solutions homework 2 - 18.06 Problem Set 2 Solution Total 100 points Section 2.5 Problem 24 Use Gauss-Jordan elimination on[U I to U 1 1 a b 1 0 0 1 c

solutions homework 2 - 18.06 Problem Set 2 Solution Total...

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18.06 Problem Set 2 Solution Total: 100 points Section 2.5. Problem 24: Use Gauss-Jordan elimination on [ U I ] to find the upper triangular 1 U 1 : 1 0 0 a b UU 1 x 1 x 2 x 3 = 0 0 0 1 0 0 1 (4 points): Row reduce [ U I ] to get [ I U 1 ] as follows (here R i stands for the i th row): 0 1 1 0 = I c . Solution 1 a b 1 0 0 ( R 1 = R 1 aR 2 ) = R 2 cR 2 ) 1 0 0 1 0 0 1 0 0 0 1 0 b ac a 0 1 c 0 1 0 ( R 2 1 0 −→ c 0 0 1 0 0 1 1 a ac b 1 ( R 1 ) ⎡ 1 0 0 1 = R 1 ( b ac ) R 3 0 1 0 0 1 0 0 1 0 0 −→ c . Section 2.5. Problem 40: (Recommended) A is a 4 by 4 matrix with 1’s on the diagonal and a, b, c on the diagonal above. Find A 1 for this bidiagonal matrix. Solution (12 points): Row reduce [ A I ] to get [ I A 1 ] as follows (here R i stands for the i th row): 1 a 0 0 1 0 0 0 0 1 b 0 0 1 0 0 0 0 1 c 0 0 1 0 0 0 0 1 0 0 0 1 ( R 1 = R 1 + aR 2 ) 1 0 ab 0 1 a 0 0 ( R 2 = R 2 + bR 2 ) 0 0 0 1 0 1 bc 0 1 b 0 −→ ( R 3 = R 3 + cR 4 ) 0 0 0 0 1 0 0 0 1 0 0 1 c ( R 1 = R 1 + abR 3 ) 1 0 0 0 1 a ab abc ( R 2 = R 2 + bcR 4 ) 0 1 0 0 0 1 b bc 0 0 1 0 0 0 1 c 0 0 0 1 0 0 0 1 −→ . Alternatively, write A = I N . Then N has a, b, c above the main diagonal, and all other entries equal to 0. Hence A 1 = ( I N ) 1 = I + N + N 2 + N 3 as N 4 = 0. Section 2.6. Problem 13: (Recommended) Compute L and U for the symmetric matrix a a a a a b b b A = . a b c c a b c d Find four conditions on a, b, c, d to get A = LU with four pivots.
̸ ̸ ̸ ̸ pset2-s10-soln: page 2 Solution (4 points): Elimination subtracts row 1 from rows 2–4, then row 2 from rows 3–4, and finally row 3 from row 4; the result is U. All the multipliers ij are equal to 1; so L is the lower triangular matrix with 1’s on the diagonal and below it. a a a a a a a a 0 0 b a b a b a b a b a b a A −→ −→ 0 0 0 0 0 b c d a d b a a a a 1 0 0 0 b a b a c b c b c a c a 0 c a 0 b a b a b a 1 1 0 0 = U, L = −→ . 0 c b c b 0 0 0 d c 1 1 1 1 The pivots are the nonzero entries on the diagonal of U . So there are four pivots when these four conditions are satisfied: a = 0, b = a , c = b , and d = c . Section 2.6. Problem 18: If A = LDU and also A = L 1 D 1 U 1 with all factors invertible, then L = L 1 and D = D 1 and U = U 1 . “The three factors are unique.” Derive the equation L 1 1 LD = D 1 U 1 U 1 . Are the two sides triangular or diagonal? Deduce L = L 1 and U = U 1 (they all have diagonal 1’s). Then D = D 1 .

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