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Homework Help: Linear algebra (trying to prepare for exam)

  1. Mar 3, 2008 #1
    Ok, here we go.

    The question asks,

    for matrix A =

    1 1 3
    3 0 1
    2 2 6

    , find a vector b such that Ax(x is a vector) = b has no solution.


    I found the rref(a) =

    1 0 1/3
    0 1 8/3
    0 0 0

    Now to, answer the question,

    I gues the equation has no solution if one of the rows has 000 = a nonzero value.

    So, is it right to say b is a with a nonzero value on the third row?
     
  2. jcsd
  3. Mar 3, 2008 #2

    Dick

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    Find a basis for the range of A. If such a b exists then dim(range(A))<3. Find a vector that's not in the range.
     
  4. Mar 3, 2008 #3
    Is there an alternate way without having to work with range -- we didn't do this in class ?
     
  5. Mar 3, 2008 #4

    Dick

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    Well, you as you've said if b has a nonzero third component then you can't solve rref(A)x=b. Fine. But that doesn't mean you can't solve Ax=b. Now you have to 'undo' your row reduction operations on e.g. b=(0,0,1). Do the opposite row operations on that b to get a vector such that Ax=b doesn't have a solution.
     
  6. Mar 4, 2008 #5

    HallsofIvy

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    ?? Surely they mentioned in class that Ax= b has a solution if and only if b is in the range of A! In fact, that is the definition of "range". You seem to be wanting to do what so many students unfortunately want- to apply a formula without having to understand what you are doing!

    Saying Ax= b, here, is the same as saying x+ y+ 3z= bx, 3x+ z= by, and 2x+ 2y+ 6z = b3 and you have shown, by row reducing the matrix, that that corresponds to x+ (2/3)z= 2bx- (1/3)b[sub[y[/sub], y+ (1/3)z)= (1/3)b- a, 0= c- 2a. Looking at the last row we see that has a solution if and only if c- 2a= 0 or c= 2a. It does not have a solution is c is not 2a.

    No, it is not. For example, b= [1, 0, 2] has a nonzero value on the third row but Ax= b for this A and b is the same as x+ y+ 3z= 1, 3x+ z= 0, 2x+ 2y+ 6z= 2. From the second equation, z= -3x. Putting that into the first equation gives x+ y- 3x= -2x+ y= 1 or y= 2x+ 1. Putting that into the third equation, 2x+ 2(2x+ 1)+ 6(-x)= 2 which is true for all x- the xs cancel on the left.

    It is true that an equation with the reduced matrix has no solution if there if the third component of b is non-zero. You should have carried the "b" values along with your row reduction to see what the right hand side should be for the original equation.
     
  7. Mar 4, 2008 #6
    "that that corresponds to x+ (2/3)z= 2bx- (1/3)b[sub[y[/sub], y+ (1/3)z)= (1/3)b- a, 0= c- 2a. Looking at the last row we see that has a solution if and only if c- 2a= 0 or c= 2a. It does not have a solution is c is not 2a."

    --> Can you please explain how you got that from the rref? I see x+ (2/3)z, but where does the 2bx come from...?
     
  8. Mar 4, 2008 #7

    HallsofIvy

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    Row reduce the augmented matrix just as you would if were solving the equation- which is, after all the whole point of the problem.
     
  9. Mar 4, 2008 #8
    Oh ok, I understand. So, the augmented matrix had only numbers and b's on the right, where did the c and a variables come from?

    This is the last thing I need to graps this. Thanks once again.
     
  10. Mar 4, 2008 #9
    can halls or anyone pick up on my last question?

    Thank you.
     
  11. Mar 4, 2008 #10

    HallsofIvy

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    For some reason, I switched from <bx, by, bz> to <a, b, c> at the minute! I may have been doing the calculations on paper using a, b, c.
     
  12. Mar 4, 2008 #11
    OK, I see. Just for closure.

    Since it is only if b_z is not 2b_x,

    2
    1
    9

    would be one such vector, right?

    Thanks!
     
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