Systems of Equations Homework Solutions

Transcription

1 Systems of Equations Homework Solutions Olena Bormashenko October 5, 2011 Find all solutions to the following systems of equations by writing the system as an augmented matrix and row-reducing it until it s in row-reduced echelon form. For each system, you should also: State which variables are dependent and which ones are independent State how many solutions the system has (sometimes the answer will be infinitely many) If there s at least one solution, provide an example and plug it into the original system to check that it works 1. Solve for [x 1, x 2 ], where x 1 + x 2 = 2 2x 1 + x 2 = R 2:R 2 2R R 2:R 2 ( 1) [ [ R 1:R 1 R x 1 = 2 x 2 = 4 Here, all the variables are dependent, and the system has the one solution [x 1, x 2 ] = [ 2, 4]. Plugging this solution back into the system of equations, ] ] so it works. x 1 + x 2 = = 2 2x 1 + x 2 = = 0 1

2 2. Solve for [a, b], where a + 2b = 2 a 2b = 2 3a b = R2:R2 R R3:R3 3R R 2:R 2/( 4) R 1:R 1 2R a = 0 b = 1 R3:R3+7R In this system, all the variables are dependent, and the only solution is [a, b] = [0, 1]. Plugging this back into the system, so it works. 3. Solve for [c, d, e], where a + 2b = = 2 a 2b = 0 2 = 2 3a b = 0 1 = 1 c + d = 0 d + e = 0 c + 2d + e = 0 2

3 R3:R3 R R 3:R 3 R 2 R 1:R 1 R 2 c e = 0 d + e = Here, the independent variable is e and the dependent variables are c and d. Expressing all the variables in terms of e, we get c = e d = e e = e Therefore, all solutions to the system are [c, d, e] = [e, e, e]. Clearly, there are infinitely many solutions. Letting e = 1, a particular solution is [1, 1, 1], and plugging this in: c + d = 1 + ( 1) = 0 d + e = ( 1) + 1 = 0 c + 2d + e = 1 + ( 2) + 1 = 0 which shows that the solution works. 4. Solve for [x 1, x 2, x 3 ], where x 1 x 3 = 2 3x 1 + x 2 + 5x 3 = 12 2x 2 + x 3 = 1 3

4 R2:R2 3R This is the system R 3:R 3+2R 2 R 3:R 3/17 R 1:R 1+R /17 R 2:R 2 8R 3 x 1 = x 2 = x 3 = / / / / /17 Here, all the variables are dependent, and there s exactly one solution. Plugging this back into the system, we get so the solution works. 5. Solve for [x 1, x 2, x 3 ], where x 1 x 3 = = = 2 3x 1 + x 2 + 5x 3 = = = 12 2x 2 + x 3 = = = 1 x 1 + x 2 x 3 = 2 3x 1 + 2x 2 + 4x 3 = 1 4

5 R 2:R 2 3R R 2:R 2 ( 1) [ [ R 1:R 1 R x 1 + 6x 3 = 3 x 2 7x 3 = 5 Here, x 3 is independent, while x 1 and x 2 are dependent. Expressing all the variables in terms of x 3, we get x 1 = 3 6x 3 x 2 = 5 + 7x 3 x 3 = x 3 Therefore, all solutions to the system are of the form [x 1, x 2, x 3 ] = [ 3 6x 3, 5 + 7x 3, x 3 ], and there are infinitely many solutions. Letting x 3 = 0, a particular solution is [ 3, 5, 0], and plugging it back into the system we get which works. 6. Solve for [x 1, x 2, x 3 ], where x 1 + x 2 x 3 = = 2 3x 1 + 2x 2 + 4x 3 = = 1 2x 1 + x 2 + 2x 3 = 2 x 1 + x 2 + x 3 = 5 3x 1 x 2 x 3 = 1 6x 1 + x 2 + 2x 3 = 0 ] ] 5

6 Putting this system in augmented matrix form and row-reducing (here, I will be doing more than one row operation per step to save space the operation that s written first is done first, so if it changes the rows, the operation after it uses the new rows): R 1:R 1 2R 2,R 3:R 3 3R 2 R 4:R 4 6R 2, Swap R 1, R 2 R 2:R 2 ( 1),R 1:R 1 R 2 R 3:R 3+4R 2,R 4:R 4+5R 2 R 3:R 3/( 4),R 1:R 1 R 3 R 4:R 4+4R / / / / Since the last equation corresponds to 0 = 16, this system has no solutions. 7. Solve for [x 1, x 2, x 3, x 4 ], where 2x 1 3x 2 + 2x 3 13x 4 = 0 4x 1 7x 2 + 4x 3 29x 4 = 0 x 1 + 2x 2 x 3 + 8x 4 = 0 Putting this system in augmented matrix form and row-reducing, with the above convention: if more than one row operation is written down, the 6

7 operation that s written first is done first, so if it changes the rows, the operation after it uses the new rows R2:R2+4R3,R1:R1 2R : Swap R 1, R 3 R 1:R 1 2R 2,R 3:R 3+7R 2 R 3:R 3/4,R 1:R 1+R 3 x 1 = 0 x 2 + 3x 4 = 0 x 3 2x 4 = 0 Here, x 4 is indepedent, while x 1, x 2, x 3 are dependent. variables in terms of x 4, we get x 1 = 0 x 2 = 3x 4 x 3 = 2x 4 x 4 = x Expressing the Therefore, all solutions to the system are of the form [x 1, x 2, x 3, x 4 ] = [0, 3x 4, 2x 4, x 4 ], and there are infinitely many solutions. Plugging in x 4 = 1 gets the particular solution [0, 3, 2, 1], and checking whether it works, 2x 1 3x 2 + 2x 3 13x 4 = = 0 4x 1 7x 2 + 4x 3 29x 4 = = 0 x 1 + 2x 2 x 3 + 8x 4 = = 0 we see that it does indeed solve the system. 8. Solve for [c 1, c 2, c 3 ], where c 1 + c 2 = 2 c 1 c 2 = 2 WARNING: Here, we have three variables, not 2, appearances notwithstanding!! 7

8 Putting this system in augmented matrix form and row-reducing (note the 3rd column which consists of all 0s!): R 2:R 2 R R 2:R 2/( 2) R 1:R 1 R c 1 = 0 c 2 = 2 Here, the independent variable is c 3 (since its column is non-pivotal) and the dependent variables are c 1 and c 2. Expressing all the variables in terms of c 3, we get c 1 = 0 c 2 = 2 c 3 = c 3 Therefore, all solutions to the system are [c 1, c 2, c 3 ] = [0, 2, c 3 ]. Clearly, there are infinitely many solutions. Letting c 3 = 1, a particular solution is [0, 2, 1], and plugging this in: c 1 + c 2 = = 2 This shows that the solution works. 9. Solve for [c 1, c 2, c 3 ], where c 1 c 2 = 0 2 = 2 c 1 + c 2 + c 3 = 2 c 1 + 2c 2 2c 3 = 2 c 1 + 4c 3 = 5 8

9 R3:R3 R1,R2:R2 R R 3:R 3+R 2,R 1:R 1 R The last equation of this system corresponds to 0 = 1, which shows that the system has no solutions. 9