Exam (elaborations) TEST BANK FOR Introduction to Linear Algebra with Applications By James DeFranza, Daniel Gagliardi (Solution Manual)-Converted

In Section 1.1 of the text, Gaussian Elimination is used to solve a linear system. This procedure utilizes three operations that when applied to a linear system result in a new system that is equivalent to the original. Equivalent means that the linear systems have the same solutions. The three operations are: • Interchange two equations. • Multiply any equation by a nonzero constant. • Add a multiple of one equation to another. When used judiciously these three operations allow us to reduce a linear system to a triangular linear system, which can be solved. A linear system is consistent if there is at least one solution and is inconsistent if there are no solutions. Every linear system has either a unique solution, infinitely many solutions or no solutions. For example, the triangular linear systems   x1 − x2 + x3 = 2 x2 − 2x3 = −1 x3 = 2 ,   x1 − 2x2 + x3 = 2 −x2 + 2x3 = −3 ,   2 x1 + x3 = 1 x2 − x3 = 2 0 = 4 have a unique solution, infinitely many solutions, and no solutions, respectively. In the second linear system, the variable x3 is a free variable, and once assigned any real number the values of x1 and x2 are determined. In this way the linear system has infinitely many solutions. If a linear system has the same form as the second system, but also has the additional equation 0 = 0, then the linear system will still have free variables. The third system is inconsistent since the last equation 0 = 4 is impossible. In some cases, the conditions on the right hand side of a linear system are not specified. Consider for example, the linear system   −x1 − x2 = a 2x1 + 2x2 + x3 = b 2x3 = c which is equivalent to −−−−−−−−−−−−−−−−→   −x1 − x2 = a x3 = b + 2a 0 = c − 2b − 4a . This linear system is consistent only for values a, b and c such that c − 2b − 4a = 0. Solutions to Exercises 1. Applying the given operations we obtain the equivalent triangular system   x1 − x2 − 2x3 = 3 −x1 + 2x2 + 3x3 = 1 2x1 − 2x2 − 2x3 = −2 E1 + E2 → E2 −−−−−−−−−−→   x1 − x2 − 2x3 = 3 x2 + x3 = 4 2x1 − 2x2 − 2x3 = −2 (−2)E1 + E3 → E3 −−−−−−−−−−−−−−→ 2 Chapter 1 Systems of Linear Equations and Matrices   x1 − x2 − 2x3 = 3 x2 + x3 = 4 2x3 = −8 . Using back substitution, the linear system has the unique solution x1 = 3, x2 = 8, x3 = −4. 2. Applying the given operations we obtain the equivalent triangular system   2x1 − 2x2 − x3 = −3 x1 − 3x2 + x3 = −2 x1 − 2x2 = 2 E1 ↔ E2 −−−−−−→   x1 − 3x2 + x3 = −2 2x1 − 2x2 − x3 = −3 x1 − 2x2 = 2 (−2)E1 + E2 → E2 −−−−−−−−−−−−−−→   x1 − 3x2 + x3 = −2 4x2 − 3x3 = 1 x1 − 2x2 = 2 (−1)E1 + E3 → E3 −−−−−−−−−−−−−−→   x1 − 3x2 + x3 = −2 4x2 − 3x3 = 1 x2 − x3 = 4 E2 ↔ E3 −−−−−−→   x1 − 3x2 + x3 = −2 x2 − x3 = 4 4x2 − 3x3 = 1 (−4)E2 + E3 → E3 −−−−−−−−−−−−−−→   x1 − 3x2 + x3 = −2 x2 − x3 = 4 x3 = −15 . Using back substitution, the linear system has the unique solution x1 = −20, x2 = −11, x3 = −15. 3. Applying the given operations we obtain the equivalent triangular system   x1 + 3x4 = 2 x1 + x2 + 4x4 = 3 2x1 + x3 + 8x4 = 3 x1 + x2 + x3 + 6x4 = 2 (−1)E1 + E2 → E2 −−−−−−−−−−−−−−→   x1 + 3x4 = 2 x2 + x4 = 1 2x1 + x3 + 8x4 = 3 x1 + x2 + x3 + 6x4 = 2 (−2)E1 + E3 → E3 −−−−−−−−−−−−−−→   x1 + 3x4 = 2 x2 + x4 = 1 +x3 + 2x4 = −1 x1 + x2 + x3 + 6x4 = 2 (−1)E1 + E4 → E4 −−−−−−−−−−−−−−→   x1 + 3x4 = 2 x2 + x4 = 1 +x3 + 2x4 = −1 x2 + x3 + 3x4 = 0 (−1)E2 + E4 → E4 −−−−−−−−−−−−−−→   x1 + 3x4 = 2 x2 + x4 = 1 x3 + 2x4 = −1 x3 + 2x4 = −1 (−1)E3 + E4 → E4 −−−−−−−−−−−−−−→   x1 + 3x4 = 2 x2 + x4 = 1 x3 + 2x4 = −1 0 = 0 . The final triangular linear system has more variables than equations, that is, there is a free variable. As a result there are infinitely many solutions. Specifically, using back substitution, the solutions are given by x1 = 2 − 3x4, x2 = 1 − x4, x3 = −1 − 2x4, x4 ∈ R. 4. Applying the given operations we obtain the equivalent triangular system   x1 + x3 = − 2 x1 + x2 + 4x3 = −1 2x1 + 2x3 + x4 = −1 (−1)E1 + E2 → E2 −−−−−−−−−−−−−−→   x1 + x3 = −2 x2 + 3x3 = 1 2x1 + 2x3 + x4 = −1 (−2)E1 + E3 → E3 −−−−−−−−−−−−−−→ 1.1 Systems of Linear Equations 3   x1 + x3 = −2 x2 + 3x3 = 1 x4 = 3 . Using back substitution, the set of solutions is given by x1 = −2 − x3, x2 = 1 − 3x3, x4 = 3, x3 ∈ R. 5. The second equation gives immediately that x = 0. Substituting the value x = 0 into the first equation, we have that y = −2 3 . Hence, the linear system has the unique solution x = 0, y = −2 3 . 6. From the second equation y = 1 and substituting this value in the first equation gives x = −4. 7. The first equation gives x = 1 and substituting this value in the second equation, we have y = 0. Hence, the linear system has the unique solution x = 1, y = 0. 8. The operation 3E2+E1 → E1 gives 5x = −1, so x = −1 5 . Substitution in equation two, then gives y = −1 5 . 9. Notice that the first equation is three times the second and hence, the equations have the same solutions. Since each equation has infinitely many solutions the linear system has infinitely many solutions with solution set S =