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Numerical Solutions of a System of Equations One way to solve the linear system of two equations x + y = 7, −x + 3y = 1, (1) is to add the two equations, obtaining 4y = 8, and hence y = 2. Substituting y = 2 into the first equation (1) yields x = 5. Our objective in these chapters is to illustrate numerical methods for solving a linear system of equations. Linear System of Equations a11 x1 + a12 x2 + · · · + a1n xn = b1 a21 x1 + a22 x2 + · · · + a2n xn = b2 .. .. . . . = .. .. .. . . . = .. an1 x1 + an2 x2 + · · · + ann xn = bn , where aij , bi , i, j = 1..n are real constants and the a0 s are the coefficients of the variables xi . The above system can be written as AX = b, where x a a . . . a1n 1 11 12 x2 a21 a22 . . . a2n , b = , x = A= .. .. .. .. . . . ... . xn an1 an2 . . . ann Inhomogeneous o AX = b Linear system b1 b2 .. . bn . / Homogeneous Methods used to solve 1-Elementary row operations AX = 0 2-Cramer’s Rule Always has at least the trivial solution 3-Gaussian elimination 1 3 If |A| = 6 0 • Inhomogeneous / Unique solution X = A−1 b Infinitly many solutions 3 + Unknowns > Equations + No solution Exercise 9 page (356) Given the linear system 2x1 − 6αx2 = 3, 3αx1 − x2 = 3/2. 1. Find values of α for which the system has no solution? 2. Find values of α for which the system has infinite number of solutions, if they exist? 3. Find values of α for which the system has unique solution if it exists?. Solution 2 −6α = 18α2 − 2 = 0 =⇒ α = ±1/3. |A| = 3α −1 • For α 6= ±1/3, the system has unique solution. • For α = −1/3, the system becomes x1 + x2 = 3/2, x1 + x2 = −3/2, and therefore it has no solution. • For α = 1/3, the system becomes x1 − x2 = 3/2, x1 − x2 = 3/2, and therefore it has infinite number of solutions. 2