Exam (elaborations) TEST BANK FOR Linear Algebra with Applications 5th Edition By Otto Bretscher (Solution Manual)-Converted

Section 1.1 1.1.1  x + 2y = 1 2x + 3y = 1  −2 × 1st equation →  x + 2y = 1 −y = −1  ÷(−1) →  x + 2y = 1 y = 1  −2 × 2nd equation →  x = −1 y = 1  , so that (x, y) = (−1, 1). 1.1.2  4x + 3y = 2 7x + 5y = 3  ÷4 →  x + 3 4y = 1 2 7x + 5y = 3  −7 × 1st equation →  x + 3 4y = 1 2 −1 4y = −1 2  ×(−4) →  x + 3 4y = 1 2 y = 2  −3 4 × 2nd equation →  x = −1 y = 2  , so that (x, y) = (−1, 2). 1.1.3  2x + 4y = 3 3x + 6y = 2  ÷2 →  x + 2y = 3 2 3x + 6y = 2  −3 × 1st equation →  x + 2y = 3 2 0 = −5 2  . So there is no solution. 1.1.4  2x + 4y = 2 3x + 6y = 3  ÷2 →  x + 2y = 1 3x + 6y = 3  −3 × 1st equation →  x + 2y = 1 0 = 0  This system has infinitely many solutions: if we choose y = t, an arbitrary real number, then the equation x + 2y = 1 gives us x = 1 − 2y = 1 − 2t. Therefore the general solution is (x, y) = (1 − 2t, t), where t is an arbitrary real number. 1.1.5  2x + 3y = 0 4x + 5y = 0  ÷2 →  x + 3 2y = 0 4x + 5y = 0  −4 × 1st equation →  x + 3 2y = 0 −y = 0  ÷(−1) →  x + 3 2y = 0 y = 0  −3 2 × 2nd equation →  x = 0 y = 0  , so that (x, y) = (0, 0). 1.1.6   x + 2y + 3z = 8 x + 3y + 3z = 10 x + 2y + 4z = 9   −I −I →   x + 2y + 3z = 8 y = 2 z = 1   −2(II) →   x + 3z = 4 y = 2 z = 1   −3(III) →   x = 1 y = 2 z = 1  , so that (x, y, z) = (1, 2, 1). 1.1.7   x + 2y + 3z = 1 x + 3y + 4z = 3 x + 4y + 5z = 4   −I −I →   x + 2y + 3z = 1 y + z = 2 2y + 2z = 3   −2(II) −2(II) →   x + z = −3 y + z = 2 0 = −1   This system has no solution. 1 Copyright c 2013 Pearson Education, Inc. Chapter 1 1.1.8   x + 2y + 3z = 0 4x + 5y + 6z = 0 7x + 8y + 10z = 0   −4(I) −7(I) →   x + 2y + 3z = 0 −3y − 6z = 0 −6y − 11z = 0   ÷(−3) →   x + 2y + 3z = 0 y + 2z = 0 −6y − 11z = 0   −2(II) +6(II) →   x − z = 0 y + 2z = 0 z = 0   +III −2(III) →   x = 0 y = 0 z = 0  , so that (x, y, z) = (0, 0, 0). 1.1.9   x + 2y + 3z = 1 3x + 2y + z = 1 7x + 2y − 3z = 1   −3(I) −7(I) →   x + 2y + 3z = 1 −4y − 8z = −2 −12y − 24z = −6   ÷(−4) →   x + 2y + 3z = 1 y + 2z = 1 2 −12y − 24z = −6   −2(II) +12(II) →   x − z = 0 y + 2z = 1 2 0 = 0   This system has infinitely many solutions: if we choose z = t, an arbitrary real number, then we get x = z = t and y = 1 2 − 2z = 1 2 − 2t. Therefore, the general solution is (x, y, z) =