Math 304–504Linear AlgebraLecture 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 sys tem is a common solution of allequations in the system.Plenty of problems in mathematics and real 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 symbo l of equival ence ⇐⇒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 simplifiedso that it is clear how to s olve it.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 oftwo linear equations in two variables.Solve the 2nd equation for y, then eliminate y fromthe 3rd equation:x = y + 2y = z − 12y + z = 4x = y + 2y = z − 12(z − 1) + z = 4Simplify:x = y + 2y = z − 13z = 6The elimination is completed. Now the system iseasily solved by 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 equations:x − y = 22x − y − z = 3x + y + z = 6Solution: (x, y , z) = (3, 1, 2)Gaussian eliminationGaussian elimination is a modi fication of theelimination method that uses only so-calledelementary 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.Proposition Any elementary operation can beundone by applying another elementary operation.Theorem Applying elementary operations to asystem of linear equations does not change thesolution set of the system.Proof: It is clear that after an elementaryoperation we do not lose any sol uti on. Since theoperation can be undone by another elementaryoperation, neither we get any garbage solutions.Operation 1: multi ply the ith equation by r 6= 0.a11x1+ a12x2+ · · · + a1nxn= b1· · · · · · · · · · · ·ai1x1+ ai2x2+ · · · + ainxn= bi· · · · · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bm=⇒a11x1+ a12x2+ · · · + a1nxn= b1· · · · · · · · · · · ·(rai1)x1+ (rai2)x2+ · · · + (rain)xn= rbi· · · · · · · · · · · ·am1x1+ am2x2+ · · · + amnxn= bmTo undo the operation, multiply the ith equation by r−1.Operation 2: add r times the i th equation to thejth equation.· · · · · · · · · · · ·ai1x1+ ai2x2+ · · · + ainxn= bi· · · · · · · · · · · ·aj1x1+ aj2x2+ · · · + ajnxn= bj· · · · · · · · · · · ·=⇒· · · · · · · · · · · ·ai1x1+ · · · + ainxn= bi· · · · · · · · · · · ·(aj1+ rai1)x1+ · · · + (ajn+ rain)xn= bj+ rbi· · · · · · · · · · · ·To undo the operation, add −r times the ithequation to the jth equation.Operation 3: interchange the i th and jth equations.· · · · · · · · · · · ·ai1x1+ ai2x2+ · · · + ainxn= bi· · · · · · · · · · · ·aj1x1+ aj2x2+ · · · + ajnxn= bj· · · · · · · · · · · ·=⇒· · · · · · · · · · · ·aj1x1+ aj2x2+ · · · + ajnxn= bj· · · · · · · · · · · ·ai1x1+ ai2x2+ · · · + ainxn= bi· · · · · · · · · · · ·To undo the operation, apply it once
View Full Document
Unlocking...