Question

7. Consider the following matrices 2 3-1 0 1 A=101-2 3 0 0 0-1 2 4 2 3 -1 B-101-2 0 0-1 2 3 -1 0 c=101-2 3 For each matrix, determine (a) The rank. (b) The number of free variables in the solution to the homogeneous system of equa- tions (c) A basis for the column space d) A basis for the null space for matrices A and HB e) Dimension of the column space (f) Nullity (g) Does your basis for the column space of B span R3? Explain (h) For each matrix, wil there be a vector b in IR3 for which there is no solution to Ax-b (and correspondingly for the other two matrices)?

Answer 1

7. The RREF of A is

1	0	0	½	21/2
0	1	0	-1	-8
0	0	1	-2	-4

The RREF of B is I₃.

The RREF of C is

1	0	5/2	-9/2
0	1	-2	3
0	0	0	0

(a). The rank of a matrix equals the no. of non-zero rows in its RREF. Hence, rank(A) = 3, rank(B) = 3 and rank(C) = 2.

(b).In case of A, if X = (x,y,z,w,u)^T, then the equation AX = 0 is equivalent to x+w/2+21u/2 = 0 or, x =             -w/2,-21u/2, y-w-8u = 0 or, y = w+8u and z-2w-4u = 0 or, z = 2w+4u so that X = (-w/2-21u/2, w+8u, 2w+4u,w,u)^T. Thus, there are 2 variables. In case of B, if X = (x,y,z)^T, then the equation BX = 0 is equivalent to x = 0,y= 0, z = 0 so that there is only a trivial solution. Thus, there are no variables. In case of C, if X = (x,y,z,w)^T, then the equation CX = 0 is equivalent to x+5z/2-9w/2 = 0 or, x = -5z/2+9w/2 and y -2z+3w = 0 or, y = 2z-3w so that X = (-5z/2+9w/2, 2z-3w,z,w)^T. Thus, there are 2 variables.

( c). It is apparent from the RREF of the 3 matrices that in case of A, the basis of col(A) is {(2,0,0)^T, (3,1,0)^T , (-1,-2,-1)^T}. In case of B, the basis of col(B) is{(2,0,0)^T, (3,1,0)^T , (-1,-2,-1)^T}. In case of C, the basis of col(C) is {(2,0,0)^T,(3,1,0)^T}.

(d). Null (A) is the set of solutions to the equation AX = 0. From part (b), we have X = (-w/2-21u/2, w+8u, 2w+4u,w,u)^T = w/2(-1,2,4,2,0)^T+u/2(-21,16,8,0,2)^T. Hence, {(-1,2,4,2,0)^T,(-21,16,8,0,2)^T } is a basis for Null(A). In case of B, the equation BX = 0 has only the trivial solution so that there is no basis for Null(B).

(e).dim(col(A)) = dim(col(B)) = 3 and dim(col(C)) = 2.

(f). Nullity of A is 2, nullity of B is 0, nullity of C is 2.

(g).The RREF of B is I₃. Hence the basis for col(B) spans R³.