Math 304–504Linear AlgebraLecture 3:Gauss-Jordan reduction.Applications of systems of linear equations.System of linear equations:a11x1+ a12x2+ ··· + a1nxn= b1a21x1+ a22x2+ ··· + a2nxn= b2·········am1x1+ am2x2+ ··· + amnxn= bmCoefficient matrix (m ×n) and column vector of theright-hand sides (m × 1):a11a12. . . a1na21a22. . . a2n............am1am2. . . amnb1b2...bmSystem of linear equations:a11x1+ a12x2+ ··· + a1nxn= b1a21x1+ a22x2+ ··· + a2nxn= b2·········am1x1+ am2x2+ ··· + amnxn= bmAugmented m × (n + 1) matrix:a11a12. . . a1nb1a21a22. . . a2nb2...............am1am2. . . amnbmSolution of a system of linear equations splits intotwo parts: (A) elimination and (B) backsubstitution.Both parts can be done by applying a finite numberof elementary operations.Since the elementary operations preserve thestandard form of linear equations, we can trace thesolution process by looking on the augmentedmatrix.In terms of the augmented matrix, the elementaryoperations are elementary row operations.Elementary row operations:(1) to multiply a row by some r 6= 0;(2) to add a row multiplied by some r ∈ R toanother row;(3) to interchange two rows.Remark. The rows are added and multiplied byscalars as vectors (namely, row vectors):a11a12. . . a1nb1a21a22. . . a2nb2...............am1am2. . . amnbm=v1v2...vm,where vi= (ai1ai2. . . ain|bi) is a row vector.Operation 1: to multiply the ith row by r 6= 0:v1...vi...vm→v1...rvi...vmOperation 2: to add the ith row multiplied by r tothe jth row:v1...vi...vj...vm→v1...vi...vj+ rvi...vmOperation 3: to interchange the ith row with thejth row:v1...vi...vj...vm→v1...vj...vi...vmThe goal of the Gaussian elimination is to convertthe augmented matrix into row echelon form:• all the entries below the staircase line are zero;• boxed entries, called pivot or lead entries, areequal to 1;• each circled star correspond to a free variable.Strict triangular form is a particular case of rowechelon form that can occur for systems of nequations in n variables:Matrix ofcoefficientsThe original system of linear equations is consistentif there is no leading entry in the rightmost columnof the augmented matrix in row echelon form.Inconsistent systemThe goal of the Gauss-Jordan reduction is toconvert the augmented matrix into reduced rowechelon form:• all entries below the staircase line are zero;• each pivot entry is 1, the other entries in itscolumn are zero;• each circled star correspond to a free variable.Example 1.x1+ 2x2+ 3x3+ 4x4= 10x2+ 2x3+ 3x4= 6Augmented matrix:1 2 3 4100 1 2 36The matrix is in row echelon form. To convert itinto reduced row echelon form, add −2 times the2nd row to the 1st row:1 0 −1 −2 −20 1 2 3 6x3and x4arefree variablesx1− x3− 2x4= −2x2+ 2x3+ 3x4= 6⇐⇒x1= x3+ 2x4− 2x2= −2x3− 3x4+ 6System of linear equations:x1+ 2x2+ 3x3+ 4x4= 10x2+ 2x3+ 3x4= 6General solution:x1= t + 2s − 2x2= −2t − 3s + 6x3= tx4= s(t, s ∈ R)Example 2.y + 3z = 0x + y − 2z = 0x + 2y + az = 0(a ∈ R)The system is homogeneous (all right-hand sidesare zeros). Therefore it is consistent(x = y = z = 0 is a solution).Augmented matrix:0 1 301 1 −201 2 a 0Since the 1st row cannot serve as a pivotal one, weinterchange it with the 2nd row:0 1 301 1 −201 2 a0→1 1 −200 1 301 2 a0Now we can start the elimination. First add −1times the 1st row to the 3rd row:1 1 −200 1 301 2 a0→1 1 −200 1 300 1 a + 20The 2nd row is our new pivotal row. Add −1 timesthe 2nd row to the 3rd row:1 1 −200 1 3 00 1 a + 2 0→1 1 −200 1 3 00 0 a − 1 0At this point row reduction is divided into two cases.Case 1: a 6= 1. In this case, multiply the 3rd rowby (a − 1)−1:1 1 −200 1 300 0 a − 10→1 1 −2 001 3 00 01 0The matrix is converted into row echelon form.We proceed towards reduced row echelon form.Add −3 times the 3rd row to the 2nd row:1 1 −200 1 3 00 0 1 0→1 1 −200 1 0 00 0 1 0Add 2 times the 3rd row to the 1st row:1 1 −200 1 000 0 10→1 1 000 1 000 0 10Finally, add −1 times the 2nd row to the 1st row:1 1 000 1 000 0 10→1 0 0 001 0 00 01 0Thus x = y = z = 0 is the only solution.Case 2: a = 1. In this case, the matrix is alreadyin row echelon form:1 1 −2 001 3 00 0 00To get reduced row echelon form, add −1 times the2nd row to the 1st row:1 1 −200 1 3 00 0 00→1 0 −5 00 1 3 00 0 0 0z is a free variable.x − 5z = 0y + 3z = 0⇐⇒x = 5zy = −3zSystem of linear equations:y + 3z = 0x + y − 2z = 0x + 2y + az = 0Solution: If a 6= 1 then (x, y, z) = (0, 0, 0);if a = 1 then (x, y, z) = (5t, −3t, t), t ∈ R.ApplicationsProblem 1 Find the point of intersection of thelines x − y = −2 and 2x + 3y = 6 in R2.x − y = −22x + 3y = 6Problem 2 Find the point of intersection of theplanes x − y = 2, 2x − y − z = 3, andx + y + z = 6 in R3.x − y = 22x − y − z = 3x + y + z = 6Method of undetermined coefficients often involvessolving systems of linear equations.Problem 3 . Find a quadratic polynomial p(x)such that p(1) = 4, p(2) = 3, and p(3) = 4.Suppose that p(x) = ax2+ bx + c. Thenp(1) = a + b + c, p(2) = 4a + 2b + c,p(3) = 9a + 3b + c.a + b + c = 44a + 2b + c = 39a + 3b + c = 4Traffic flow450 400610 640520 600Problem. Determine th e amount of trafficbetween each of the four intersections.Traffic flowx1x2x3x4450 400610 640520 600x1=?, x2=?, x3=?, x4=?Traffic flowA BCDx1x2x3x4450 400610 640520 600At each intersection, the incoming traffic has tomatch the outgoing traffic.Intersection A: x4+ 610 = x1+ 450Intersection B: x1+ 400 = x2+ 640Intersection C : x2+ 600 = x3Intersection D: x3= x4+ 520x4+ 610 = x1+ 450x1+ 400 =
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