Can anyone help me with thisworksheet? For 1b please post a legit pic of the graph because I amhaving trouble getting it on my graphing calculator. Also need helpwith part e on this. Any help would be much appreciated!

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Hubert KochLv2

25 Jul 2019

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I am having trouble getting the image to appear apparently, I can see it when I submit the problem. If you can not see the image any suggestions would be appreciated to help me out.

Problems

Consider the discretized square plate in Figure 1.

Figure 1: Discretized Square Plate

Write a system of linear equations for this system and clearly identify the matrix A in the matrix equation Ax=b.

Compute the determinant of A.

Compute the classical adjoint matrix (i.e., the adjugate matrix) of A.

Using your answers from problems 2 and 3, solve for A^-1 and x.

Use Cramerâs rule to check your answer by solving for just x₁ and x₄.

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orangescorpion4

I am having trouble with number 3... if I put in 72 def for Tm in both parts of the equation Iâm getting k as -0.03/45.... which doesnât seem right, can you please help me understand where I am going wrong?

Hello there. I just had one question dealing with binary operationsin abstract algebra. I need to see if this operation has anidentity element for all x in R. The operation is as follows:x*y=|x-y| This is what I did. I basically broke it up since thereis an absolute value. x*e=|x-e| = x-e if x> 0 or -(x-e) if xe,then x*e= x-e. We need to find e for x*e=x. So x-e=x and e=0. Nowmy first question is this: 1) If x>0, then why for thisequation, x-e=x, can x be values that are less than 0. It seemslike a contradiction to me. Is e=0 only valid for x>0? Thenafter that I tried out x*0= |x|= x if x>0 and -x if x<0.However, is it just a coincidence that it is x if x>0 and alsowe chose x*0 to be x-0 for x>0? I mean, it seems like either way( whether you use the |x| way or the (x-e) way, you get x as ananswer). My next question is that if we choose x<0 for the abovex*0=..., then doesn't e change? I am confused about this. I mean inone case we have |x|= -x if x<0, but if x<0, then e changes,doesn't it? Please just help me answer these questions and guide mestep by step through this problem. Thank you.

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