In Exercise 17–36, use the Gauss-Jordan elimination method to find all solutions of the system of linear equations.
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
Finite Mathematics & Its Applications (12th Edition)
- 3. Solve the linear system of equations x1 – *2 + 2x3 - -2, -201 +x2 – 03 – 2, 4x1 – x2 + 203 – 1, using 3 digit rounding arithmetic and Gaussian elimination with partial pivoting.arrow_forwardFind the LU factorization of A = A = -12 -11 X1 = x2 = and use FORWARD SUBSTITUTION AND BACKWARD SUBSTITUTION to solve the system -3 -2 -2 3 3 -2 x3 = 6 X1 x2 -3 x3 ය 3 -2 -2 3 -2 -12 -11 6 9 -14 53arrow_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
- 6. Use Cramer’s Rule to solve for x3 of the linear system 2x1 + x2 + x3 = 63x1 + 2x2 − 2x3 = −2x1 + x2 + 2x3 = −4arrow_forwardIn Exercises 1 through 12, find all solutions of the equations with paper and pencil using Gauss–Jordan elimination. Show all your work.arrow_forward2. Use Gauss elimination with back substitution to solve the system of linear equations: &x, +x2 +4x3 +8x, = 5 x1 - 7x, – 2x, – 7x4 : 1 7x, - 2x2 + 7x3 +2x4 =-5 X1 +x, +2x3 – 6x, = -5 Round-off to 5 significant figures.arrow_forward
- Please show your work for all questions. Thank you in advance!arrow_forwardConsider the system of linear equations in x and y. ax+by=ecx+dy=f Under what conditions will the system have exactly one solution?arrow_forwardConsider the difference equation an+2+an+1+an = 0. (a) Write down the associated discrete linear system n+1 = An- (b) Find a formula for A" and use this to give an explicit expression for an in terms of ao and a₁. [Hint: Note that A" has only 3 values.]arrow_forward
- Section 2.2 2.1. Solve the following difference equations: (a) Yk+1+ Yk = 2+ k, (b) Yk+1 – 2yk = k³, (с) ук+1 "Yk = 0, (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 + 2k – k², (g) Yk+1+ Yk = k + 2 · 3k, (h) Yk+1 – Yk = ke“, Yk = Bak*, = cos (bk), (k) Yk+1 + Yk = (-1)*, Yk – k. ,2k (i) Ук+1 (j) Yk+1 – aYk (1) Yk+1 k+1arrow_forwardFind a basis and dimension of the solution set to the system 2.x1 + x2 + x3 + x4 = 0 2.x1 – 3x2 – 3.x3 – 9x4 = 0 -2.x1 – 2x2 + 5x3 – x4 = 0 - - %3D - 5x1 + x2 – 3x3 – x4 = 0 - %3Darrow_forward
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:CengageElementary Linear Algebra (MindTap Course List)AlgebraISBN:9781305658004Author:Ron LarsonPublisher:Cengage LearningAlgebra and Trigonometry (MindTap Course List)AlgebraISBN:9781305071742Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage Learning
- College AlgebraAlgebraISBN:9781305115545Author:James Stewart, Lothar Redlin, Saleem WatsonPublisher:Cengage Learning