MATH 304Linear AlgebraLecture 1:Systems of linear equations.Linear equationThe equation 2x + 3y = 6 is called linearbecause its solution set is a straight 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 sy s tem is a commo n solution of al lequations in the system.Plenty of problems in mathematics and r eal worldrequire solving systems of linear equations.Problem Find the point of intersection 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 intersect 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 the equation used in theelimination, and return to step (1).The algorithm reduces the number of variables (aswell as the number of equations), hence it stopsafter a finite number of steps.After the algorithm stops, the system is simpli fiedso that it should be clear how to complete solution.Example.x − y = 22x − y − z = 3x + y + z = 6Solve the 1st equation for x:x = y + 22x − y − z = 3x + y + z = 6Elimi nate 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 the 2nd equation for y:x = y + 2y = z − 12y + z = 4Elimi nate 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 the 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 equatio ns: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 the 1st equation for x:x = −y + 2z + 1y − z = 3−x + 4y − 3z = 14Elimi nate 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 the 2nd equation for y:x = −y + 2z + 1y = z + 35y − 5z = 15Elimi nate 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 equatio ns: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 straight line in R3passing through the point (−2, 3, 0) in the direction(1, 1,
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