Stitz-Zeager_College_Algebra_e-book

# The product of a and b denoted ab is the matrix dened

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Unformatted text preview: 2 and E 3 (E 2) x2 − 3x3 − 2x4 = 0 −− − − − − − − − − −→ 2 (E 3) 1 x2 − x3 − 3 x4 = 0 3 460 Systems of Equations and Matrices 1 (E 1) x1 + 1 x2 + 3 x4 = 2 3 (E 2) x2 − 3x3 − 2x4 = 0 2 1 (E 3) 3 x2 − x3 − 3 x4 = 0 1 (E 1) x1 + 3 x2 + 1 x4 = 2 3 Replace E 3 (E 2) x2 − 3x3 − 2x4 = 0 −−−−−→ −−−−− 1 with − 3 E 2 + E 3 (E 3) 0=0 Equation E 3 reduces to 0 = 0,which is always true. Since we have no equations with x3 or x4 as leading variables, they are both free, which means we have a consistent dependent system. We parametrize the solution set by letting x3 = s and x4 = t and obtain from E 2 1 that x2 = 3s + 2t. Substituting this and x4 = t into E 1, we have x1 + 3 (3s + 2t) + 1 t = 2 3 which gives x1 = 2 − s − t. Our solution is the set {(2 − s − t, 2s + 3t, s, t) : −∞ < s, t < ∞}.13 We leave it to the reader to verify that the substitutions x1 = 2 − s − t, x2 = 3s + 2t, x3 = s and x4 = t satisfy the equations in the original system. Like all algorithms, Gaussian Elimination has...
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