Suggested problems solutions Solving systems using Gauss-Jordan elimination P: Whih of the matries below are in redued row ehelon form? If a matrix is not in redued row ehelon form, explain where it fails, and use Gauss Jordan elimination to transform it to redued row ehelon form. (a) 0 3 0 4 0 0 5 Not in redued row ehelon form - the leading one in the seond row needs a zero below it. Perform R2 + R3 R3: 0 3 0 4 0 0 9 Now the third row needs a leading. Perform R3 R3: 0 3 0 4 0 0 9 And now that there s a leading one in the third row, there need to be zeros in the olumn above it. Perform R3 + R2 R2 and R3 + R R: 0 0 6 0 0 5 0 0 9 NOW, it s in redued row ehelon form.
(b) 2 4 0 Not in redued row ehelon form. First, every row should have a leading one, so perform 2 R2 R2: 2 2 2 5 The leading ones should progress inward from the top down; swap rows. R R2: 2 2 2 5 The leading one in the seond row needs a zero above it in the fourth olumn. 2R2 + R R: 2 2 0 9 NOW, it s in redued row ehelon form. () 0 0 0 7 0 0 0 0 0 0 0 This matrix is in redued row ehelon form - every row starts with a one, the ones progess inward, and every leading one has zeros above and below it. (d) 0 0 7 0 4 0 0 0 0 2 0 Not in redued row ehelon form; needs a leading one in the third row. 2 R3 R3: 0 0 7 0 4 0 0 0 0 0 And needs a zero above the leading one. 4R3 + R2 R2: 0 0 7 0 0 0 0 0 0 0 NOW, it s in redued row ehelon form.
P3: (Yes, P2 is missing) Solve using Gauss-Jordan elimination: Augmented matrix: x 2x 2 = 2x 3x 2 = 2 2 3 Leading one in the first row, get a zero below it. 2R + R2 R2: 2 0 5 Leading one in the seond row, get a zero above it. 2R2 + R R: And turn bak into a system. Instant solution: 0 2 0 5 x = 2 x 2 = 5
P4: Solve using Gauss-Jordan elimination: Augmented matrix: x 2 + 3x 3 = 2x + 2x 3 = 4 3x + x 2 2x 3 = 2 0 3 2 0 2 4 3 2 2 Divide down row two. 2 R2 R2: 0 3 0 2 3 2 2 And swap. R2 R: 0 2 0 3 3 2 2 Zero in the third row, first olumn. 3R + R3 R3: 0 2 0 3 0 5 4 Zero in third row, seond olumn. R2 + R3 R3: 0 2 0 3 0 0 5 Leading one in row 3. R3 R3: 0 2 0 3 0 0 5 Zeros above the leading one in the third olumn. 3R3 + R2 R2 and R3 + R R: 0 0 0 0 7 0 0 5 Solution: x = x 2 = 7 x 3 = 5
P5: Consider the system of equations ax + bx 2 = x + dx 2 = 0 (Assume a, b,, and d are all non-zero.) Use Gauss-Jordan elimination to transform the augmented matrix for the system to redued row ehelon form, and express the solutions for x and x 2 in terms of a, b,, and d. You have now obtained a formula that will produe solutions for all systems in this form. Augmented matrix: a b d 0 Get a one in the first row pivot position by dividing by a. Zero below the leading one. R + R2 R2: b a a d 0 a R R: For this one, I ll show the srath work (note the LCD... d = ad a... to ombine): b a a ad a 0 0 ad b a a b a a 0 ad b a a To get a leading one in the seond row, you need to multiply through by the reiproal. ( a )R2 R2 ad b (Note ( a ad b )( a ) = ad b ): b a a 0 ad b Almost there - need a zero above that one. a b R2 + R R: Srath: b a(ad b) + a = 0 b a b a(ad b) b a a 0??? b (ad b) b + ad b + = = a(ad b) a(ad b) a(ad b) 0 a b b a(ad b) b a a 0 d ad b ad a(ad b) = d ad b
0 d ad b 0 ad b Phew. Done. The solution is d x = ad b x 2 = ad b Test it out on the system x + 2x 2 = 3x + x 2 = 0 by obtaining x and x 2 from the formula, and plugging bak in to hek: x = 3 2 = 5 x 2 = 3 3 2 = 3 5 Chek: 5 + 2(3 5 ) = 5 5 = 3( 5 ) + 3 5 = 0 Yeah! Ideally, to derive a more general formula, I should have had you solve a b e d f The algebra on that would have been outrageous, though. The point here is that Gauss-Jordan is Gauss-Jordan in terms of the proess; you just get some nasty srath work if it s all symboli. Being able to solve these in the abstrat will allow us to drawn some onlusions about what it takes to have a solution. You ll notie I started by assuming none of the oeffiients were zero (to avoid potential divide by zero errors). The solution proess reveals that something else an t be zero - the quantity ad b. In other words, if ad = b, there s a problem with the solution.