Review• We have a standardized recipe to find al l solutions of systems such as:3x2−6x3+6x4+4x5= −53x1−7x2+8x3−5x4+8x5= 93x1−9x2+12x3−9x4+6x5= 15• The c omputational part is t o start with t h e augmented matrix0 3 −6 64 −53 −7 8 −5 8 93 −9 12 −9 6 15,and to calculate its reduced echelon form (which is u nique!). Here:1 0 −2 3 0 −240 1 −2 2 0 −70 0 0 0 1 4.• pivot variables (or basic variabl es): x1, x2, x5free variables: x3, x4• solving each equation for the pivot variables in terms of the free variables:x1−2x3+3x4= −24x2−2x3+2x4= −7x5= 4x1= −24 + 2x3− 3x4x2= −7 + 2x3− 2x4x3freex4freex5= 4Ar min [email protected] of existe nce and uniquenessThe question whether a system has a solution and whether it is uniq ue, is easier toanswer than t o determine the solution set.All we need is an ec helon form of the augmented matrix.Example 1. Is th e following system consistent? If so, does it have a unique solution?3x2−6x3+6x4+4x5= −53x1−7x2+8x3−5x4+8x5= 93x1−9x2+12x3−9x4+6x5= 15Solution. In the course of an e arlier examp le, we obtaine d the echelon form:3 −9 12 −96 150 2 −4 4 2 −60 0 0 0 1 4Hence, it is consiste nt (imagine doing back-substitution to get a solution).Theorem 2. (Existence and uniqueness theorem) A linear system is consistent ifand only if an echelon form of t he augmented matrix has no row of the form[0 ...0 b],where b is nonzero.If a linear system is consistent, then the solutions consist of either• a uniq ue solution (when there are no free v ariables) or• infinitely many solutions (when there is at least one fre e variable).Example 3. For wh at values of h will the following system be consistent?3x1−9x2= 4−2x1+6x2= hSolution. We perform row reduction to find an echelon form:3 −9 4−2 6 h>R2→R2+23R1"3 −9 40 0 h +83#The system is consistent if and only if h = −83.Ar min [email protected] summary of what we learn ed s o far• Each linear system corresponds to an augmente d matrix .• Using Gaussian elimination (i.e. row reduction to echelon form) on the augmentedmatrix of a linear system, we ca n◦ read off, whether the system has no, one, or infinitely many solutions;◦ find all sol utions by back-substitution.• We can continue row reduction to the reduced echelon form.◦ Solutions to the linear system can now be just read off.◦ This form is unique!Note. Besides for solving linear systems, Gaussian elimination has other important uses,such as computi ng determina nts or inverses of matrices.A recipe to solve linear systems (Gauss–Jordan elimination)(1) Write the augmented matrix of the system.(2) Row reduce to obtain an equivalent augmented matrix in echelon form.Decide whether the syste m is consistent. If not, stop; otherwi se go to the next step.(3) Continue row reduction to obtain the red uced echelon form.(4) Express this fin al matrix as a system of equ ations.(5) Decl are the free variables and state the solution in terms of these.Ques tions to check our understanding• On an exam, you are asked to find all solutions to a system of linear equations. Youfind exactly two solutions. Sho uld you be worried?Yes, because if there is more than one solution, there have to be infinit ely many solutions. Canyou see how, given two s olutions, one can construct infinitely many more?• True or false?◦ There is no more than one pivot in any row.True, because a pivot is the first nonzero entry in a row.◦ There is no more than one pivot in any column.True, because in echelon form (that’s where pivots are defined) the entrie s below a pivothave to zero.◦ There cannot be more free variables than pivot variables.False, consider, for instance, the augmented matrix [1 75 3].Ar min [email protected] geometry of linear equationsAdding a nd scaling vectorsExample 4. We have a lready encountered matrices such as1 4 2 32 −1 2 23 2 −2 0.Each column is wh at we call a (column) vector.In this example, each column vector has 3 entries and so lies inR3.Example 5. A fundamental property of vectors is that vectors of the same kind can beadded an d sca led.123+4−12=515, 7 ·x1x2x3=7x17x27x3.Example 6. (Geometric description of R2) A vectorx1x2represents the point(x1, x2) in the plane.Givenx =13and y =21, graph x, y, x + y, 2y.01234012340123401234Ar min [email protected] a nd scaling vectors, the most general thing we can do is:Definition 7. Given vectors v1, v2,, vmin Rnand scalars c1, c2,, cm, the vectorc1v1+ c2v2++ cmvmis a linear co mbination of v1, v2,, vm.The scalarsc1,, cmare the coefficients or weights.Example 8. Linear combinations of v1, v2, v3include:• 3v1− v2+ 7v3,• v2+ v3,•13v2,• 0.Example 9. Express15as a linear combination of21and−11.Solution. We have to find c1and c2such thatc121+ c2−11=15.This is the same as:2c1−c2= 1c1+c2= 5Solving, we find c1= 2 and c2= 3.Indeed,221+ 3−11=15.Note th at the augmented matrix of the linear system is2 −1 11 1 5,and that this example provides a new way of think about t his system.Ar min [email protected] row and column pictureExample 10. We can thi nk of the linear system2x − y = 1x + y = 5in two different geometric ways.Row picture.Each equation defines a li ne in R2.Which points lie on the intersection of these lines?Column picture.The system can be wri tte n as x21+ y−11=15.Which l inear combinations of21and−11produce15?This ex ample has the unique solution x = 2, y = 3.• (2, 3) is the (only) intersection of the two lines 2x − y = 1 and x + y = 5.• 221+ 3−11is the (only) linear combination producing15.Ar min
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