anonymous
  • anonymous
obtain the row rank and parametrically represented solution of the system. 3x-y+3z=5 x+2y+2z-3w=-1 2x+5y+4z+2w=10
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
augment the coeff matrix
amistre64
  • amistre64
row rank = colA = dimA = number of pivot points
anonymous
  • anonymous
i didn't understand any of that.

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amistre64
  • amistre64
then you really need to review your material, or ask more probing questions
anonymous
  • anonymous
what exactly is row rank?
amistre64
  • amistre64
make sure each equation HAS the same variables; whatevers missing put a 0 to it so that it is included 3x -1y+3z+0w=5 1x+2y+2z-3w=-1 2x+5y+4z+2w=10 strip off everything thats NOT a number 3 -1 3 0 5 1 2 2 -3 -1 2 5 4 2 10 now this is a coeff matrix that you can augment
amistre64
  • amistre64
row rank is the number of nonzero rows that you end up with
anonymous
  • anonymous
i was about to say this........
amistre64
  • amistre64
if we format this for the wolf, life gets a bit easier :) rref{{3, -1, 3, 0, 5},{1, 2, 2, -3, -1},{2, 5, 4, 2, 10}}
anonymous
  • anonymous
ok i got that. what's next?
amistre64
  • amistre64
http://www.wolframalpha.com/input/?i=rref%7B%7B3%2C+-1%2C+3%2C+++0%2C+++5%7D%2C%7B1%2C++2%2C++2%2C+-3%2C++-1%7D%2C%7B2%2C++5%2C++4%2C+++2%2C++10%7D%7D
amistre64
  • amistre64
there are 3 nonzero rows; so row rank = 3 the last column of the augmented matrix is a constant vector the next to last is the a free variable column
amistre64
  • amistre64
our parametric is based on these last 2 columns in this case; our constant vector and our free vectors
amistre64
  • amistre64
\[\vec x=\begin{pmatrix}x\\y\\z\\w \end{pmatrix}=\begin{pmatrix}109/3\\12\\-92/3\\0\end{pmatrix}+w\begin{pmatrix}-73/3\\-8\\65/3\\1 \end{pmatrix}\]
anonymous
  • anonymous
in this typa of questions, i was taught to suppose an arbitary constant and then calculate values of x,y,z,w for a particular value of arbitary constant.
amistre64
  • amistre64
yes, in this case the free variable is what you are calling an arbitrary constant
amistre64
  • amistre64
the augmented matrix produces our vectors for us; by augmenting with the "=" column vector to begin with we produce a particular value vector; and all thos coumns that dont stand on a "1" are called free variables ... arbitrary constants
amistre64
  • amistre64
it means that the rest of it can be defined using just those column vectors

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