Suggested problems solutions

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1 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) Not in redued row ehelon form - the leading one in the seond row needs a zero below it. Perform R2 + R3 R3: Now the third row needs a leading. Perform R3 R3: 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: NOW, it s in redued row ehelon form.

2 (b) Not in redued row ehelon form. First, every row should have a leading one, so perform 2 R2 R2: The leading ones should progress inward from the top down; swap rows. R R2: The leading one in the seond row needs a zero above it in the fourth olumn. 2R2 + R R: NOW, it s in redued row ehelon form. () 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) Not in redued row ehelon form; needs a leading one in the third row. 2 R3 R3: And needs a zero above the leading one. 4R3 + R2 R2: NOW, it s in redued row ehelon form.

3 P3: (Yes, P2 is missing) Solve using Gauss-Jordan elimination: Augmented matrix: x 2x 2 = 2x 3x 2 = Leading one in the first row, get a zero below it. 2R + R2 R2: Leading one in the seond row, get a zero above it. 2R2 + R R: And turn bak into a system. Instant solution: x = 2 x 2 = 5

4 P4: Solve using Gauss-Jordan elimination: Augmented matrix: x 2 + 3x 3 = 2x + 2x 3 = 4 3x + x 2 2x 3 = Divide down row two. 2 R2 R2: And swap. R2 R: Zero in the third row, first olumn. 3R + R3 R3: Zero in third row, seond olumn. R2 + R3 R3: Leading one in row 3. R3 R3: Zeros above the leading one in the third olumn. 3R3 + R2 R2 and R3 + R R: Solution: x = x 2 = 7 x 3 = 5

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

6 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 5 Chek: 5 + 2(3 5 ) = 5 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.