MATH 235 Final: MATH 235 UMass Amherst solution-practice-final06
Document Summary
Spring 2008: (20 points) you are given below the matrix a together with its row reduced echelon form. 0: determine the rank of a, dim(ker(a)), and dim(im(a)). Explain how these are de- termined by the matrix b. Answer: rank(a) = number of pivots in b = 3. dim(ker(a)) = number of free variable = 6 3 = 3. dim(im(a)) = rank(a) = 3: find a basis for the kernel ker(a) of a. 2x3 x4 x5: find a basis for the image im(a) of a. Answer: the variables x3, x4, and x5 are free. Expressing the basic variables in terms of the free variables, we get that the general solution is: x3~v1 + x4~v2 + x5~v3. The vectors ~v1, ~v2, ~v3 are clearly linearly independent, and so a basis of ker(a). x3 x4 x5. , a2 = a1 = : does the vector b := . Answer: the pivot columns of a are the rst, second, and sixth, so computations.
