To determine whether a set of n vectors from R" is independent, we can form a matrix A whose columns are the vectors in the set. If the vectors are linearly independent, what will we see in the reduced row echelon form of A? O None of these O An identity matrix O A row of all zeros O A column of all zeros

Let A = 3 - 3 12 - 12 does have a solution. and b = b2 Show that the equation Ax = b does not have a solution for some choices of b, and describe the set of all b for which Ax = b How can it be shown that the equation Ax = b does not have a solution for some choices of b? A. Find a vector x for which Ax=b is the identity vector. B. Row reduce the augmented matrix A b to demonstrate that [a b] has a pivot position in every row. C. Find a vector b for which the solution to Ax=b is the identity vector. D. Row reduce the matrix A to demonstrate that A does not have a pivot position in every row. E. Row reduce the matrix A to demonstrate that A has a pivot position in every row. Describe the set of all b for which Ax=b does have a solution. The set of all b for which Ax = b does have a solution is the set of solutions to the equation 0 = (Type an integer or a decimal.) b₁ + b₂.

h 9 a ³1² ²₂/+1² 2+1² c d f Express Select one: This cannot be done for all values of a, b, c, d, e, f, g, h as there 8 of these but only 4 unknowns for the entries of the single matrix Oh+a b+e c+d d+fl Oh+2a g+b+e 3c 2d+2f Oh+a_g+b+e C d+f e as a single determinant in a manner that works for all a, b, c, d, e, f, g, h = R. We use | | to denote the determinant. None of the others apply

3 4 3 3. Compute AB if A = -2 and B = -3 -1 1 -6 (b) Show that First Row of AB is linear combination or of rows of B c) Show that Second column of AB is linear combination of clomns of A