13.2 Solving Larger Systems by Gaussian Elimination

Transcription

1 . Solving Larger Sstems b Gaussian Elimination Now we want to learn how to solve sstems of equations that have more that two variables. We start with the following definition. Definition: Row echelon form A sstem is in row echelon form when it has a stair-step pattern with leading coefficients of. Here is an eample of a sstem that is in row echelon form. The reason we need row echelon form, is that when a sstem is in row echelon form it is ver eas to solve. We simpl use a technique called back substituting. To do this we simpl take the variables and substitute the value one at a time into the equation above it. We illustrate this technique in the following eample. Eample : Solve the sstem. Solution: So we need to back substitute to solve the sstem. Since we can see from the last equation that =, we can substitute this value into the middle equation and solve for. We get Now that we have both and, we can substitute those values into the top equation and solve for. We get So now we have, and. We put the solution as an ordered triple of the form,,. So the solution to our sstem is,,. So we can see that having a sstem in row echelon form can be a ver useful thing in seeking the solution to a sstem of equations. With that in mind we make the following definition.

2 Definition: Equivalent sstems Two sstems are called equivalent if the have the same solution set. So if we are able to put a sstem into row echelon form in a wa that keeps a sstem equivalent, then we could easil solve the sstem b back substitution. Fortunatel for us, we do have such a technique. We use something we call the elementar row operations. Elementar Row Operations The following operations on a sstem of linear equations produces an equivalent sstem:. Interchange two equations. Multipl one equation b a non-ero constant. Add a multiple of one equation to another to replace the latter equation Using the row operations to solve a sstem is called solving b Gaussian Elimination. Eample : Solve b Gaussian Elimination. a. b. Solution: a. To solve a sstem b Gaussian elimination we need to use our row operations to put the equation into row echelon form. Then we can easil back substitute and solve. Remember, row echelon form has a stair step pattern with leading coefficients of one. With this in mind, we can see that we should leave the first equation on top, since it alread has leading coefficient of. So, we need to find a wa to make the other two equations not have an term. We use row operation. That is, we can multipl the top equation b - and add it to the second equation to get a new second equation. To simplif the matter we call the top equation E, the middle equation E and the bottom equation E. So, we are going to do -E + E. This gives So, is our new E. So our sstem now looks like Now we do the same thing to eliminate the from E. We do -E + E. This will become our new E. We get So our sstem is now

3 Now, remember, to get row echelon form, we need the leading coefficients to be. So we can multipl equation b ¼ to make the leading coefficient. So we write ¼E and we get So our sstem is now We almost have row echelon form. We need onl get rid of the in E. Just like we did for, we can use the third row operation. This time we will use E. So we use -E + E. We get This now becomes our new E. So our sstem is To finish getting it into row echelon form we simpl go E. We get Now we simpl need to back substitute to solve. We proceed as follows.,, So the solution is. b. Again, we need to start b getting the sstem into row echelon form. We can see clearl that if we do E + E, then we will eliminate one value relativel easil. So we get In fact, this operation eliminated several steps. So this will become our new E. We will also, go ahead and insert E just to simplif the matter. Our sstem is now

4 Now we can go ½E to get the coefficient to be. We get Clearl now we will do -E + E to replace E. We get So our sstem becomes Notice that it is alread in row echelon form. Thus, we onl need to back substitute to solve. We get,, So the solution is. In a similar wa to sstems of two variables, we have the possibilit of having one solution, no solution or infinite solutions. However, this time we are dealing with one more variable. So geometricall, these tpes of equations are planes in three dimensions. So here we are talking about the different was two planes can intersect. The following shows different was that three planes can intersect. One solution Infinite solutions No solutions Just like in the last section, it is usuall fairl clear when we have one of these situations. We either end up with a nonsense statement (like =, for eample) which means no solution, or we end up with a statement that is alwas true (like = ). Eample : Solve b Gaussian Elimination. a. b. Solution:

5 a. Again we will solve b getting the sstem into row echelon form. Notice that we alread have somewhat of a stair step pattern. So we will start b doing ½E. We get the sstem Now we will do -E + E to replace E. We get So our sstem is Lets multipl E b. We get Now we do E + E. We get Whenever we see a statement of this form we know that the sstem must have no solution. b. Again, we will start b doing -E + E to replace E. We get So our sstem is now Notice that the last two equations are multiples. Thus lets go ahead and perform the operation -E + E. We get So, again, when we get a statement of this form we know that the sstem has infinite solutions. Lastl, we want to see some applications of larger sstems of equations. Man of the applications are similar to what we have done earlier in this tet. The onl difference is we can

6 now set them up using as man variables as needed. This, in turn can prove to streamline the entire problem. Thus the following eample is an eample of a problem we have not et encountered. Eample : Find the constants a, b, and c so that the parabola = a + b + c passes through the three points (, -), (-, ) and (, ). Solution: First we need to set up a sstem of equations. Since we know that each of the given points is on the graph, the each must satisf the equation. Therefore, we can substitute the values of and into the equation for each given ordered pair. We get For (, -): For (-, ): For (, ): - = a () + b () + c = a (-) + b (-) + c = a () + b () + c - = a + b + c = a - b + c = a + b + c So this is our sstem of equations. It has variables a, b, and c. a b c a b c a b c We solve the sstem b Gaussian Elimination as follows. a b c a b c a b c E E a b c b a b c E E a b c b b c E E a b c b c E and E a b c b c So now we simpl back substitute to solve. We get So, a, b, and c. a a a. Eercises Solve b Gaussian Elimination

7

8 . w w w w. w w w w. w w w. w w w w Find the constants a, b, and c so that the parabola = a + b + c passes through the given points.. (, ), (, -), (, ). (, ), (, ), (, -). (-, -), (, ), (, ). (-, -), (, ), (, -) Find the constants D, E, and F so that the circle + +D +E + F = passes through the given points.. (, ), (, ), (, ). (, -), (-, ), (, ). (-, ), (, ), (, ). (, ), (, ), (, ) Find the position equation s = ½at + v t + s for an object that has the indicated values.. s = feet at t = second. s = feet at t = second s = feet at t = seconds s = feet at t = seconds s = feet at t = seconds s = feet at t = seconds. s = feet at t = second. s = feet at t = second s = feet at t = seconds s = feet at t = seconds s = feet at t = seconds s = feet at t = seconds. A recent basic model of the XRT sport coupe had a price of $,. With automatic transmission and power windows the price was $,. With air conditioning and power windows the price was $,. With air conditioning and automatic transmission the price was $,. What was the price of air conditioning, automatic transmission, and power windows individuall?. Frank Smith has three rock polishers. When the are all working, Frank can polish rocks in a week. When onl the first two are working, Frank can polish onl rocks in a week. When onl the last two are working, Frank can onl polish rocks in a week. How man rocks can each machine polish in a week?. John, Andrew and Ethan work for compan that makes Christmas ornaments. Working together, the can make ornaments a da. On the das that Ethan is off, the can make ornaments a da. On the das Andrew is off, the can make ornaments a da. How man can each person make alone in a da?. Emma, Jennifer and Li volunteer for a local church stuffing envelopes. In one da, together the can stuff envelopes. Emma and Jennifer together can stuff envelopes in a da. Jennifer and Li together can stuff envelopes in a da. How man envelopes can each person stuff in a da working alone?

9 . In a recent basketball game, the College of the Sequoias basketball team scored a total of points. This consisted of -pointers, pointers and foul shots worth one point. All together the team made baskets and more pointers than pointers. How man of each kind of shot was made?. A standard par golf course is made up of three tpes of holes: par holes, par holes and par holes. There is the same number of par holes as par holes, and there are two more par holes than the total of all the par and par holes. How man of each tpe of hole is on a standard hole golf course?. Matt has a bunch of change in the ash tra of his car. Since Matt hates pennies, he onl has nickels, dimes and quarters in the ash tra. The total value of the cash in his ash tra is $.. There are twice as man dimes as quarters and there coins in all. How man of each time of coin does Matt have in his ash tra?. The change from a coin operated washing machine is collected once a week. The machine onl accepts nickels, dimes and quarters. One particular week the machine had coins totaling $.. There were less nickels than twice the number of dimes. How man of each tpe of coin was in the machine?. In the card game Cribbage there are three tpes of card runs: a run of - worth points, a double run of - worth points, and a double-double run of - worth points. If a plaer had total card runs which gave him points and had more double runs than doubledouble runs, how man of each tpe of card runs did the plaer have?. In the card game Cribbage ou can score points for a pair, three of a kind and four of a kind. A pair is worth points, a three of a kind is worth points and a four of a kind is worth points. If a plaer had total pairs, three of a kinds and four of a kinds, that gave her points, how man pairs did the plaer have?