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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

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augment the coeff matrix
row rank = colA = dimA = number of pivot points
i didn't understand any of that.

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then you really need to review your material, or ask more probing questions
what exactly is row rank?
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
row rank is the number of nonzero rows that you end up with
i was about to say this........
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}}
ok i got that. what's next?
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
our parametric is based on these last 2 columns in this case; our constant vector and our free vectors
\[\vec x=\begin{pmatrix}x\\y\\z\\w \end{pmatrix}=\begin{pmatrix}109/3\\12\\-92/3\\0\end{pmatrix}+w\begin{pmatrix}-73/3\\-8\\65/3\\1 \end{pmatrix}\]
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.
yes, in this case the free variable is what you are calling an arbitrary constant
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
it means that the rest of it can be defined using just those column vectors

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