Midterm_solutions

# Midterm_solutions - MAE 107 Computational Methods Summer...

This preview shows pages 1–3. Sign up to view the full content.

MAE 107 – Computational Methods Summer Session II 2009 Midterm – Solutions Wednesday, August 19 4 problems + 5 short questions – 40 points Problem 1 – 10 points Consider the matrix A = 3 2 0 - 3 - 3 2 6 4 4 . 1. The ﬁrst step of the Forward Elimination procedure aims at canceling the coeﬃcients below the pivot coeﬃcient a 11 = 3 in the ﬁrst column of A . To do so, we deﬁne λ 21 = a 21 a 11 = - 1 and λ 31 = a 31 a 11 = 2. The following row operations are then performed: - (2) (2) - λ 21 (1) (2) + (1) - (3) (3) - λ 31 (1) (3) - 2 × (1) The result of this operation can be written as a decomposition for A : A = 1 0 0 - 1 1 0 2 0 1 3 2 0 0 - 1 2 0 0 4 We observe that the ﬁrst step cancelled all the subdiagonal coeﬃcients and therefore no second step is necessary to remove the subdiagonal coeﬃcient of the second column. The λ ij are the subdiagonal coeﬃcients of L . The LU decomposition of A is therefore simply: A = L · U , with L = 1 0 0 - 1 1 0 2 0 1 and U = 3 2 0 0 - 1 2 0 0 4 . 2. Using the fundamental property of the determinant | A | = | LU | = | L |·| U | . L and U are triangular matrices: their determinant is simply the product of their diagonal coeﬃcients. | L | = 1 and | U | = - 12, therefore | A | = - 12 . Using the deﬁnition of the determinant and expanding over the ﬁrst row of A (since it has a zero coeﬃcient): | A | =3 × ± ± ± ± - 3 2 4 4 ± ± ± ± - 2 × ± ± ± ± - 3 2 6 4 ± ± ± ± + 0 × ± ± ± ± - 3 - 3 6 4 ± ± ± ± =3( - 12 - 8) - 2( - 12 - 12) = - 60 + 48 = - 12 | A | 6 = 0 therefore the square system Ax = b has a unique solution. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
3. To compute Y = A - 1 we need to solve 3 linear systems for the column vectors y (1) , y (2) and y (3) , which are the columns of Y : Ay (1) = 1 0 0 , Ay (2) = 0 1 0 , Ay (3) = 0 0 1 . Using the LU decomposition of A we can replace these three systems by three pairs of triangular systems: ± Lz ( k ) = b Uy ( k ) = z ( k ) for 1 k 3 . For each k = 1, 2 and 3, we can solve successively Lz ( k ) = b for z ( k ) using Forward Substitution and Uy ( k ) = z ( k ) using Backward Substitution. -
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 10/11/2011 for the course MAE 107 taught by Professor Rottman during the Summer '08 term at UCSD.

### Page1 / 6

Midterm_solutions - MAE 107 Computational Methods Summer...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online