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Problem 2 (a) Find the LU factorization of the following matrix, then verify your answer by computing LU -1 4 5 a) 6 2 -4 1 -21 (b) Find the determinants of the following matrices. Show all your calculations and steps: [-1 4 51 a)6 2 -4b) 0 6 8 2 -4 3 3 2 6 8 10

math   Statistics-And-Probability   
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Answer #1

A)

Solution: Lu decomposition: If we have a matrix A, then an upper triangular matrix U can be obtained without pivoting under Gaussian Elimination method, and there exists lower triangular matrix L such that A=LU. 1 4 Here A6 2 -4 Using Gaussian Elimination method 1 4 5 0 26 26 1 -2-1 R3 - R3- (-1) xR1 | L3,1-1 1 45 - 0 26 26 R3 - K313 1 4 5 -0 26 261 4 5 U0 26 26 L is just made up of the multipliers we used in Gaussian elimination with 1s on the diagonal -6 1 0 -1 13 . LU decomposition for A is 1 0 0 1 4 5 1 4 5 xl 0 26 26 |=LU 13

B)Your matrix Sign 114 6 +26 2 4 3 1 -2 1 Eliminate elements in the 1st column under the 1st element Sign | | A1 | A2 | A3 1 1 4 6 +20 26 32 Eliminate elements in the 2nd column under the 2nd elementSign | | Al | A2 | A3 + 2 0 26 32 3 00 33/13 Multiply the main diagonal elements Sign | | A1 | A2 | A3 1 1 4 + 20 26 32 3 0 0 33/13 Δ =-1 x 26 x 33/13-_66

2nd)Your matrix 1 3 1 5 1 2 4 0 6 8 3 24 3 3 4 26 8 10 Eliminate elements in the 1st column under the 1st element Sign | | A1 | A2 | A3 | A4 1 31 2 | 0 | 4/3 1-2/3 | -20/3 3 0 -10/3 -1/3 11/3 4 0 20/3 14/3 32/3 -1Eliminate elements in the 2nd column under the 2nd element Sign 1 31 2 0 4/3 -2/3 -20/3 3 0 02 -13 Eliminate elements in the 3rd column under the 3rd element Sign 1 31 5 0 | 4/3 1-2/3 | -20/3 3 0 0-2 -13 4 0Multiply the main diagonal elements Sign 4 1 3 1 -1 0 | 4/3 1-2/3 | _20/3 3 0 0 213 4 00 -8 -3 x 4/3 x (-2) x (-8)-64

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