2.2 - Math 2940 Solutions Fall 2011 Section 2 2 9 We need b...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

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

Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 2 . 2 9 ) We need b 1 b 2 to be a linear combination of the columns 3 6 and- 2- 4 . Both columns are multiples of 1 2 so any linear combination is such a multiple. So we must have b 2 = 2 b 1 . When there is one solution, there will be infinitely many. Just add any multiple of 2- 3 to any solution. 11 ) (a) If ( x, y, z ) and ( X, Y, Z ) are two solution, so is their average, (( x + X ) / 2 , ( y + Y ) / 2 , ( z + Z ) / 2). In fact, for any real number t ( tx + (1- t ) X, ty + (1- t ) Y, tz + (1- t ) Z ) is a solution. The average used t = 1 / 2. (b) They meet the at entire line containing the two points. 13 ) Adding- 2 time the first equation to the second and- 1 time the first equation to the third gives 2 x- 3 y = 3, y + z = 1, 2 y- 3 z = 2. Adding- 2 time the new second equation to the new third gives- 5 z = 0. So z = 0 y = 1 and x = 3. 21 ) Adding- 1 / 2 times the first equation to the second eliminates x to give 2 x + y = 0 , (3 / 2) y + z = 0...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

2.2 - Math 2940 Solutions Fall 2011 Section 2 2 9 We need b...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online