MAT223H1 Final: Exam note.docx
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Solve for variables (x1, x2, . xn) column the last last entry of the equation has a solution. If the last column last entry of the equation is just a zero > free variable > inf. solution. Transformation if: u,verm x2 v2 u2 u3],v=[ v1 v3] U1+u2+u3]=ct ([u1 u3]) =ct(u) u2 ( u1+u2 +u3)+( v1+v2+v3)]=t[u1 (2u1+3u3)+(2v1+3v3) u3]+t[v1 v3]=t (u )+t (v) u2 v2. Cannot be this: variable in every row of an echelon form of. Linear transformation t: rm >rn leading variable a is called 1 1 in every if for all u,verm, Ab ii) ab=0 doesn"t mean a=b=0 case of n doesn"t x equal n matrix: 1: place the augmented matrix in ref to get. C] where b = in and a 1 = c there exists invertible the an n x m matrix b such and onto. Ker(t): t: rm >rn and t(x) = 0 ax=t(x) zero in vector subspace and and if:
