Linear algebra - general solutions to Ax=b

  • Thread starter Villhelm
  • Start date
  • #1
37
0
Could someone check my working for this?

Homework Statement



(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.

Homework Equations


Basic linear algebra?

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?
 

Answers and Replies

  • #2
35,285
7,129
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>.
 
  • #3
37
0
Thanks for checking it :cool:
 

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