Question

Question 3) (8 points) Consider the following matrix: A= ſi 4 0 0 28 3 12 2 11 -5 5 6 0 8 1 (a) Find a basis for the Rowspace(A). Then state the dimension of the Rowspace(A). (b) Find a basis for the Colspace(A). Then state the dimension of the Colspace(A). (e) Find a basis for the Nullspace(A). Then state the dimension of the Nullspace(A). (d) State and confirm the Rank-Nullity Theorem for this matrix.

Answer 1

Solution : 4 2 1 A = 0 -5 5 2 6 3 12 8 1 Reducing the matrix A into Reduced Row Echelon Form (RREF) by elementary row operations 2 ܚܝ 4 2 1 ܟ - 5 5 -ss R3 R3 -2Rq 6 Ro Rq-3 R 2 8 O 2 -2 0 3 12 2 .2 ܝ 1 4 2 27 1 q 2 1 0 5 5 Pr 3 R2 1 0 0 2 - -2 -2 o 2-2 09 2 2 -2 - 9 . 2 & 2 2 ܝܝ O 1 -1 R3 R3 - 2 R2 Rq. Rq. - 2 R2 D 2 O 0 -2 -2 o 2 0 0 0

(a) The row operations do not change the row space. The non-zero rows in RREF are linearly independent. Therefore, a basis for row space iy given by {(1,4,2,1), 0,0,1,-1)} Angwer. dimension of Row space (A) = 2 (6) From the RREF of A, we see that first and third conlumns are pivert columns. Hence first and third columns of the original matrix A forrm basis four column space, ib. 1 2 -5 { A Anywhe dimension of colspace (A) =2

(c) het x = € Null space of A, then 558 xq AX= | 4 2 1 OG SO G 24 26, +4252 + 2x2 + 364 50 xz - Xq=0 OC3 - oce and G + 4 X + 2 xq + 4 =0 x + 4 X2 + 3 xq=0 = G = - 4X2-30 -4.32 -3.369 3-2- 1. X2 +0.xq X=0. J + 1. xq 24 = 0.32 +2.32

.& 3 il . 588 X + X4 Mence, a basis for Nullspace (A) is given by 01:07] Arswer. dimension of Nullsbace (A) = 2 (d) Rank- nullity Theorem for a matrix A Rank (A) + nullity (A) = n Here A is AX & matrix is. Rank (A)=2 and nullity (A) = 2 is. Rank (A) + nullity (A) = 2+2=4 n=q