06win-m1sols

# 06win-m1sols - MATH 51 MIDTERM 1 SOLUTIONS February 2 2006...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MATH 51 MIDTERM 1 SOLUTIONS February 2, 2006 1. Find all solutions of the following system: x 1 + 2 x 2 + x 3 + x 4 = 7 x 1 + 2 x 2 + 2 x 3 +- x 4 = 12 2 x 1 + 4 x 2 + 6 x 4 = 4 Solution: 1 2 1 1 | 7 1 2 2- 1 | 12 2 4 6 | 4 → 1 2 1 1 | 7 1- 2 | 5- 2 4 |- 10 → 1 2 3 | 2 1- 2 | 5 | . The last augmented matrix corresponds to the system x 1 + 2 x 2 + 3 x 4 = 2 x 3- 2 x 4 = 5 = . So: x 1 = 2- 2 x 2- 3 x 4 x 3 = 5 + 2 x 4 (where x 2 and x 4 can take any values), or x 1 x 2 x 3 x 4 = 2 5 + - 2 1 x 2 + - 3 2 1 x 4 2(a). What three conditions must a set V of vectors in R n have in order to be a linear subspace? Solution: It must contain the zero vector, it must be closed under addition, and it must be closed under scalar multiplication. 2(b,c) . Determine which of those three properties each of the following sets has: 1 2(b) . The set V of vectors x in R n such that each coordinate of x is an integer. (The integers are the numbers 0, 1,- 1, 2,- 2, etc.) Solution: Contains , closed under addition, but not closed under scalar multipli- cation. (For example, e 1 ∈ V , but (1 / 2) e 1 / ∈ V .) 2(c) . The set W of vectors v = x y z in R 3 such that xyz = 0. Solution: It contains the zero vector. Also, it is closed under scalar multiplication. But it is not closed under addition. (For example, 1 1 and 1 are both in W , but 1 1 + 1 = 1 1 1 is not in W .) 3. Consider the system: 1 2 3 3 7 10 2 5 7 1 3 4...
View Full Document

{[ snackBarMessage ]}

### Page1 / 6

06win-m1sols - MATH 51 MIDTERM 1 SOLUTIONS February 2 2006...

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

View Full Document
Ask a homework question - tutors are online