T(x1,x2,x3,x4)=(x3+x4,x2+x3,x2+x3,0) a. Is the linear transformation one-to-one? A. T is one-to-one because the column vectors are not scalar multiples of each other. B. T is not one-to-one because the columns of the standard matrix A are linearly independent. C. T is not one-to-one because the standard matrix A has a free variable. D. T is one-to-one because T(x)=0 has only the trivial solution. b. Is the linear transformation onto? A. T is onto because the standard matrix A does not have a pivot position for every row. B. T is not onto because the columns of the standard matrix A span R4. C. T is not onto because the fourth row of the standard matrix A is all zeros. D. T is onto because the columns of the standard matrix A span R4