Numerical Analysis Questions & Answers  
Question by Student 201529190
Dear Professor，for the work times of diagonal matrix. To turn the number to zero from bottom. At $row_{N}$ no work so, $work_{N=0}=o$. At $row_{N-1}$,we need turn $A_{N-1,N}$ to 0. It use 4 works (2moct+2add).so, $ work_{N-1}=4$ \begin{bmatrix} .. &.. & .. & .. & ..\\ ..& ..& ..&.. &.. \\ ..& .. & .. &.. &.. \\ ..& .. & A_{N-1,N-1} & A_{N-1,N} &X_{N-1} \\ ..& ..& .. & A_{N,N} & X_{N} \end{bmatrix} At $row_{N-2}$, we need turn $A_{N-2,N-1}$ and $A_{N-2,N}$ to 0. It use 8 works 2*(2moct+2add).so $ work_{N-2}=8$ \begin{bmatrix} .. &.. & .. & .. & ..\\ ..& ..& ..&.. &.. \\ ..&A_{N-2,N-2}& A_{N-2,N-1} &A_{N-2,N} &X_{N-2} \\ ..& .. & A_{N-1,N-1} & 0 &X_{N-1} \\ ..& ..& .. & A_{N,N} & X_{N} \end{bmatrix} then \begin{bmatrix} .. &.. & .. & .. & ..\\ ..& ..& ..&.. &.. \\ ..&A_{N-2,N-2}& 0 &0 &X_{N-2} \\ ..& .. & A_{N-1,N-1} & 0 &X_{N-1} \\ ..& ..& .. & A_{N,N} & X_{N} \end{bmatrix} so $ work_{N-3}=12$, $ work_{N-n}=4n$ total work$ =\sum_{m=1}^{N-1} 4\times (N-n) =2*(N-1)^{2}\propto N^{2}$. then C2 = 2
10.23.17
This is a very good explanation. There is only a small problem with it: you should have written $B$ instead of $X$ within the last column. 3 points bonus boost.
Question by Student 201700278
Dear Professor,
For Question 1 in Assignment 3, may I know is Gaussian decomposition means Gaussian elimination? I tried to search it online but the results are mostly showing either Gaussian Elimination or LU decomposition. I am confused which method should we use in that question?
10.27.17
Fixed. Good observation. 1 point bonus boost.
Question by Student 201427116
Professor, I have a question about what we studied at last class. Matrices that used in last class are below : $$A= \begin{bmatrix} -2&2&-1 \\6&-6&7 \\3&-8&4 \end{bmatrix} P_{12} = \begin{bmatrix} 0&1&0 \\ 1&0&0 \\ 0&0&1 \end{bmatrix} m _{1} = \begin{bmatrix} 1&0&0 \\ \frac{1}{3}&1&0 \\-2&0&1 \end{bmatrix} \\ P _{23} = \begin{bmatrix} 1&0&0 \\0&0&1 \\0&1&0 \end{bmatrix} $$ With only Gauss Elimination, we can’t handle the “zero” pivot. We must use permutation matrix, as a result, the Upper triangular matrix transfomed from A is below: $$ P_{23}m_{1}P_{12}A = U = \begin{pmatrix} 6&-6&7 \\0&-5&\frac{1}{2}\ \\0&0&\frac{4}{3} \end{pmatrix} $$ And the next step is to get the Lower triangular matirx, L. At first, I thought that L had to be $(P_{23}m_{1}P_{12})^{-1}$. Because A can be expressed as $A = (P_{23}m_{1}P_{12})^{-1}$ and we used these same method in A = LU decomposition to get Lower triangular matrix. But you added more processes and concluded that $L = (P_{23}m_{1}P_{23})^{-1}$. $$ \\ $$ I wonder why L must be $(P_{23}m_{1}P_{23})^{-1}$, not $(P_{23}m_{1}P_{12})^{-1}$.
10.28.17
Well, try it. Calculate $(P_{23}M_{1}P_{12})^{-1}$ and see if that is lower triangular.. If not, you answered your question.
Question by Student 201427152
Dear Professor, in last class, When you explained "why put the minus (-) next to RHS", You wrote the 3rd row is $$L[0][0]*V[0]+L[0][1]*V[1]+L[0][2]*V[2] = B[2]$$ So, you wrote $$L[0][2]*V[2] = B[2] - L[0][0]*V[0]+L[0][1]*V[1]$$ But, I think it is not 3rd row. because it is $B[0]$ and 3rd row is $$L[2][0]*V[0]+L[2][1]*V[1]+L[2][2]*V[2] = B[2]$$ Isn't there any wrong in your notation?
10.29.17
Is that when I wrote the code using the online C IDE? I may have made a mistake when explaining the logic on the blackboard. Of course, the third row should read $$ L[2][0]*V[0]+L[2][1]*V[1]+L[2][2]*V[2] = B[2] $$ Good observation. 1 point bonus.
Question by Student 201627131
Professor, I wonder about partical pivoting. In partical pivoting, I learned to interchange rows to put largest possible magnitude number within column on pivot. But, I think if interchanging row to put smallest absolute number on pivot, calculation process is more easy because some numbers can be eliminated by pivot multiply integer. Is there a reason to use largest number on pivot?
11.04.17
I can't understand what you mean. Why would putting the smallest number on the pivot result in less computing effort?
Question by Student 201327139
Professor, In Q.2, I found \( A_n \) , using Jacobian,

\( A_n \)=\( \begin{pmatrix} 4x_n^3& 1 \\ y_n & x_n+1.5y_n^\left( 0.5 \right) \end{pmatrix} \),

and A[0][1]=1. I was writing C code for this matrix and trying to make A[1][0]=0 (define double A[2][2],x,y and \( x_n \)=1.0, \( y_n \)=1.0),

\( A_0 \)=\( \begin{pmatrix} 4& 1 \\ 0 & 2.25 \end{pmatrix} \).

But, when I defined double A[1][1],

\( A_n \) matrix became \( A_n \)=\( \begin{pmatrix} 4& 0 \\ 0 & 2.5 \end{pmatrix} \).

Why A[0][1]=0 and A[1][1]=0 when I define double A[1][1]?
11.09.17
I don't understand... Why are you defining "double A[1][1]"? You should rather define A only once as "double A[N][N]".
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