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by: Miriam Kim

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# example trial

Miriam Kim
IIT

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COURSE
PROF.
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Class Notes
PAGES
3
WORDS
KARMA
25 ?

## 1

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"I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Miriam!!!"
Kaela

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This 3 page Class Notes was uploaded by Miriam Kim on Monday February 15, 2016. The Class Notes belongs to at Illinois Institute of Technology taught by in Spring 2016. Since its upload, it has received 39 views.

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## Reviews for example trial

I'm pretty sure these materials are like the Rosetta Stone of note taking. Thanks Miriam!!!

-Kaela

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Date Created: 02/15/16
COMPUTER ASSIGNMENT #1 Problem 1 [Problem 1.1] a)Let B be the 4x4 identity matrix. Then identifying each step by a different matrix, it can then be represented as the product of 8 matrices by placing all of the operations that involved rows (B2, B3 and B5) to the left side of B and all of the column operations (B1, B4, B6, and B7) on the right side of B. >>B=eyes(4); >> B5=[1 -1 0 0; >> B1=[2 0 0 0; 0 1 0 0; 0 1 0 0; 0 -1 1 0; 0 0 1 0; 0 -1 0 1]; 0 0 0 1]; >> B6=[1 0 0 0; >> B2=[1 0 0 0; 0 1 0 0; 0 1 0 0; 0 0 1 1; 0 0 1/2 0; 0 0 0 0]; 0 0 0 1]; >> B7=[0 0 0; >> B3=[1 0 1 0; 1 0 0; 0 1 0 0; 0 1 0; 0 0 1 0; 0 0 1]; 0 0 0 1]; >> B5*B3*B2*B*B1*B4*B6*B7 >> B4=[0 0 0 1; ans = 0 1 0 0; -1.0000 0.5000 0.5000 0 0 1 0; 1.0000 0 0 1 0 0 0]; -1.0000 0.5000 0.5000 -1.0000 0 0 b) It can be split into 3: (B5*B3*B2)*(B)*( B1*B4*B6*B7) >> B5*B3*B2 >> B1*B4*B6*B7 ans = ans = 1.0000 -1.0000 0.5000 0 0 0 0 0 1.0000 0 0 1 0 0 0 -1.0000 0.5000 0 0 1 1 0 -1.0000 0 1.0000 0 0 0 Problem 2 >> A=[2 1 1; 4 3 2; 3 2 2]; (to verify solution): >> b=[4; 9; 7]; >> A*x >> x=A\b ans = x = 4 1 9 1 7 1 Problem 3 >> C=B5*B3*B2*B*B1*B4*B6*B7 >> norm(C,2) C = ans = -1.0000 0.5000 0.5000 2.1358 1.0000 0 0 >> norm(C,inf) -1.0000 0.5000 0.5000 ans = -1.0000 0 0 2 >> norm(C,'fro') >> norm(C,1) ans = ans = 2.2361 4 Hand calculations for the norms: 1-norm (max col. Sum), which is the first column, so 1+1+1+1=4 2-norm: multiplying (C*)C gives: 4.0000 -1.0000 -1.0000 -1.0000 0.5000 0.5000 -1.0000 0.5000 0.5000 Then the eigenvalues are 0.0000, 0.4384, and 4.5616. The largest is 4.5616, so taking the root of it gets 2.2361 Infinity-norm (max row sum), which is either the first or third row, so 1+0.5+0.5=2 2 2 2 2 2 2 2 2 (1 0.5 0.5 )(1 00)(1 0.5 0.5 )(1 00)  5  2.2361 Frobenius norm is the Problem 4 >> A=[1 2; 2 2; 2 1]; >> [U, S, V]=svd(A) U = -0.5145 -0.7071 0 0 -0.5145 0.7071 0.4851 V = 0.4851 S = -0.7071 -0.7071 -0.6860 0 -0.7276 4.1231 0 -0.7071 0.7071 0 1.0000 Problem 5 >> x=-1:0.05:1; >> y1=exp(x); >> y2=3.*exp(-10.*x.^2); >> plot(x,y1,x,y2) >> title('Graph of Exponential Functions') >> xlabel('x values') >> ylabel('y values') >> legend('e^x', '3e^(-10x^2)')

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