The column space of A is the set of all vectors that are linear combinations of the columns of A. 3 -2 1] 1. Let A =|-2 -5 7 3 and u -8 6] 4 -2 and v= 1 3 7 a) Determine if u is in Nul (A). Could u be in Col A?

Consider the following vectors: 3 9. 6 -2 -5 -6 a = 8 1 1 9. For each of the following vectors, determine whether it is in span{a, 6. c}. If so, express it as a linear combination using a, b, and c as the names of the vectors above. 9. V1 = 9. -7 -15 13 -8 -17 18 -10 16

Suppose a1, a2, a3, a4, and ag are vectors in R°, A = (a1 | a2 | as | a4 | a5), and -5 4 rref(A) = 0 1 0 0 0 1 -2 2 1 1 a. Select all of the true statements (there may be more than one correct answer). OA. span{a1, a2, ag} = R OB. {a1, a2, a3} is a basis for R Oc. {a1, a2, a3, a4} is a basis for R3 OD. span{a1, a2, ag, a4} = R³ OE. {a1, a2, a3, a4} is a linearly independent set OF. {a1, a2, az} is a linearly independent set OG. {a1, az} is a linearly independent set OH. {a1, a2} is a basis for R3 OI. span{a1, a2} b. If possible, write a, as a linear combination of a, a2, and ag; otherwise, enter DNE. a4 = c. The dimension of the column space of A is and the column space of A is a subspace of d. Find a basis for the column space of of A. If necessary, enter a1 for a,, etc., or enter coordinate vectors of the form or . Enter your answer as a comma separated list of vectors. A basis for the column space of A is { e. The dimension of the null space of A is and the null space of A is a…

Let A = 10 - 6 03 - 4 and b = -39 2 11 - 1 - 28 Denote the columns of A by a₁, a2, a3, and let W = Span {a₁, a₂, a3}. a. Is b in {a₁, a₂, a3}? How many vectors are in {a₁, a2, a3}? b. Is b in W? How many vectors are in W? c. Show that a2 is in W. [Hint: Row operations are unnecessary.] a. Is b in {a₁, a₂, a3}? Select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice. A. Yes, b is in {a₁, a2, a3} since b=a (Type a whole number.) B. No, b is not in {a₁, a2, a3} since b is not equal to a₁, a2, or a3. C. Yes, b is in (a₁, a2, a3} since, although b is not equal to a₁, a₂, or a3, it can be expressed as a linear combination of them. In particular, b = (Simplify your answers.) аз. D. No, b is not in {a₁, a2, a3} since it cannot be generated by a linear combination of a₁, a2, and ) a₁ + a₂ + ( ) a3.