There's some problems in your question formulation: you need to make sure the math is surrounded by \$signs. Please post again below with correct typesetting.  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}$. then C2 = 2 This is a very good explanation. There is only a small problem with it: you should have written$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?  10.28.17 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$(P_{23}m_{1}P_{12})^{-1}$. Well, try it. Calculate$(P_{23}M_{1}P_{12})^{-1}$and see if that is lower triangular.. If not, you answered your question.  Previous 1 ... 8 , 9 , 10 ... 14 Next • PDF 1✕1 2✕1 2✕2 $\pi\$