MATH 235 Midterm: MATH 235 UMass Amherst midterm1-spring05-solution
Document Summary
Answer: the row reduction takes 5 elementary operations: 0: find the general solution for the system. Answer: x3 and x5 are free variables. x1 x2 x3 x4 x5. 2x3 + x5 + 2 x3 x5 + 1 x3. 1 x5: (18 points) let u1 = . Answer: let a be the matrix, whose columns are the vectors ui. For which value of h does the matrix a have a pivot in every row? . Row reduction yields that a is row equivalent to the following matrix in echelon form: 0 h 2 h 3 1 h (1) If h 6= 3, then we get a pivot in the (3, 3) position (third row and third column). If h = 3, then we get a pivot in the (3, 4) position. Thus, for every value of h, we get a pivot in every row. Consequently, the set {u1, u2, u3, u4} span the whole of.
