Question

_Determine which of these sets is a basis for R3. 1 - 1 ^ {[:] 7] [8]} - {[7] | c{[7] [8] [8]} » {[![10]} Determine the dimension of the subspace W = span {V1, V2, V3, V4} where A. dim(W)=1 B. dim(W) = 2 C. dim(W) = 3 D. dim(W) = 4 8 Determine both the rank and nullity of the matrix A= [1 0 1 | 2 -3 4 -2 -2 2 -4 1 0 3 -4 2 2 1 0 1 2 1 1 -1 0 0 A. rank(A) = 1 and null(A) = 5 C. rank(A) = 4 and null(A) = 3 E. rank(A) = 3 and null(A) = 3 B. rank(A) = 4 and null(A) = 2 D. rank(A) = 3 and null(A) = 2 F. rank(A) = 3 and null(A) = 4

Answer 1

solutions 5. For {(8). (),(9) s hat abertas met a (9)-(2)+($) can war we can express c) as a linear combination canapel). of others. Fศ 13 facile {(1).(+), (3)} ..4(1) + ( 1 ) +3 (3) - let for scalars a a, (2, C3 / 2 To W Then reitg + 2 63=0 4-202 = 0 . 20 +2+363 =0 Therefore the above set hence it is a basis of , the only sopa B 4,292 = (2=0. is linearly independent, R². Fore. { l'), (2), (3)} We have ? (12) 42 (2) = (36). Hence it is nola Dans Fors. {(:), (*). (1) are linearly dependents as so it does not form a basis 7. Wa span{ (1), (4), (7), (3)}. Among these forces vectors the first one can be written as a linear combination of the others as (-)= ( 1-4 (71)+0(1)

But the remaining weetors are linearly independent since the corresponding determinant is II- 121 T T - 110+8)+1(0+2)+2(8-2) 2 8 0 8+2+12=2270 Hence dimension of wis 3 ce dim (W) = 3. o' A= D 4 -4 -4 0 -1 1 1 -2 1 2 1 0 12 -2 0 2 2 0 Lave to find the rank of A we apply row operation on it 11 - 3 2 3 1 li / Toinen -3 2 3 1 1 1 4 -4 -4 0-1 R3-K3-Ki Lo 4 -4 -4,0 -1 11 -2 2 1 ó e lo 1 -1 -1 0 -1 662-2 o 22 o J R = Rq-22 to 4 -4 -4 0 -2 IRQ = RZ w am 3. 11 Telese -3 2 3 11 o 1 -1 0-44, R3-13 *2 1 0 1 -1 -1 0 -14 0 0 0 0 0 -3/4 / f o l 0 1 -1 -1 0 - ooooi - R&= R14 R2 o 4 -4 -4 0-2 1 - 4 / R3 = R' 1 1 - 3 2 3 1 iloise I l Rå=RAAR2 / 1-3 79.1 Loo dioo CoIw ܝܢ ܘ_ܘ_ܘ Number of non zero rows is 3. so rank (1) 33. since A is 4x6 matrix by rank- nulity theorem 6 = rank (49 + neitity () a Nality (A) = 6-3=3. © tank (A) = 3 and null (A)= 3.