are to be used.
diag does. (We already learned about ones, zeros, and sum, but if you are unsure, look them up as well.) Using sum and diag, find the sum of the diagonal of a matrix. (For example, A in the next question.)A=magic(6). What expression will give you the \(2\times2\) submatrix of elements in the upper left corner? How about lower right? Can you write an expression that will also work for any other matrix A, for example A=magic(10)?A whose both coordinates are odd.A "flipped" right to left. Up to down.sum, diag and the colon (:) notation.x = [2 5 1 6]. Add 3 to just the odd-positioned elements (resulting in a 2-vector). Now write the expression that adds 3 to the odd-positioned elements and puts the result in the even positions of the original x vector.y = [4 2 1 3]. Think of y as a specific reordering (permutation) of the numbers \(\{1, 2, 3, 4\}\). "4 goes to 1, 2 remains, 1 goes to 3 and 3 goes to 4." Use y to reorder the elements of x in the same manner. (The result should be [6 5 2 1].)y is a permutation z that, when combined with y (in either order) gives the original (non-permuted) elements. Given a permutation vector (as y is in the previous question), find the vector z which corresponds to the inverse of y.b(1:3,1:2:4)=1. Make a "checkerboard" matrix: an 8-by-8 matrix whose cell \(a(i,j)\) equals 1 if \(i+j\) is odd and 0 if it is even. (Of course, do not use loops or if statements!) If you like, use the function spy or pcolor to "see" the checkerboard that you created. (Tricky. Do it in two commands, not one. If you want to do this with one command you may use ones. Also possible using mod and reshape.)A) can also be referenced using a single coordinate: A(3). Remind yourself how this coordinate is related to the original matrix.diag.) Do the same to extract the "anti-diagonal".