This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*
**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