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Math Help - The Matrix Equation Ax=b

  1. #1
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    The Matrix Equation Ax=b

    I am having trouble answering this hw problem on section 1.4 in my linear algebra book by David C. Lay the updated 3rd edition. Anyhow here's the question

    Let A= [2 -6(vertical column), -1 3 (vertical also) ] and b= [b1, b2] (vertical column)

    Show that the equation Ax=b does not have a solution for all possible b, and describe the set of all b for which Ax=b does have a solution.

    Thanks!
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  2. #2
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    Have you noticed that the matrix A is singular?
    Have you looked at the row reduction of
     \left[ {\begin{array}{rrcc}<br />
   2 & { - 1} & \| &  {b_1 }  \\<br />
   { - 6} & 3 &\| &  {b_2 }  \\<br />
\end{array}} \right]?
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  3. #3
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    yes! Thank you
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  4. #4
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    Same Problem : Matrix Equation Ax=b

    this is what I had for the reduction, I am not sure if it's correct...

    column 1 [1 0] column 2 [-1/2 0] column 3 [b1/2 b2/3]

    the answer in the back of the book gave the following response:

    The equation Ax=b is consistent when 3b1+b2 is nonzero.(I need to show my work to show why it's not consistent). The set of b for which the equation is consistent is a line through the origin- the set of all points (b1,b2) satisfying b2=-3b1.

    I don't know what any of this means. Could please explain?
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  5. #5
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    Quote Originally Posted by googoogaga View Post
    I am having trouble answering this hw problem on section 1.4 in my linear algebra book by David C. Lay the updated 3rd edition. Anyhow here's the question
    One of my favorite math books! Everything is very clear and easy to understand. I'm sure you'll agree with me after your completion of the book.
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  6. #6
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    Quote Originally Posted by googoogaga View Post
    this is what I had for the reduction, I am not sure if it's correct...

    column 1 [1 0] column 2 [-1/2 0] column 3 [b1/2 b2/3]

    the answer in the back of the book gave the following response:

    The equation Ax=b is consistent when 3b1+b2 is nonzero.(I need to show my work to show why it's not consistent). The set of b for which the equation is consistent is a line through the origin- the set of all points (b1,b2) satisfying b2=-3b1.

    I don't know what any of this means. Could please explain?
    Now to answer your question. This is quite easy.

    Did you row-reduce the matrix?

    So, we have the following (I'll use Maple syntax for showing the Matrix; if someone could convert it to "pretty print" I'd appreciate it):

    Let A = [[2, -1],[-6, 3]] and let b = [[b_1], [b_2]]

    By row reducing the AUGMENTED matrix of the system, we have the following:

    [[2, -1, b_1],[-6, 3, b_2]] ~ (Multiply the top row by -3 and add to 2nd row):

    [[2, -1, b_1],[0, 0, 3b_1 + b_2]]

    Now, we can see that there is a PIVOT in the augmented column. Also, we see that if:

    3b_1 + b_2 != 0, then Ax = b does NOT have a solution.

    Further, we note that for all b such that 3b_1 + b_2 = 0, then the system DOES have a solution (consistent).

    Does this make sense?
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by AfterShock View Post
    I'll use Maple syntax for showing the Matrix; if someone could convert it to "pretty print" I'd appreciate it
    Here you go:

     \left[ {\begin{array}{rrcc}<br />
   2 & { - 1} & \| &  {b_1 }  \\<br />
   { - 6} & 3 &\| &  {b_2 }  \\<br />
\end{array}} \right]

     \left[ {\begin{array}{rrcc}<br />
 1 & { - \frac {1}{2}} & \| &  { \frac {b_1}{2} }  \\<br />
0 & 0 &\| &  {3b_1 + b_2 }  \\<br />
\end{array}} \right]

    this means that in the 2 x 1 solution matrix x, x_1 = \frac {b_1}{2} + \frac {1}{2}x_2 and x_2 = t for some parameter t. This means that t = 3b_1 + b_2. Now, if t is anything other than 0, the matrix is inconsistent, since it would mean that 0x_1 + 0x_2 = something other than zero, which makes no sense! So we set t=0, so 3b_1 + b_2 = 0 \Rightarrow b_2 = -3b_1. If we plot this solution on a set of axis, we would get a line through the origin, since b_2 would be analagous to y and b_1 would be analagous to x, we would essentially be plotting the line y = -3x, which is a line through the origin.


    I'm not that good at coding structures like matrices in LaTex, I just stole Plato's code and modified it

    EDIT: In retrospect i think to say that t = 3b_1 + b_2 is incorrect, i was trying to make some other connection and somehow confused myself along the way. pretty much everything i said after that is correct though. that is if 3b_1 + b_2 \neq 0 then it would mean 0x_1 + 0x_2 = something other than zero ... and the argument follows from there
    Last edited by Jhevon; May 30th 2007 at 10:36 AM.
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  8. #8
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    Thank You all for the help!

    It is much appreciated!
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