1. The augmented matrix for this system is " " " # (remember, I can't draw the V Ç V ß #V V Ä V ß $V V Ä V

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1 MATH 339, Fall 2017 Homework 1 Solutions Bear in mind that the row-reduction process is not a unique determined creature. Different people might choose to row reduce a matrix in slightly different ways. So, your method of getting to the solution set for a system of equations might look a little different than mine, but here is one way to get to a solution for these problems. Ô " " " 1. The augmented matrix for this system is " " " (remember, I can't draw the $ " % % Ø vertical line seperating the last column representing the constants on the right hand side of the equations with my mathematical word processing program so you either need to draw it in yourself or just get used to imagining it is there). Applying row operations matrix to V Ç V ß V V Ä V ß $V V Ä V " " " should bring your Ô " " " Ô " " " " " & $ " & ÞNow applying V Ä V gives 1, then % " Ø $ % " Ø %V V$ Ä V$ produces Ô " " " Ö " & 1 Ù which is in row echelon form and allows us to solve uniquely for Bß Cß D. Last ( "% 0 Ø row indicates that ( Dœ "%, so Dœ. Second row says C " Dœ & and since Dœ this gives & C œ, so Cœ". First row says B C Dœ so B " œ giving Bœ ". Ô B Ô " In column vector form, our unique solution is C œ ". D Ø Ø

2 2. The augmented matrix for this system is " " " " " " &. Applying " " $ " % Ø V V ÄV and V V ÄV brings us to V Ç V " " $ % Þ Applying " " $ Ø brings us to " " $. % Ø This is row echelon form and you can solve the system at this point choosing BßBßB % ' as free variables (variables corresponding to columns without pivots; pivots are in columns "ß $ß &) starting with the last row, working your way up and back substituting as you go. Or you can proceed with row operations to bring the matrix to reduced row echelon form. I will do it this way to show the benefits of this procedure (more row operations as opposed to the back substitution process). " Take the matrix where we left off. Apply V ÄV and V$ ÄV$ to arrive at V V Ä V and " " $. Applying " " Ø V V Ä V $ $ " " gives Ô " " " " $ " $ &. Applying " " Ø V V Ä V " " gives Ô " " 0 " $ " $ &. " " Ø This matrix is now in reduced row echelon form (pivots are now all "'s and they are the only non-zero elements in their columns). We can now solve directly for dependent variables BßBßB in terms of free variables BßBßB without any back substitution. " $ & % ' Last row is B& B' œ so B& œ B'. Second row is B$ B% $B' œ & so B$ œb% $B' &. First row is B B B œ so B œ B $B B. " % ' " % '

3 In column vector form our solution set is all vectors of the form Ô B B% $B' Ô Ô B Ô B% Ô $B' B B & B% $B' & B% $B' œ œ B % B% Ö Ù Ö Ù Ö Ù Ö Ù Ö Ù B' B' B Ø Ø Ø Ø B Ø ' Ô Ô " Ô " Ô $ " & $ B B% B' where B ßB% ßB' are real numbers, the solution set " Ö Ù Ö Ù Ö Ù Ö Ù " Ø Ø Ø " Ø expressed as a sum of scalar multiples of fixed column vectors. ' & ) $ 3a. The augmented matrix for this system is. Two operations brings this " $ " " to row echelon form: V" Ç V brings us to, then V" V ÄV gives us. " " I'll solve this one with the back substitution idea. No pivots in columns $ and % so B$ and B% are free. Second row corresponds to B B$ B% œ or B œ B$ B% (solving for dependent variable B in terms of free variables B and B Ñ. $ % First row corresponds to B" B $B$ B% œ ". Solving for B" ß we have B" œ" B $B$ B%. But from above we know that B œ B$ B%, so we substiute to arrive at B œ" a B B b $B B œ" % %B B $B B œ& B B " $ % $ % $ % $ % $ % Ô & B$ B% B$ B% Our general solution in vector form then looks like Ö Ùwhich can be rewritten B$ B Ø % Ô & Ô " Ô " " as Ö Ù B$ Ö Ù B% Ö Ù " Ø Ø " Ø

4 3b. There are several ways to approach this idea of trying to express the solution set in terms of different free variables. To me the simplest is to just rearrange how the variables appear in your equations. We can rewrite the original system as B $B B B œ " % $ " $B )B B &B œ % $ " " $ " " Which in matrix form is (note I rearranged is such a way $ ) & to conveniently have a " in the row 1, column 1 positioin. Now performing $V" V Ä V brings us to " $ " ", V gives us " $ " " Ä V and " " " $ " " " $ V V" ÄV" brings us to a reduced row echelon form of " " ). Our pivots are now in columns " and so variables associated " " " $ with columns $ and % are now chosen to be free. Now, these variables are B" and B according to the rearranged system. In our rearranged system, the second row corresponds to B B B œ $ so B œ $ B B. The first row corresponds to B B B œ ) so B œ ) B B % " % " $ " $ " Note that we immediately got to solve for dependent variables in terms of free variables since the extra steps were taken to get my matrix in reduced row echelon form. We can now express the general solution as Ô B% Ô ) B" B Ô ) Ô Ô " B B" B $ " " Ö Ù œö Ù œö Ù B" Ö Ù B Ö Ùor if we rearrange the B" B " " B Ø B Ø Ø Ø " Ø variables back to their original order: Ô B " Ô Ô " Ô B " Ö Ù œ Ö Ù B" Ö Ù B Ö Ù. B " " B Ø ) Ø Ø " Ø %

5 Ô " " $ + 4. The matrix associated with this system is $ &,. Row reduce.: " -Ø Ô " " $ + V" V Ä V and V" V$ Ä V$ gives us " ", +, then - " " - + Ø Ô " " $ + V V$ ÄV$ brings us to row echelon form of " ", + where the +, -Ø / lement in row $ column $ is the result of computing a, + b - +œ, + - +œ+, -. While reduced row echelon form is a convenient form for actually solving a system, row echelon form is sufficient to determing the nature of the solution set. In our case, if +, -Áßthe last row of our matrix indicates an inconsistent system. So, in order for our system to be consistent, we must have that: +, -œ. Ô " " 5 5. The matrix associated with this system is $ 5 $5. Applying $V" V ÄV and 5 $ * Ø Ô " " 5 5V" V$ ÄV$ gives us 5 $, then V V$ ÄV$ brings us to a row $ 5 * 5 Ø Ô " " 5 echelon form of 5 $. The terms 5 $ and * 5 are trying to serve as * 5 Ø pivots in this system, but fail to be pivots if they take on a value of. The term 5 $ is when 5œ$ while the term * 5 is if 5œ$ or $. So these values should be investigated. Ô " " $ If 5œ$, our row echelon form becomes, a matrix with two columns without Ø pivots, indicating free variables and a 2-parameter family of solutions Ô " " $ If 5œ $, we have ', a matrix with just one column without a pivot, Ø indicating " free variable and a 1-parameter family of solutions. If is anything other that or, and are both non-zero so our matrix has a 5 $ $ 5 $ * 5 pivot in each column indicating a unique solution.

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