Praktikum Metode Komputasi (Vector Spaces)

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1 Praktikum Metode Komputasi (Vector Spaces) Vina Apriliani January 17, 016 Soal Latihan MATLAB Bab 3 Buku Leon Aljabar Linear Berikut 1 Soal Latihan MATLAB Bab 3 Buku Leon Aljabar Linear yang saya ambil untuk ditulis dengan LaTeX dan harus dikerjakan ( digit terakhir NRP modulo jumlah soal di Latihan) yaitu Soal Nomor (): Rank-Deficient Matrices In this exercise we consider how to use MATLAB to generate matrices with specified ranks. a In general, if A is an m x n matrix with rank r, then r <= min(m,n). Why? Explain. If the entries of A are random numbers, we would expect that r = min(m,n). Why? Explain. Check this out by generating random 6 x 6, 8 x 6, and 5 x 8 matrices and using the MATLAB command rank to compute their ranks. Whenever the rank of an m x n matrix equals min(m,n), we say that the matrix has full rank. Otherwise, we say that the matrix is rank deficient. Jika A adalah matriks m x n dengan rank r, maka r <= min(m,n) karena rank r adalah dimensi dari ruang baris sehingga rank r <= m, sedangkan dimensi ruang baris = dimensi ruang kolom sehingga rank r <= min(m,n). (Teorema) Jika entri dari matriks A adalah bilangan acak, kita dapat menduga bahwa r=min(m,n) karena antar baris tidak pernah merupakan kelipatan satu dengan yang lain. Generating random 6 x 6, 8 x 6, 5 x 8 matrices and compute their ranks using MATLAB command: A = rand(6,6) A =

2 rank(a) = 6 B = rand(8,6) B = rank(b) = 6 C = rand(5,8) C = rank(c) = 5 b MATLAB s rand and round commands can be used to generate random m x n matrices with integer entries in a given range [a,b]. This can be done with a command of the form For example, the command A = round((b-a)*rand(m,n))+a A = round(4*rand(6,8))+3 will generate a 6 x 8 matrix whose entries are random integers in the range from 3 to 7. Using the range [1,10], create random integer 10 x 7, 8 x 1, and 10 x 15 matrices and in each case check the rank of the matrix. Do these integer matrices all have full rank? Random integer 10 x 7 matrix in the range [1,10]: A1 = round(9*rand(10,7))+1

3 A1 = rank(a1) = 7 This integer matrix has full rank because r = min(10,7) = 7. Random integer 8 x 1 matrix in the range [1,10]: A = round(9*rand(8,1)) A = rank(a) = 8 This integer matrix has full rank because r = min(8,1) = 8. Random integer 10 x 15 matrix in the range [1,10]: A3 = round(9*rand(10,15))+1 3

4 A3 = rank(a3) = 10 This integer matrix has full rank because r = min(10,15) = 10. c Suppose that we want to use MATLAB to generate matrices with less than full rank. It is easy to generate matrices of rank 1. If x and y are nonzero vectors in R m and R n, respectively, then A = xy T will be an m x n matrix with rank 1. Why? Explain. Verify this in MATLAB by setting x = round(9*rand(8,1))+1, y = round(9*rand(6,1))+1 and using these vectors to construct an 8 x 6 matrix A. Check the rank of A with the MATLAB command rank. Jika x dan y adalah vektor-vektor taknol masing-masing di R m dan R n, maka A = xy T akan menjadi matriks m x n dengan rank 1 karena setiap baris pada matriks A (baris ke- sampai baris ke-m) merupakan kelipatan baris pertama, dapat dilihat dari bentuk eselon baris tereduksi dari matriks A. Construct an 8 x 6 matrix A: x = round(9*rand(8,1))+1 4

5 5 8 x = y = round(9*rand(6,1))+1 5 y = A = x y = d In general, rank(a) = 1 rank(ab) <= min(rank(a),rank(b)) (1) (See Exercise 8 in Section 6.) If A and B are noninteger random matrices, the relation (1) should be an equality. Generate an 8 x 6 matrix A by setting X = rand(8,), Y = rand(6,), A = X*Y What would you expect the rank of A to be? Explain. Test the rank of A with MATLAB. X = rand(8,) 5

6 X = Y = rand(6,) Y = A = X Y = Menurut dugaan saya, rank(a) = karena matriks X dan matriks Y adalah matriks yang berisi bilangan acak takbulat sehingga rank(x)= dan rank(y)=rank(y )=. Oleh karena itu, persamaan (1) berlaku sehingga rank(a)=rank(x*y )=min(rank(x),rank(y ))=min(,)=. Test the rank of A with MATLAB: rank(a) = rank(x) = rank(y) = rank(y ) = e Use MATLAB to generate matrices A, B, and C such that (i) A is 8 x 8 with rank 3. (ii) B is 6 x 9 with rank 4. (iii) C is 10 x 7 with rank 5. 6

7 A = rand(8,3)*(rand(8,3)) A = rank(a) = 3 B = rand(6,4)*(rand(9,4)) B = rank(b) = 4 C = rand(10,5)*(rand(7,5)) C = rank(c) = 5 Berikut Modifikasi Soal Nomor () menjadi soal untuk dikerjakan dalam Scilab: Rank-Deficient Matrices In this exercise we consider how to use SCILAB to generate matrices with specified ranks. a In general, if A is an m x n matrix with rank r, then r <= min(m,n). Why? Explain. If the entries of A are random numbers, we would expect that r = min(m,n). Why? Explain. Check this out by generating random 7

8 6 x 6, 8 x 6, and 5 x 8 matrices and using the SCILAB command rank to compute their ranks. Whenever the rank of an m x n matrix equals min(m,n), we say that the matrix has full rank. Otherwise, we say that the matrix is rank deficient. Jika A adalah matriks m x n dengan rank r, maka r <= min(m,n) karena rank r adalah dimensi dari ruang baris sehingga rank r <= m, sedangkan dimensi ruang baris = dimensi ruang kolom sehingga rank r <= min(m,n). (Teorema) Jika entri dari matriks A adalah bilangan acak, kita dapat menduga bahwa r=min(m,n) karena antar baris tidak pernah merupakan kelipatan satu dengan yang lain. Generating random 6 x 6, 8 x 6, 5 x 8 matrices and compute their ranks using SCILAB command: A = rand(6,6) A = rank(a) = 6 B = rand(8,6) B = rank(b) = 6 C = rand(5,8) C = rank(c) = 5 8

9 b SCILAB s rand and round commands can be used to generate random m x n matrices with integer entries in a given range [a,b]. This can be done with a command of the form For example, the command A = round((b-a)*rand(m,n))+a A = round(4*rand(6,8))+3 will generate a 6 x 8 matrix whose entries are random integers in the range from 3 to 7. Using the range [1,10], create random integer 10 x 7, 8 x 1, and 10 x 15 matrices and in each case check the rank of the matrix. Do these integer matrices all have full rank? Random integer 10 x 7 matrix in the range [1,10]: A1 = round(9*rand(10,7)) A1 = rank(a1) = 7 This integer matrix has full rank because r = min(10,7) = 7. Random integer 8 x 1 matrix in the range [1,10]: A = round(9*rand(8,1))+1 9

10 A = rank(a) = 8 This integer matrix has full rank because r = min(8,1) = 8. Random integer 10 x 15 matrix in the range [1,10]: A3 = round(9*rand(10,15)) A3 = rank(a3) = 10 This integer matrix has full rank because r = min(10,15) =

11 c Suppose that we want to use SCILAB to generate matrices with less than full rank. It is easy to generate matrices of rank 1. If x and y are nonzero vectors in R m and R n, respectively, then A = xy T will be an m x n matrix with rank 1. Why? Explain. Verify this in SCILAB by setting x = round(9*rand(8,1))+1, y = round(9*rand(6,1))+1 and using these vectors to construct an 8 x 6 matrix A. Check the rank of A with the SCILAB command rank. Jika x dan y adalah vektor-vektor taknol masing-masing di R m dan R n, maka A = xy T akan menjadi matriks m x n dengan rank 1 karena setiap baris pada matriks A (baris ke- sampai baris ke-m) merupakan kelipatan baris pertama, dapat dilihat dari bentuk eselon baris tereduksi dari matriks A. Construct an 8 x 6 matrix A: x = round(9*rand(8,1)) x = y = round(9*rand(6,1))+1 5 y = A = x y = rank(a) = 1 11

12 d In general, rank(ab) <= min(rank(a),rank(b)) (1) (See Exercise 8 in Section 6.) If A and B are noninteger random matrices, the relation (1) should be an equality. Generate an 8 x 6 matrix A by setting X = rand(8,), Y = rand(6,), A = X*Y What would you expect the rank of A to be? Explain. Test the rank of A with SCILAB. X = rand(8,) X = Y = rand(6,) Y = A = X Y = Menurut dugaan saya, rank(a) = karena matriks X dan matriks Y adalah matriks yang berisi bilangan acak takbulat sehingga rank(x)= dan rank(y)=rank(y )=. Oleh karena itu, persamaan (1) berlaku sehingga rank(a)=rank(x*y )=min(rank(x),rank(y ))=min(,)=. Test the rank of A with SCILAB: 1

13 rank(a) = rank(x) = rank(y) = rank(y ) = e Use SCILAB to generate matrices A, B, and C such that (i) A is 8 x 8 with rank 3. (ii) B is 6 x 9 with rank 4. (iii) C is 10 x 7 with rank 5. A = rand(8,3)*(rand(8,3)) A = rank(a) = 3 B = rand(6,4)*(rand(9,4)) B = rank(b) = 4 C = rand(10,5)*(rand(7,5)) C = rank(c) = 5 13