Geometry of Span (continued) The Plane Spanned by u and v

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Geometric Description of Span Geometr of Span (contined) 2 Geometr of Span (contined) 2 Span {} Span {, } 2 Span {} 2 Geometr of Span (contined) 2 b + 2 The Plane Spanned b and If a plane is spanned b ectors and, then all possible ectors b located in this plane can be epressed as b + where and are arbitrar scalars. We sa that b is comprised of a linear combination of the ectors and. b + Can Be Epressed as a Matri Vector Prodct b + + + +

This Eqialence Can Be Used to Determine if Some Other Vector c Lies in the Plane Spanned b and c c c -2 - - Eample - 8 Soltion: Redce The Agmented Matri piot - - piot -2-8 - 2 - -8 This eqation poses the qestion: Can coefficients and be fond sch that + c? c Note: This sstem has three eqations in onl two nknowns, and. The third eqation has no soltion, so c is not in the plane! - - 2-2 Let s Retrn to the Epression of b + as a Matri Vector Prodct b + In plain langage, this means that a matri-ector prodct is eqialent to a linear combination of the colmns of the matri. + + + With the Matri Vector Prodct A b Defined B m rows of matri A a a 2 a n a 2 a 22 a 2n a m a m2 a mn a a 2 a n n colmns of matri A b b 2 b m n b We hae the Powerfl Theorem: Matri eqation A b has the same soltion as the ector eqation a + a 2 + + n a n b which has the same soltion as the sstem of linear eqations whose agmented matri is [a a 2 a n b]

Homogeneos Sstems Definition: A homogeneos sstem can be written as A. Soltions The triial soltion: (all entries of are ero). This soltion alwas eists for this case. The non-triial soltion: ( can hae some ero entries as long as not all of its entries are ero). Sch a soltion ma not alwas eist for this case. Eample : Determine if the Following Homogeneos Sstem A Has a Nontriial Soltion + - - - 2 + 6 + - 8 - - -2 6-8 Strateg: Row-redce the agmented matri to redced echelon form. piot Row-Redction Process - - - -2 6-8 -9 - - piot 86 piot End of the Row-Redction Process - - There is Indeed a Nontriial Soltion: free Eample 2: Determine if the Following Single Homogeneos Eqation Has a Nontriial Soltion - - 2 free - 89 A nontriial soltion eists: an scalar mltiple times the ector [/,, ]. This is a line in -D space that passes throgh the origin (i.e., when ), which in fact is the triial soltion. Strateg: Sole for the basic ariable in terms of and, which are treated as free ariables.

Sole for in Terms of the Free Variables and. +.2. +.2..2 + + is the plane throgh the origin in -D spanned b the ectors and. Note that and are an two real nmbers s and t. Parametric Vector Form of the Soltion The soltion s + t is the parametric ector eqation of a plane throgh the origin. t 2 s 2 s + t Generaliation The soltion set of the homogeneos eqation A can alwas be epressed as the Span{, 2,..., p } for a sitable set of ectors. The triial soltion is the ector. If a nontriial soltion eists and the sstem has onl one free ariable, then the soltion set is a line throgh the origin. For two free ariables, the nontriial soltion is a plane throgh the origin. Eample : Repeat Eample, Bt for A b (Nonhomogeneos Sstem) + - 7 - - 2 + - - 7 - -2 - piot Row-Redction Process - 7 - -2-6 -8 - - 7 6-9 -8 piot piot End of the Row-Redction Process 7 - - - 2 2 6 + - 8-6 -8 - was [,, ] T in Eample Strateg: Again row-redce the agmented matri to redced echelon form. 7-2 - 7 6 - + 2 free - - 2 97

The Soltion Is: Parametric Vector Form General Soltion w of the Arbitrar Nonhomogeneos Sstem A b - + 2-2 p + - 2 + t p a soltion of the inhomogeneos sstem A b w p + h an soltion of the homogeneos sstem A This soltion is the same line as that of Eample with the addition of the constant ector p [-, 2, ]. p + t, where t is a real nmber. assming that A b has at least one nonero soltion. To Verif That w Soles A b, Jst Sbstitte: A w A ( p + h ) A w A p + A h p is a soltion of A b h is a soltion of A b + b