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Unformatted text preview: 18.06 Problem Set 2 Solution Due Wednesday, 18 February 2009 at 4 pm in 2106. Total: 155 points. Problem 1: What threee elimination matrices E 21 , E 31 , and E 32 put A into upper triangular form E 32 E 31 E 21 A = U ? Using these, compute the matrix L (and U ) to factor A = LU . A = 1 1 1 2 4 5 0 4 0 Solution (10 points) The Gaussian elimination process is as follows: 1 1 1 2 4 5 0 4 0 ; 1 1 1 0 2 3 0 4 0 ; 1 1 1 0 2 3 0 0 6 . As we can see from this process, the matrix E 21 corresponds to subtracting twice of the first row from the second row; the matrix E 31 is trivial; the matrix E 32 corresponds to subtracting twice of the second row from the third row. In other words, E 21 = 1 0 0 2 1 0 0 1 , E 31 = 1 0 0 0 1 0 0 0 1 , E 32 = 1 1 2 1 . Hence, L = E 1 21 E 1 31 E 1 32 = 1 0 0 2 1 0 0 0 1 1 0 0 0 1 0 0 2 1 = 1 0 0 2 1 0 0 2 1 ; U = 1 1 1 0 2 3 0 0 6 . Problem 2: Suppose we have a 3 × 3 lowertriangular L matrix of the form L = 1 ‘ 21 1 ‘ 31 ‘ 32 1 . 1 (a) When you do the usual Gaussianelimination steps on L , what matrix do you get? (b) When you do the same elimination steps to I , what matrix do you get? (Hint: you can write the answer in terms of L very simply.) (c) When you apply the same steps to a matrix A = LU , what matrix do you get (write your answer in terms of L , U , and/or A ). (It is possible to answer this question without doing any calculations.) Solution (15 points = 5+5+5) (a) We get the identity matrix if we apply the usual Gaussian elimination, be cause Gaussian elimination puts zeros below the pivots while leaving the pivots (= 1 here) unchanged.. (b) In part (a), we said that doing Gaussian elimination to L gives I —that is, EL = I where E is the product of the elimination matrices (multiplying on the left since these are row operations). But EL = I means that E = L 1 . Hence, doing the same elimination steps to I gives EI = E = L 1 = 1 l 21 1 l 21 l 32 l 31 l 32 1 . [Note to grader: the student need not compute L 1 explicitly as was done here. (c) If we do the same elimination steps to A = LU , this corresponds to multiply ing A on the left by the elimination matrices E , so we get EA = ELU = ( EL ) U = U , using the fact that EL = I from (a). Equivalently, from (b), E = L 1 so we get L 1 A = L 1 LU = U . REMARK: More generally, we applying the same elimination steps to a matrix A , we will get EA = L 1 A . Problem 3: Without computing A or A 1 or A 2 or A 2 explicitly, compute A 1 x + A 2 y , where you are given the following LU factorization A = LU : L = 1 0 0 1 1 0 0 1 1 , U = 1 0 1 0 1 1 0 0 1 , x = 1 1 , y =  1 1 1 ....
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 Spring '08
 Strang
 Matrices, Invertible matrix, P AP

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