Homework #1: Solutions to select problems

Transcription

1 Homework #: Solutions to select problems Linear Algebra Summer 20 Math S200X (3) Corrin Clarkson June 6, 20 Section Exercise (42) Find a system of linear equations with three unknowns whose solutions are the points on the line through (,,) and (3,5,0) Solution The line through (,,) and (3,5,0) can be parametrized bu the following three equations: x = 2t + y = 4t + z = t + Solving the last equation for t gives t = z + We can then plug this in for t in the first two equations to get x = 2( z + ) + y = 4( z + ) + Finally, we put this linear system in standard form x + 2z = 3 y + 4z = 5 We check this solution by first noting the both of the given points satisfy these equations Then we check that the two equations describe two different planes in R 3 This is is shown by finding a point that satisfies one of the

2 equations, but not the other; (, 0, ) is such a pointthus we have two distinct plains intersecting in R 3 so we know their intersection is a line Because (,,) and (3,5,0) are on both planes, the intersection line must be the line through the given points Section 2 Exercise (48) For an arbitrary positive integer n 3, find all solutions x, x 2,, x n of the simultaneous equations x 2 = (x 2 + x 3 ) x 3 = (x x 4 ) x n = (x 2 n 2 + x n ) Solution We begin by considering the case when n = 3 This will not give a complete solution, but will provide insight into the general case When n = 3, there is only one equation in the linear system x 2 = 2 (x + x 3 ) In standard form, this equations becomes and solving for x gives The parametrized solution is then 2 x x x 3 = 0 x = 2x 2 x 3 x = 2t s x 2 = t x 3 = s The solution has one leading variable and two free variables Before solving the system for a general n, we will consider one more special case: n = 4 In this case the linear system consists of the two equations x 2 = 2 (x + x 3 ) and x 3 = 2 (x 2 + x 4 ) 2

3 In standard form, these equations become 2 x x x 3 = 0 and 2 x 2 x x 4 = 0 From these equations, we can determine the augmented matrix of the linear system [ ] To row reduce this matrix, multiply both rows by 2 and then add twice the second row to the first row The reduce row echelon form is then [ ] This gives the following parametrized solution: x = 3t + 2s x 2 = 2t + s x 3 = t x 4 = s Now we will consider the system for a general n The basic strategy will be the same: find the augmented matrix of the system, row reduce it and then find the parametrized solution First, we put the linear equations in standard form 2 x 2 + x 2 3 = x 3 + x 2 4 = 0 2 n 2 x n + x 2 n = 0 The augmented matrix of this system is then ()

4 We begin row reducing this matrix by multiplying all the rows by 2 There is a leading one in each row and in the first n 2 columns We begin to clear the columns with leading ones by adding twice the second row to the first row At this point the first two columns are done; they each contain a single leading one and no other nonzero entries The third column can be cleared by adding the thrice the third row to the first row and twice the third row to the second row Continuing to apply row operations in this manner will put the matrix in reduced row echelon form The complete sequence of row operations can be summarized as follows: row = 2(r + 2r 2 + 3r (n 2)r n 2 ) row 2 = 2(r 2 + 2r 3 + 3r (n 3)r n 2 ) row n 3 = 2(r n 3 + 2r n 2 ) row n 2 = 2(r n 2 ) Here r i indicates the i th row We can check that these row operations have the desired affect on the middle columns by considering their affect on three consecutive rows m [ ] (m + ) [ ] + (m + 2) [ ] [ m m + 0 m 3 m + 2 ] 4

5 The resulting row reduced matrix is then n n n n (2) The following parametrized solution describes all possible solutions to the general linear system x = (n )t + (2 n)s x 2 = nt + ( n)s x n 3 = 3t 2s x n 2 = 2t s x n = t x n = s Exercise (56) Find the conics through the points (0,0), (,0), (0, ) and (, -) Draw a rough sketch of your solutions curve(s) Solution We are given that a conic is described by an equation of the form c + c 2 x + c 3 y + c 4 x 2 + c 5 xy + c 6 y 2 = 0 Plugging each of the points into this equation gives the linear system c = 0 c + c 2 + c 4 = 0 c + c 3 + c 6 = 0 c + c 2 c 3 + c 4 c 5 + c 6 = 0 The augmented matrix of this linear system is

6 We begin to row reduce this matrix by subtracting the first row from each of the rows below it Next, we subtract the second from the forth row and add the third row to the forth row Finally, we multiply the fourth row by to achieve reduced row echelon form The parameterized solution of the linear system is c = 0 c 2 = t c 3 = s c 4 = t c 5 = 2s c 6 = s Thus the conics that pass through the given points are described by equations of the form tx sy + tx 2 + 2sxy + sy 2 = 0 Figure is a graph of a few of these conics 6

7 Figure : Conics through (0,0), (,0), (0, ) and (, -) 7