Question

1. For the matrix A given below, find col(A) and Nul(A). Also determine if the given vector is in the column space, null space, both or neither. A = -2 -5 1 3 3 11 1 7 8 -5 -19 -13 0 1 7 5 -171 5 1 -3 1 5 1

Answer 1

Solution -2 o -17 A- 1 -5 8 3 - 5 -19 7 -13 5 s & - 3 7 5 -3 I) find Col (A) By row echelon form R₂ R₂ + 1 Ri Qк, - kg +3e, g eqək - 10, 2 2 - 5 Ú - 8 Y2 7/2 - 7 -9 -17 -7/2 -49/2 -23/ 7 5 Rzes R₃ -722 & Ruc Ru - 9 R2 -2 8 -5 12 0 - 1 0 0 -7/2 0 0 0 20 0 1-4 Ru - 5 8 -17. Y2 - 1 -712 0 0 J 0 20 0 0 0 0 So, The matrin has 3 pivots in Column 1, 2 & 4. So, these columns in the original mālsun define the column space of matron Lo, Column (A) 2 a )

2 Null (A) Nollspace of A= Solution set of Ax=0 Ax=0 -2 -5 co o 17. 24 - 3 11 5 U2 3 -5 -19 -13 0 0 - - 7 5 1 t 113 nu -3 15 o 60) By applying reduced echelen forom - ₂-RI 2 -4 0 17 2 INM= 5 3 -5 -19 -13 t 7 5 -3 REJ R - R ; R2 E-> R₂ -3R & Rę co Re - Re 5 -4 0 12 / -1 12 72 U - 72 -49 7 9/ -7 -9 5 - 23 R & Ro-52 0 -5 26 0 1/2 1 - -7/2 D ta 7 7/ % 0 -9 5 1/42-

R₂ to R₂ R3 - 7R2 & Ruc o Ru-9 R2 1 - 5 26 -7% -1 0 0 0 0 -4 20 R₂ ES 2R2 & Re Ry 5 - 26 -7 0 -2 2 0 -4 20 0 o 0 Ro 3 R₂ 4 1-5 O -2 2 26 - 7 -5 0 0 oo 0 RES Rt5R3 & R₂ R₂ - 2 0 0 -2 0 o 3 -5 0 0 o from (2) 2(B) 0 0 -2. 6 0 0 -5 13 0 0 0 o o nu My

4+2 +2=0 24 -2g - 25 12-2213 +385 = 0 X2 = 223 - 325 Ry - 50 =0 = xy = 529 There are two free variables Bu So, a Null A= (-12-15 조 2213-375 -3 Uz + ng M 0 5 505 ng 0 Noll A 2 2 2 -wowa I Determine the guien vector is in the column space, Wall Spaces, both or neether. int 2. Boy - 5 By definition, the column space of a maturx is the space of its columns; since matrin has 4 rows & vector has 5 Rows. So, it can't be expressed as a linear combination ie; An=1 8 Auxs asxi = v sxi { which can't be possible?

for a vector to be noll space. Av=o & we get put ß eq 0 1 0 - - 2 2 -1 o O O 0 -5 o 0 5 000 (a) -2 +0+1+0+1 = 0 -1-2+3= 0 5- 520 As (a) 91 o is in noll space. (a) Satisfies Av=o