Math 311-504Topics in Applied MathematicsLecture 6:Row echelon form (continued).Linear independence.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. . . amnb1b2...bmSystem of linear equations:a11x1+ a12x2+ · · · + a1nxn= b1a21x1+ a22x2+ · · · + a2nxn= b2· · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bmAugmented m × (n + 1) matrix:a11a12. . . a1nb1a21a22. . . a2nb2...............am1am2. . . amnbmSolution 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 operations for systems of linear equations:(1) to multiply an equation by a nonzero scalar;(2) to add an equation multiplied by a scalar toanother equation;(3) to interchange two equations.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...vmOperation 2: to add the ith row multiplied by r tothe jth row:v1...vi...vj...vm→v1...vi...vj+ rvi...vmOperation 3: to interchange the ith row with thejth row:v1...vi...vj...vm→v1...vj...vi...vmThe 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 leading entries,are nonzero;• each circled star corresponds 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 back substitution is to go from rowechelon form to reduced row echelon form (orsimply reduced form):• all entries below the staircase line are zero;• each leading entry is 1, the other entries in itscolumn are zero;• each circled star corresponds to a free variable.Example.x − y = 22x − y − z = 3x + y + z = 61 −1 022 −1 −131 1 16Row echelon form (also strict triangular):x − y = 2y − z = −13z = 61 −1 0 201 −1 −10 03 6Reduced row echelon form:x = 3y = 1z = 21 0 0 30 1 0 10 0 1 2Another example.x + y − 2z = 1y − z = 3−x + 4y − 3z = 11 1 −210 1 −13−1 4 −31Row echelon form:x + y − 2z = 1y − z = 30 = −131 1 −2 101 −1 30 0 0−13Reduced row echelon form:x − z = 0y − z = 00 = 11 0 −1 00 1 −1 00 0 0 1Yet another example.x + y − 2z = 1y − z = 3−x + 4y − 3z = 141 1 −210 1 −13−1 4 −314Row echelon form:x + y − 2z = 1y − z = 30 = 01 1 −2 101 −1 30 0 00Reduced row echelon form:x − z = −2y − z = 30 = 01 0 −1 −20 1 −1 30 0 0 0New example.x2+ 2x3+ 3x4= 6x1+ 2x2+ 3x3+ 4x4= 10Augmented matrix:0 1 2 361 2 3 410To obtain row echelon form, interchange the rows:1 2 3 4 1001 2 3 6The system is consistent. There are two freevariables: x3and x4.To obtain reduced row echelon form, add −2 timesthe 2nd row to the 1st row:1 2 3 4 1001 2 3 6→1 0 −1 −2 −201 2 3 6x1− x3− 2x4= −2x2+ 2x3+ 3x4= 6⇐⇒x1= x3+ 2x4− 2x2= −2x3− 3x4+ 6General solution:x1= t + 2s − 2x2= −2t − 3s + 6x3= tx4= s(t, s ∈ R)(x1, x2, x3, x4) = (t + 2s − 2, −2t − 3s + 6, t, s) == t(1, −2, 1, 0) + s(2, −3, 0, 1) + (−2, 6, 0, 0).a11x1+ a12x2+ · · · + a1nxn= b1a21x1+ a22x2+ · · · + a2nxn= b2· · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bmThe system is consistent if there is no leading entryin the rightmost column of the reduced augmentedmatrix. In this case, the general solution is(x1, x2, . . . , xn) = t1v1+ t2v2+ · · · + tkvk+ v0,where viare certain n-dimensional (row) vectorsand tiare arbitrary scalars.k = n − (# of leading entries)Definition. A subset S ⊂ Rnis called a hyperplane(or an affine subspace) if it has a parametricrepresentation t1v1+ t2v2+ · · · + tkvk+ v0,where viare fixed n-dimensional vectors and tiarearbitrary scalars.Hyperplanes are solution sets of systems of linearequations.The number k of parameters may depend on arepresentation. The hyperplane S is called ak-plane if k is as small as possible.Example. Suppose v2= rv1, where r ∈ R.Then t1v1+ t2v2+ v0= (t1+ rt2)v1+ v0.Hence t1v1+ t2v2+ v0and tv1+ v0are differentrepresentations of the same hyperplane.0-plane is a point.tv1+ v0is a 1-plane if v16= 0.t1v1+ t2v2+ v0is a 2-plane if vectors v1and v2arenot parallel.Thus 1-planes are lines, 2-planes are planes.Definition. Given vectors v1, v2, . . . , vk∈ Rnandscalars t1, t2, . . . , tk, the vectort1v1+ t2v2+ · · · + tkvkis called a linear combination of vectors v1, . . . , vk.The linear combination is called nontrivial if thecoefficients t1, . . . , tkare not all equal to zero.Proposition The following conditions are equivalent:(i) the zero vector is a nontrivial linear combinationof v1, . . . , vk;(ii) one of vectors v1, . . . , vkis a