In Exercises 1–24, use Gaussian elimination to find the complete solution to each system of equations, or show that none exists.
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- In Exercises 7–10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system. 1 7 3 -4 1 -4 1 -1 3 7. 8. 1 7 1 1 -2 0 -4 0 -7 1 -1 1 -3 9. 1 -3 -1 4arrow_forwardIn Exercises 15–16, solve each system using matrices. 15. (2x + y = 6 13x – 2y = 16 x - 4y + 4z = -1 2х — у + 52 16. -x + 3y - z =arrow_forwardIn Exercises 1–4, determine if the system has a nontrivial solution. Try to use as few row operations as possible.arrow_forward
- 2. Solve for x in the given matrix equalitiesarrow_forwardFind the values of C1, C2 and C3 with Gaussian elimination.arrow_forwardSolve the following linear equations using the 5 methods: (Gaussian Elimination, Gauss-Jordan Elimination, LU Factorization, Inverse Matrix and Cramer's Rule). Show your complete solutions. b. 2x1 — 6х, — Хз 3D — 38 -3x1 – x2 + 7x3 = -34 -8x1 + x2 – 2x3 = -20arrow_forward
- 5. By using the matrix methods to solve the following linear system: I1 + 12 – 13 = 5, 3r1 +x2 – 2r3 = -4, -I1 + 12 - 2r3 = 3;arrow_forwardCalculate the values of the variables in the following sets of equations using the Gaussian elimination method.arrow_forwardSection 2.2 2.1. Solve the following difference equations: (a) Yk+1+Yk = 2+ k, (b) Yk+1 – 2Yk k3, (c) Yk+1 – 3 (d) Yk+1 – Yk = 1/k(k+ 1), (e) Yk+1+ Yk = 1/k(k+ 1), (f) (k + 2)yk+1 – (k+1)yk = 5+ 2* – k2, (g) Yk+1+ Yk = k +2 · 3k, (h) Yk+1 Yk 0, Yk = ke*, (i) Yk+1 Bak? Yk (j) Yk+1 ayk = cos(bk), (k) Yk+1 + Yk = (-1)k, (1) - * = k. Yk+1 k+1arrow_forward
- 1–28, (a) graph each system so that approximate real number solutions (if there are any) can be predicted, and (b) solve each system using the substitution method or the elimination-by-addition method. (x+y=4) (x2+y2=10) Kaufmann, Jerome E.; Schwitters, Karen L.. Intermediate Algebra (p. 536). Cengage Learning. Kindle Edition.arrow_forwardSolve the following linear system by Gaussian eliminationarrow_forwardThree components are connected to form a system as shown in the accompanying diagram. Because the components in the 2–3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2–3 subsystem. The experiment consists of determining the condition of each component [S (success) for a functioning component and F (failure) for a nonfunctioning component]. (Enter your answers in set notation. Enter EMPTY or ∅ for the empty set.) There us a graph shown in the pictures. Questions are posted on the pictures too.arrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage