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Unformatted text preview: MATH 245 Linear Algebra 2, Solutions to Assignment 1 1: Let p = 2 1 2 1 , u 1 = 1 2 1 3 , u 2 = 3 4 1 5 , q = 1 2 3 , v 1 = 1 1 4 2 and v 2 = 1 2 1 1 . Find the point of intersection of the plane x = p + t 1 u 1 + t 2 u 2 and the plane x = q + s 1 v 1 + s 2 v 2 . Solution: We need to try to find values of t 1 , t 2 , s 1 and s 2 such that p + t 1 u 1 + t 2 u 2 = q + s 1 v 1 + s 2 v 2 , that is, t 1 u 1 + t 2 u 2 s 1 v 1 s 2 v 2 = q p . Equivalently, we can write this as the matrix equation Ax = q p where A is the matrix with columns u 1 ,u 2 ,v 1 ,v 2 , and x is the vector with entries t 1 ,t 2 , s 1 , s 2 . We reduce the augmented matrix ( A q p ) . 1 3 1 1 2 4 1 2 1 1 4 1 3 5 2 1 1 1 5 1 ∼ 1 3 1 1 0 2 1 4 0 4 5 2 0 4 1 4 1 1 6 4 ∼ 1 3 1 1 0 1 1 2 2 0 4 5 2 0 4 1 4 1 3 2 6 4 ∼ 1 0 1 2 5 0 1 1 2 2 0 0 3 6 0 0 1 4 7 2 3 2 2 ∼ 1 0 1 2 5 0 1 1 2 2 0 0 1 2 0 0 1 4 7 2 3 2 2 ∼ 1 0 0 6 0 1 0 3 0 0 1 2 0 0 0 6 7 2 3 2 2 ∼ 1 0 0 6 0 1 0 3 0 0 1 2 0 0 0 1 7 2 3 2 1 3 ∼ 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 3 2 1 2 2 3 1 3 The solution to the equation Ax = q p is given by t 1 = 3 2 , t 2 = 1 2 , s 1 = 2 3 and s 2 = 1 3 . Thus the point of intersection is the point p + t 1 u 1 + t 2 u 2 = 2 1 2 1 + 3 2 1 2 1 3  1 2 3 4 1 5 = 2 2 1 . (You can check that this point is also equal to q + s 1 v 1 + s 2 v 2 ). 2: Let A = 2 1 4 3 2 5 3 2 5 4 3 7 1 2 1 3 4 2 2 3 5 6 1 , R = 1 0 3 1 5 0 1 2 0 1 2 0 0 1 1 1 0 0 and P = 1 2 2 2 1 1 1 1 1 1 1 1 4 3 . Given that ( A y ) is row equivalent to ( R Py ) , do the following. (a) Find a basis U for Null(A). Solution: Solving Ax = 0 as usual, by solving the equivalent equation Rx = 0, yields U = { u 1 ,u 2 ,u 3 } , where u 1 =  3 2 1 , u 2 = 1 1 1 1 , u 3 =  5 2 1 1 ....
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This note was uploaded on 12/08/2011 for the course MATH 245 taught by Professor New during the Fall '09 term at Waterloo.
 Fall '09
 NEW
 Linear Algebra, Algebra

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