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Linear algebra - general solutions to Ax=b

  1. Jul 25, 2010 #1
    Could someone check my working for this?

    1. The problem statement, all variables and given/known data

    (Assuming that t is Real).

    For a matrix, A =
    1 0 3 t
    0 1 t 0
    1 0 2 -2t
    2 0 -2 t2

    and a vector, b = [5 2t 1 t]T

    i) Find the rank of A and A|b for all values of t.
    ii) Find conditions on t so that Ax=b is consistent, Ax=b has a unique solution.
    iii) Find the general solution to Ax=b when t=-22, stating the dimension of the nullspace and a basis for it.

    2. Relevant equations
    Basic linear algebra?

    3. The attempt at a solution
    i) row reducing A|b ->
    [1 0 3 t .........|5]
    [0 1 t 0 .........|2t]
    [0 0 1 3t .......|4]
    [0 0 0 t(t+22) |t+22]

    rank(A) = 4 when t(t+22) =/= 0, so t =/= -22 and t =/= 0 and rank(A) = 3 if t = -22 or t = 0
    rank(A|b) = 4 when t =/= -22 and rank(A|b) = 3 if t = -22 or t = 0

    ii) A|b is consistent when t =/= 0
    and A|b has a unique solution when t =/= -22 (when t=-22, solution will be parametric) and t =/= 0 (t=0 is inconsistent)

    iii) For t=-22, A|0 =
    [1 0 3 -22 .|0]
    [0 1 -22 0 .|0]
    [0 0 1 -66 .|0]
    [0 0 0 0 ....|0]

    So, the dimension of the null space is 1 and:

    -> x3= 66x4
    -> x2= 22x3
    -> x1= 22x4 - 3x3

    let x4 = 1/22
    -> x = [-8 66 3 1/22]T is a basis for the null space of A.

    Is the above correct/enough working for an exam question?
     
  2. jcsd
  3. Jul 25, 2010 #2

    Mark44

    Staff: Mentor

    Looks fine to me. The only thing I would do differently is in part iii - I would write the basis for the nullspace as <-176, 1452, 66, 1>.
     
  4. Jul 26, 2010 #3
    Thanks for checking it :cool:
     
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