Solutions to ExercisesCollege of the RedwoodsMathematics DepartmentMath 45—Linear AlgebraQuiz #2—Linear AlgebraDavid ArnoldCopyrightc 2000 [email protected] Revision Date: September 11, 2001 Version 1.002Exercise 1. Use elimination and back substitution to solve the fol-lowing system of equations.x1+2x2− 3x3= −122x1+6x2+ x3= −7−x1+2x2− x3= −8Exercise 2. What matrices E21, E31,andE32put the coefficientmatrix in Exercise 1 into triangular form U? Multiply those E’s toget one matrix M that does elimination: MA = U.Solutions to Exercises 3Solutions to ExercisesExercise 1. Using the coefficients and the answers of the rows tocreate an augmented matrix,A=12−3 −12261 −7−12−1 −8Use the pivot in row 1 to cancel out the pivots in rows 2 and 3. First,subtract 2 times row 1 from row 2, then add 1 times row 1 to row 3:R2− 2R1, R3+ R1.12−3 −1202 7 1704−4 −20Solutions to Exercises 4Now, use the pivot in row 2 to cancel out the pivot in row 3: R3−2R2.12 −3 −1202 7 1700−18 −54This augmented matrix represents the following system of equations.x1+2x2− 3x3= −122x2+7x3=17−18x3= −54Use back substitution to solve for x1,x2,x3. First, solve the thirdequation above for x3.−18x3= −54x3=3Solutions to Exercises 5Next, plug x3into equation 2 and solve for x2.2x2+7x3=172x2+ 7(3) = 172x2+21=172x2= −4x2= −2Now, plugging in x2and x3into equation 1, we can solve for x1.x1+2x2− 3x3= −12x1+2(−2) − 3(3) = −12x1− 4 − 9=−12x1=1Exercise 1Solutions to Exercises 6Exercise 2. The elementary matrices associated with the elementaryrow operations in part 1 areE21=100−210001,E31=100010101, and E32=1000100 −21Multiply these matrices together to get the elimination matrix M .M = E32E31E21=1000100 −21100010101100−210001=1000100 −21100−210101=100−2105 −21Exercise
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