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 January 24th, 2007, 12:18 PM #1 Newbie   Joined: Jan 2007 Posts: 18 Thanks: 0 linear algebra On this problem, I think I'm more confused about the possible answers than the problem itself. How would you suggest I go about it? Solve the system {x+2y-2z=-1 {x+3y+z=10 {2x+6y+2z=20 solutions: a. (8c-23, -3c+10, c) for any real number c. b. (8c-22, -3c+10, c) for any real number c. c. (8c-22, -3c+11, c) for any real number c. d. (8c-23, -3c+11, c) for any real number c.
 January 24th, 2007, 01:51 PM #2 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 I would use Gaussian Elimination to get the following result after manipulation to get the first equation in terms of two variables: {x-8z=-23 {y+3z=11 {0=0 What happens is that one of the equations inevitably becomes 0=0, which means that you are left with a line. From here, you would simply substitute c for z and solve for x and y to get (8c-23, -3c+11, c). Note that the first step does not require Gaussian Elimination. There are several other ways to solve a system of three three-variable equations.
 January 24th, 2007, 09:17 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 Re: linear algebra The third equation is simply twice the second. In a slightly more difficult case, the third equation might turn out to be a less obvious linear combination of the first two equations. In either case, the third equation is redundant and the form of the solutions you have to choose from gives away that fact. If you weren't sure whether the third equation is redundant, you could calculate the determinant of the coefficients matrix, which will be zero if the third equation is a linear combination of the first two. If the determinant turned out not to be zero, the solution of the system of equations would consist of specific numerical values for x, y and z, which could be found by methods you should already know. Subtracting the first equation from the second gives y + 3z = 11, so y = -3z+11. Substituting for y in the first equation gives x - 8z = -23, so x = 8z - 23. There are other ways of getting these results from the first two equations, but it's generally best to make it clear exactly what you did, since you might get some credit for so doing even if you make a slip in the arithmetic or at some later stage. If you were expected to use a matrix method, do so (ask if you don't see how). Obviously, it makes no difference whether the answers are presented in terms of z or in terms of some new unknown c. Strictly speaking, z doesn't have to be real (unless it was stated somewhere that all the variables mentioned are real). Test your understanding: what can you conclude about a system of n linear equations in n unknowns (where n is an integer greater than 1) if the determinant of the coefficients matrix is zero? For n = 3, explain the situation in geometrical terms (assuming the variables are all real).
January 25th, 2007, 08:55 AM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: linear algebra

Hello, gretchen

Quote:
 Solve the system: [color=beige]. . . [/color]x + 2y - 2z[color=beige] .[/color]=[color=beige] .[/color]-1 [color=beige]. . . [/color]x + 3y + z[color=beige] . [/color]=[color=beige] .[/color]10 [color=beige]. . [/color]2x + 6y + 2z[color=beige] .[/color]=[color=beige] .[/color]20 Solutions: [a] (8c-23, -3c+10, c) for any real number c [b] (8c-22, -3c+10, c) for any real number c [c] (8c-22, -3c+11, c) for any real number c [d] (8c-23, -3c+11, c) for any real number c

Here's a back-door solution . . .

Since this is a multiple-choice question,
[color=beige]. . [/color]you can plug in all four answer-choices and see which one works.

Test: [a][color=beige] .[/color]x = 8c - 23,[color=beige] .[/color]y = -3x + 10,[color=beige] .[/color]z = c

Substitute into the first equation:
[color=beige]. . [/color](8c - 23) + 2(-3c + 10) - 2(c)[color=beige] .[/color]=[color=beige] .[/color]-1

and we get:[color=beige] .[/color]-3[color=beige] .[/color]=[color=beige] .[/color]-1 . . . not true . . . so [a] is not the answer.

See?

 January 25th, 2007, 11:50 AM #5 Senior Member   Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 That method works for multiple choice equations, but it is better to be able to use some other method, because, in all likelyhood, you will eventually be asked to solve a similar type of problem, and will not be given a list of possible answers. For example, would you be able to solve this if this question had been phrased as "find x and y in terms of z," without giving any choices?
 January 25th, 2007, 12:52 PM #6 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Hello, roadnot taken! Of course, I agree completely. But if they insist on giving multiple-choice questions, [color=beige]. . [/color]this is one of the lazier methods to use. To tell the truth, I gave a detail (legitimate) solution at another site.

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