{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam1soln

# exam1soln - 18.06 Quiz 1 Solutions October 5 2009 1 2 3...

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

18.06 Quiz 1 Solutions October 5, 2009 Problem 1. Consider the matrix A = 1 2 3 1 1 1 0 1 3 . (a) Find the factorization A = LU . (b) Find the inverse of A . (c) For which values of c is the matrix 1 2 3 1 1 1 0 1 c invertible? Solution (a) We row reduce A by subtracting row 1 from row 2 ( E 12 ) and then add row 2 to row 3 ( E 23 ) to ﬁnd the upper triangular matrix U = 1 2 3 0 - 1 - 2 0 0 1 . Since we can reverse this process and subtract row 2 from row 3 in U , followed by adding row 1 to row 2 to obtain A , we see that the lower triangular matrix is the product: 1 0 0 1 1 0 0 0 1 1 0 0 0 1 0 0 - 1 1 = 1 0 0 1 1 0 0 - 1 1 . Hence we ﬁnd A = 1 0 0 1 1 0 0 - 1 1 1 2 3 0 - 1 - 2 0 0 1 .

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

View Full Document
A - 1 = ( LU ) - 1 = U - 1 L - 1 . We explicitly compute U - 1 and ﬁnd: 1 2 3 | 1 0 0 0 - 1 - 2 | 0 1 0 0 0 1 | 0 0 1 1 2 0 | 1 0 - 3 0 - 1 0 | 0 1 2 0 0 1 | 0 0 1 1 0 0 | 1 2 1 0 1 0 | 0 - 1 - 2 0 0 1 | 0 0 1 . Similary, we compute L - 1 = 1 0 0 - 1 1 0 - 1 1 1 . Therefore A - 1 = 1 2 1 0 - 1 - 2 0 0 1 1 0 0 - 1 1 0 - 1 1 1 = - 2 3 1 3 - 3 - 2 - 1 1 1 . (c) We row reduce to ﬁnd : 1 2 3 1 1 1 0 1 c 1 2 3 0 - 1 - 2 0 0 c - 2 . Note that A is invertible if and only if it has 3 nonzero pivots. Thus A is invertible when c 6 = 2. Problem 2.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

exam1soln - 18.06 Quiz 1 Solutions October 5 2009 1 2 3...

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

View Full Document
Ask a homework question - tutors are online