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MATH 107.01 HOMEWORK #1 SOLUTIONS Problem 1.1.2. Solve the system 2x + y − 2z = 0 2x − y − 2z = 0 x + 2y − 4z = 0 Solution. Note that 2 rref 2 1 1 −2 0 1 −1 −2 0 = 0 2 −4 0 0 0 1 0 0 0 1 0 0 0 Hence the only solution is 0 x y = 0 . 0 z Problem 1.1.8. Solve the system x1 + x2 − x3 + 2x4 = 1 x1 + x2 − x3 − x4 = −1 x1 + 2x2 + x3 + 2x4 = −1 2x1 + 2x2 + x3 + x4 = 2 Solution. Note that 1 1 1 1 rref 1 2 2 2 1 0 0 0 −1 2 1 0 1 0 0 −1 −1 −1 = 1 2 −1 0 0 1 0 1 1 2 0 0 0 1 11/3 −10/3 2/3 2/3 Hence the only solution is x1 11/3 x2 −10/3 = x3 2/3 . x4 2/3 Problem 1.1.15. Solve the system 2x1 − x2 − x3 + x4 + x5 = 0 x1 − x2 + x3 + 2x4 − 3x5 = 0 3x1 − 2x2 − x3 − x4 + 2x5 = 0 Solution. Note that 2 −1 rref 1 −1 3 −2 1 0 0 7 −1 1 1 0 1 2 −3 0 = 0 1 0 9 0 0 1 4 −1 −1 2 0 1 −4 0 −5 0 −4 0 2 MATH 107.01 HOMEWORK #1 SOLUTIONS Hence the solutions are given by 4x5 − 7x4 x1 x2 5x5 − 9x4 x3 = 4x5 − 4x4 . x4 x4 x5 x5 Problem 1.1.17. Determine conditions on a, b, and c so that the system 2x − y + 3z = a x − 3y + 2z = b x + 2y + z = c has a solution. Solution. The row reduction 2 −1 3 a 1 −3 2 b 1 2 1 c WV PQ 2R2 → R2 2R3 → R3 R2 − R1 → R2 R3 − R1 → R3 WV PQ 2 / 2 2 2 / 0 0 2 / 0 0 R3 +R2 →R3 −1 3 −6 4 4 2 a 2b RS 2c UT −1 3 a −5 1 2b − a RS 5 −1 2c − a UT −1 3 a −5 1 2b − a 0 0 −2a + 2b + 2c shows that the rref of the system is consistent if and only if −2a + 2b + 2c = 0. Hence the system has a solution if and only if a − b − c = 0. Problem 1.1.18. Determine conditions on a, b, and c so that the system x + 2y − z = a x + y − 2z = b 2x + y − 3z = c has a solution. Solution. The row reduction R1 → R2 1 1 2 −1 a RR2 −−2R 3 1 → R3 / 0 1 1 −1 b 2 1 −3 c 0 WV PQ R3 −3R2 →R3 1 / 0 0 2 −1 −1 −2 −3 −1 a b−a RS c − 2a UT 2 −1 a −1 −2 b−a 0 5 a − 3b + c MATH 107.01 HOMEWORK #1 SOLUTIONS 3 shows that the rref of the system is consistent. Hence the system has a solution for all values of a, b, and c. Problem 1.1.19. Determine conditions on a, b, c, and d so that the system x1 + x2 + x3 − x4 = a x1 − x2 − x3 + x4 = b x1 + x2 + x3 + x4 = c x1 − x2 + x3 + x4 = d has a solution. Solution. The row reduction R2 − R1 → R2 1 1 1 −1 a R3 − R1 → R3 1 −1 −1 R4 − R1 → R4 1 b 1 1 1 1 c 1 −1 1 1 d WV PQ R4 −R2 →R4 WV PQ R3 ↔R4 1 0 / 0 0 1 1 −1 a −2 −2 2 b − a 0 0 2 c − a −2 0 2 d − a ml RS 1 0 / 0 0 1 1 −1 a −2 −2 2 b−a 0 0 2 c−a 0 2 0 d − 2a − b ml RS 1 0 / 0 0 1 1 −1 a −2 −2 2 b−a 0 2 0 d − 2a − b 0 0 2 c−a shows that the rref of the system is consistent. Hence the system has a solution for all values of a, b, c, and d. Problem 1.1.22. Determine if the homogeneous system √ 99x1 + πx2 − 5x3 = 0 2x1 + sin (1)x2 + 2x4 = 0 3.38x1 − ex3 + ln (2)x4 = 0 has a nontrivial solution. Solution. This is a homogeneous system in 3 linear equations in 4 variables. Since 3 < 4, Theorem 1.1 in the book implies that this system has infinitely many nontrivial solutions. Problem 1.1.23. Determine if the homogeneous system x−y+z =0 2x + y + 2z = 0 3x − 5y + 3z = 0 has a nontrivial solution. 4 MATH 107.01 HOMEWORK #1 SOLUTIONS Solution. Note that 1 rref 0 0 −1 1 0 1 1 2 0 = 0 −5 3 0 0 0 1 0 1 0 0 0 0 . 0 That is, the system is consistent and has free variable x3 . Hence the system has infinitely many nontrivial solutions. Problem 1.1.26. Give an example of a system of linear equations with more equations than variables that illustrates each of the following possibilities: (a) Has exactly one solution. (b) Has infinitely many solutions. (c) Has no solution. Solution. (a) The system x=0 2x = 0 has more equations than variables and has x = 0 as a unique solution. (b) The system x+y =0 2x + 2y = 0 3x + 3y = 0 has more equations than variables. Furthermore, 1 1 1 1 0 rref 2 2 0 = 0 0 0 0 3 3 0 0 0 . 0 That is, the system is consistent and has free variable y. Hence the system has infinitely many solutions. (c) The system x=0 x=2 has more equations than variables. Moreover, it has no solutions since 0 6= 2.