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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...
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 '05
 HUI
 Math, Linear Algebra, Algebra, Vector Space, Linear combination, ﬁrst equation

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