TOP IN CONTEMP MATH II
TOP IN CONTEMP MATH II MATH 166
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This 4 page Class Notes was uploaded by Evert Christiansen on Wednesday October 21, 2015. The Class Notes belongs to MATH 166 at Texas A&M University taught by D. Lackey in Fall. Since its upload, it has received 12 views. For similar materials see /class/226012/math-166-texas-a-m-university in Mathematics (M) at Texas A&M University.
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Date Created: 10/21/15
Math 166 Lecture Notes 44 Section 44 Systems of Linear Equations Undetermined and Overdetermined Theorem a Ifthe number of equations is greater than or equal to the number of variables in a linear system then one of the following is true i The system has no solution ii The system has exactly one solution iii The system has infinitely many solutions b lfthere are fewer equations than variables in a linear system then either i The system has no solution or ii The system has infinitely many solutions In other words there is not one unique solution Look back at the three cases for systems oftwo equations in two variables Compare each outcome solution to its rowreduced matrix Example 1 3x 2y 7 2x 4y 10 Solution 3 1 the lines intersect in a single point 103 011 This gives us a distinct solution Example4 X2y 3 2x 3y 8 x 4y 9 The resulting rowreduced Example 2 3x 2y 6 6x 4y 12 In nite Solutions this is really the same line 1 667 2 0 0 0 This leaves us with 1 equation in 2 variables Note It is NOT the row of zeros that leads us to our nal conclusion 1 0 1 matrix is 0 1 2 0 0 0 Example 3 3x 2y 6 6x 4y 8 No Solution these are parallel lines 1 667 0 0 0 1 This gives us a row of zeros with a nonzero number at the end Note This type of row does lead us to our conclusion lfthere is a row in the augmented matrix containing all zeros to the left and right of the line then it is as if that row could be completely ignored as it represents 0 0 which by itself is inconclusive Whether or not there are solutions depends on the remaining rows Note that if you ignore the last row the matrix is gives us the solution x 1 and y 2 Example 5 3x 9y 62 12 X 3y 22 4 2x 6y 42 8 1 3 2 4 The resulting rowreduced matrix is 0 0 0 1 0 0 0 0 Ifthere is a row in the augmented matrix containing all zeros to the left ofthe vertical line and a nonzero entry to the right of the line then the system has no solution In this example that row would represent the equation 0 1 which is FALSE The row of all zeros could be ignored and would be inconclusive On the other hand the row with the zeros and the 1 at the end is conclusive there is no solution Example 6 X2y4z2 x y22 1 1 0 0 0 The resultIng rowreduced matrIx Is 0 1 2 1 Since we have more variables than equations there is no unique solution We have a situation with in nite solutions How can we express this in a helpful way In this example the first row tells us that 1X 0 The second row tells us that y 22 1 Using our known value of x and the relationships from the second equation we de ne a pattern for the infinite solutions Note We can let the parameter t represent the y value and get the pattern 0 t Note We can let the parameter t represent the z valueand get the pattern 0 1 2t t Example 6I continued Use the general pattern forthe solutions 0 1 2t t to find the missing portions of these speci c solutions a 4 b 0 0 7 Example 7 x 2y z 3 2x y 22 2 X 3y 32 5 1 0 6 2 The resulting rowreduced matrix is 0 1 8 16 0 0 0 0 In this example removing the third row would leave you with fewer equations than variables so we know that there is not a unique solution There is no row to indicate a no solution situation 80 then we must have a situation with infinite solutions How can we express the solutions in a helpful way First translate the rows back into equations and then solve for all the other variables in terms of one of them Option 1 Express the parametric solution in the form t ft gt The result is t Option 2 Express the parametric solution in the form ft t gt The result is t Option 3 Express the parametric solution in the form ft gt t The result is t Example 8 Mr and Mrs Garcia have a total of 100000 to be invested in stocks bonds and a money market account The stocks have a rate of return of 12 per year while the bonds and the money market account pay 8 and 4 per year respectively They have stipulated that the amount invested in stocks should be equal to the sum of the amount invested in bonds and 3 times the amount invested in the money market account How should the Garcias allocate their resources if they require an annual income of 10000 from their investments Assign X y and z to the amount invested in stocks bonds amp the money market respect Writing the equations from the problem we get and after reformatting Xyz100000 X y z 100000 xy3z Xy3z0 12X 08y 042 10000 12X 08y 042 10000 1 0 1 50000 The resulting rowreduced matrix is 0 1 2 50000 0 0 0 0 After ignoring the row of all zeroes we have fewer equations 2 than variables 3 Therefore the solution will not be unique There is either no solution or infinite solutions Since there is no row indicating no solution there must be infinite solutions Translating the rows back into equations we have X z 50000 y 22 50000 If we let 2 t which seems to be the easiest of the options and solve for X and y we get the general solution 50000 t 50000 2t t Now in this reallife application we realize some restrictions We know they not investing negative amounts of money Therefore t must be nonnegative t 2 0 and 50000 2t must be nonnegative t 5 25000 So 0 s t 5 25000 There are in nitely many solutions so we couldn t list them all but we can give some examples But we must be careful to choose tso that 0 s t 5 25000 Suppose we choose 25000 for t What amounts would be invested in each category Amount in stocks Amount in bonds Amount in money market Suppose we choose 12000 for t What amounts would be invested in each category Amount in stocks Amount in bonds Amount in money market above
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