Question

4. Consider the matrix [1 0 01 A- 1 0 2-1and the vector b2 (a) Construct the augmented matrix [Alb] and use elementary row operations to trans- form it to reduced row echelon form. (b) Find a basis for the column space of A. (c) Express the vectors v4 and vs, which are column vectors of column 4 and 5 of A, as linear combinations of the vectors in the basis found in (b) (d) Find a basis for the null space of A (e) Find a particular solution to the linear system Ax b. (f) Find the general solution to the linear system Ax b. (g) Can you find a vector c such that Ax - c has no solution? Give your reason

Answer 1

Augmented matrix [A/b] = [1 2 0 1 1 0 0 1 1 2 3 1 - 1 1 1 2 5] for Row Reduced Echelon form, the element which is circled represents the pivot element Row 1, Row 2 and Row 3 are referred to a R_1, R_2, R_3 here on First pivot element-1 [1 1 2 0 1 1 0 1 1 1 2 3 1 - 1 1 1 2 5] R_2 = R_2 - R_1 R_3 = R_3 - 2 R_1 Rightarrow [1 0 0 0 1 1 0 0 1 1 1 1 1 - 25 - 1 1 1 3] Second pivot element-1 R_3 = R_3 - R_2 R_1 = R_1 - 0. R_2 [1 0 0 0 1 0 0 0 1 1 1 0 1 - 2 + 1 1 1 2] Third pivot element-1 R_` = R_1 - 0.R_3 R_2 = R_2 - 0.R_3 [1 0 0 0 1 0 0 0 1 1 1 0 1 - 2 1 1 1 2] This is Row reduced Echelon form \r\n The Row reduced Echelon form has pivot elements in column 1, 2, 3 so the corresponding columns of the original matrix represents basis for column space Basis for column space are [1 1 2], [0 1 1], [0 0 1]