MATH 311Topics in Applied MathematicsLecture 1:Systems of linear equations.Linear equationThe equation 2x + 3y = 6 is called linearbecause its solution set is a line in R2.A solution of the equation is a pair of numbers(α, β) ∈ R2such that 2α + 3β = 6.For example, (3, 0) and (0, 2) are solutions.Alternatively, we can write the first solution asx = 3, y = 0.xy2x + 3y = 6General equation of a line: ax + by = c,where x, y are variables and a, b, c are constants(except for the case a = b = 0).Definition. A linear equation in variablesx1, x2, . . . , xnis an equation of the forma1x1+ a2x2+ · · · + anxn= b,where a1, . . . , an, and b are constants.A solution of the equation is an array of numbers(γ1, γ2, . . . , γn) ∈ Rnsuch thata1γ1+ a2γ2+ · · · + anγn= b.System of linear equationsa11x1+ a12x2+ · · · + a1nxn= b1a21x1+ a22x2+ · · · + a2nxn= b2· · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bmHere x1, x2, . . . , xnare variables and aij, bjareconstants.A solution of the system is a common solution of allequations in th e system.Plenty of problems in mathematics and real worldrequire solving syste ms of linear equations.Problem Find the point of int ersection of the linesx − y = −2 and 2x + 3y = 6 in R2.x − y = −22x + 3y = 6⇐⇒x = y − 22x + 3y = 6⇐⇒x = y − 22(y − 2) + 3y = 6⇐⇒x = y − 25y = 10⇐⇒x = y − 2y = 2⇐⇒x = 0y = 2Solution: the lines inte rsect at the point (0, 2).Remark. The symbol of equivalence ⇐⇒means that two systems have the same solutions.xyx − y = −22x + 3y = 6x = 0, y = 2xy2x + 3y = 22x + 3y = 6inconsistent system(no solutions)xy4x + 6y = 122x + 3y = 6⇐⇒ 2x + 3y = 6Solving systems of linear equationsElimination method always works for systems oflinear equations.Algorithm: (1) pick a variable, solve one of theequations for it, and eliminate it from the otherequations; (2) put aside t he equation used in theelimination, and retu rn to step (1).The algorithm reduces the number of variables (aswell as the number of equations), hence it stopsafter a fin ite number of steps.After the algorithm stops, the system is simplifiedso that it should be clear how to complete solution.Example.x − y = 22x − y − z = 3x + y + z = 6Solve t he 1st equation for x:x = y + 22x − y − z = 3x + y + z = 6Eliminate x from the 2nd and 3rd equations:x = y + 22(y + 2) − y − z = 3(y + 2) + y + z = 6Simplify:x = y + 2y − z = −12y + z = 4Now the 2nd and 3rd equations form the system of two linearequations in two variables.Solve t he 2nd equation for y:x = y + 2y = z − 12y + z = 4Eliminate y from the 3rd equation:x = y + 2y = z − 12(z − 1) + z = 4Simplify:x = y + 2y = z − 13z = 6The elimination is completed. Now the system is easily solvedby back substitution.That is, we find z from the 3rd equation, thensubstitute it in th e 2nd equation and find y, thensubstitute y and z in the 1st equation and find x.x = y + 2y = z − 1z = 2x = y + 2y = 1z = 2x = 3y = 1z = 2System of linear equations:x − y = 22x − y − z = 3x + y + z = 6Solution: (x, y , z) = (3, 1, 2)Another example.x + y − 2z = 1y − z = 3−x + 4y − 3z = 14Solve t he 1st equation for x:x = −y + 2z + 1y − z = 3−x + 4y − 3z = 14Eliminate x from the 3rd equations:x = −y + 2z + 1y − z = 3−(−y + 2z + 1) + 4y − 3z = 14Simplify:x = −y + 2z + 1y − z = 35y − 5z = 15Solve t he 2nd equation for y:x = −y + 2z + 1y = z + 35y − 5z = 15Eliminate y from the 3rd equations:x = −y + 2z + 1y = z + 35(z + 3) − 5z = 15Simplify:x = −y + 2z + 1y = z + 315 = 15The elimination is completed.Here z is a free variable. It can be assigned anarbitrary value. Then y and x are found by backsubstitution.z = t, a parameter;y = z + 3 = t + 3;x = −y + 2z + 1 = −(t + 3) + 2t + 1 = t − 2.System of linear equations:x + y − 2z = 1y − z = 3−x + 4y − 3z = 14General solution:(x, y , z) = (t − 2, t + 3, t), t ∈ R.In vector form, (x, y, z) = (−2, 3, 0) + t(1, 1, 1).The set of all solutions is a line in R3passingthrough the point (−2, 3, 0) in the direction(1, 1, 1).Gaussian eliminationGaussian elimination is a modification of theelimination method that uses only so-calledelementary operations.Elementary operations for systems of linear equations:(1) t o multiply an equation by a nonzero scalar;(2) t o add an equation multiplied by a scalar toanother equation;(3) t o interchange two equations.Theorem Applying elementary operations to asystem of linear equations does not change thesolution set of t he system.Example.x − y = 22x − y − z = 3x + y + z = 6Add −2 times the 1st eq uation to the 2nd equation:x − y = 2y − z = −1x + y + z = 6Add −1 times t he 1st equation to the 3rd equation:x − y = 2y − z = −12y + z = 4Add −2 times the 2nd equation to the 3rd equation:x − y = 2y − z = −13z = 6The elimination is completed, and we can solve thesystem by back substitution. However we may aswell proceed with elementary operations.Multiply t he 3rd equation by 1/3:x − y = 2y − z = −1z = 2Add the 3rd equation to the 2nd equ ation:x − y = 2y = 1z = 2Add the 2nd e quation to the 1st equation:x = 3y = 1z = 2System of linear equations:x − y = 22x − y − z = 3x + y + z = 6Solution: (x, y , z) = (3, 1,
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