MATH 2050 Assignment 6 Fall 2018 Due: Thursday, November 1. x + y + 2z = 2 x + y + z = c 4x + 2z = 2

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1 MATH 5 Assignment 6 Fall 8 Due: Thursday, November [5]. For what value of c does have a solution? Is it unique? x + y + z = x + y + z = c 4x + z = Writing the system as an augmented matrix, we have c R R+R R3 R3 4R 3 c + R3 R3+R 3 c (c + ) Thus, the system has a solution only when 6 + (c + ) = or c =, so that the last equation is consistent. The solution will not be unique, however, since there is a free variable in the row echelon form. []. Write all solutions of the following linear systems in vector form. (a) 3 4 R3 R3 R 3 x + x x 3 + 3x 4 = 4 x + x 4 = x + x x 4 = R3 R3+R 3 4 Thus, we see that x 4 = t is a free variable, giving x 3 = + t, x = t, and x = 4 x + x 3 3x 4 = + 3t. Writing this in vector form, we have x 3 x x 3 + t x 4 (b) x + y + 4z = 3 x + y + 6z = 5 x + 3y + 5z = 4

2 (c) R R R R3 R3 R R R3 This gives z =, y =, and x = 3 y 4z =, so x + 5y z = 3x + 5y 6z = 6 x 5y + z = R R 3R R3 R3+R x y z 5 So, both y and z are free variables. Taking y = s and z = t, we have x =. (d) s + t, giving R R+R R4 R4 R3 R5 R5 R4 x y z + s t x x x 3 x 4 x 5 R3 R3+R R5 5R

3 Solving, this gives x 5 = 5, x 4 = 4, x 3 = 3, x =, and x =, or x x x 3 x x 5 5 [5] 3. Find conditions on a, b, and c (if any) such that the system x + z = x y = y + z = 4 ax + by + cz = 3 has (i) no solution, (ii) a unique solution, and (iii) infinitely many solutions. Working from the augmented matrix of the system, we have R R R R4 R4 ar a b c 3 b c a 3 + a R R R3 R3+R R4 R4+bR This is row echelon form. 4 (c a) b (3 + a) + 4b R3 R 4 (c a) b (3 + a) + 4b (i) If c a b = and 3 + a + 4b, the third equation says = 3 + a + 4b, which is a contradiction. There is no solution in this case. (ii) If c a b, there is a unique solution: z = 3+a+4b, y = 4 z, x = z. c a b (iii) If c a+b = and 3+a+4b =, then the row echelon form is so z = t is a free variable, and there are infinitely many solutions. 4,. [] 4. Find all solutions of Does x + 3y = x + 4y + 5z = y + z = x + 3y = π x + 4y + 5z = 7/9 y + z = e

4 have a unique solution? First we put the system matrix in row echelon form R R R 5 R3 R3 R 5 Since this has only pivot columns, we introduce z = t as a free variable, giving y = 5t, and x = 5t. Thus, solutions are of the form x 5 y t 5 z For the non-homogeneous system, we know that if x, y, and z solve the given system, then so do x + 5t, y 5t, z + t for any value of t. Thus, it does not have a unique solution. Indeed, this system has NO solution when we use Gaussian-elimination to solve it. [5] 5. Show that the vectors u = are linearly independent. 3, v = 6 4, and w = 6. The vectors are linearly independent if and only if the only solution of the homogeneous system, A x =, with these vectors as columns of the matrix, A, is the trivial solution. So, we put A into row echelon form: R R 3R R3 R3+R R4 R4+R R R R4 R4 4R 4 Since the row echelon form of the matrix has 3 pivot columns, there are no free variables and, thus, the unique solution of A x = is x =. This means that the vectors that make up the columns of A are linearly independent. [5] 6. For what values of x (if any) are the vectors u = w = 3x 3 linearly independent? 3, v = x 3, and The vectors are linearly independent if and only if the only solution of the homogeneous system, A x =, with these vectors as columns of the matrix, A, is the

5 trivial solution. So, we put A into row echelon form: R R R R3 R3 3R 3 R3 R3 xr 3 x 3 3x 3 x 3x 6 3 3x + 3x 6 This system will have 3 pivot columns only when 3x + 3x 6 and, consequently, the vectors will be linearly independent when 3x + 3x 6. Solving 3x + 3x 6 = we have x = and x =. Thus, the vectors are linearly independent for all values of x except for x = and x =. [5]