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)
- Please help. This problem involves finding the augmented matrix and using back substitution. Thank you.arrow_forwardIn Exercises 7–10, the augmented matrix of a linear system hasbeen reduced by row operations to the form shown. In each case,continue the appropriate row operations and describe the solutionset of the original systemarrow_forwardIn Exercises 7–10, determine the values of the parameter s for which the system has a unique solution, and describe the solution.arrow_forward
- 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_forward6. Use Cramer’s Rule to solve for x3 of the linear system 2x1 + x2 + x3 = 63x1 + 2x2 − 2x3 = −2x1 + x2 + 2x3 = −4arrow_forward
- In Exercises 13–17, determine conditions on the bi ’s, if any, in order to guarantee that the linear system is consistent. 13. x1 +3x2 =b1 −2x1 + x2 =b2 15. x1 −2x2 +5x3 =b1 4x1 −5x2 +8x3 =b2 −3x1 +3x2 −3x3 =b3 14. 6x1 −4x2 =b1 3x1 −2x2 =b2 16. x1 −2x2 − x3 =b1 −4x1 +5x2 +2x3 =b2 −4x1 +7x2 +4x3 =b3 17. x1 − x2 +3x3 +2x4 =b1 −2x1 + x2 + 5x3 + x4 = b2 −3x1 +2x2 +2x3 − x4 =b3 4x1 −3x2 + x3 +3x4 =b4arrow_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_forward1. Find a basis for the solution space of the system 3x1 X2 + x4 = 0 X1 + x2 + X3 + X4 =0.arrow_forward
- Find 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_forward4. Use Gaussian elimination with backward substitution to solve the following linear system: 2x1 + x2 – x3 = 5, x1 + x2 – 3x3 = -9, -x1 + x2 + 2x3 = 9;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
- Algebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage