#Matrices
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lofty nebula
Yes, but they are asking you to put x3 in terms of any 'free variables'
free variables can be any real number, so you are technically correct
but there is a specific way they want you to word it
because columns 3 and 4 have no pivots (that is, they don't have the first nonzero number in any row), the variable that corresponds to that column in the system is 'free'
note that they tell you, that if x3 and x4 are free variables, then let x3 = s and x4 = t
you also know that the system is not inconsistent, because where a row is entirely zeroes, the solution part (to the right of the bar) is also zero
and 0 = 0
now because column 3 doesn't have pivots
x3 is free
and thus x3 = s
if they were asking for something that wasn't x3, say x2, then you would have to solve x2 - 9x3 - 6x4 = 20
where x3 = s and x4 = t
again, free variables
so x2 in that case would be x2 = 20 + 9s + 6t
solving this entire system for all variables would result in
[\left(\begin{matrix}x_1\x_2\x_3\x_4\end{matrix}\right)=\left(\begin{matrix}20\20\0\0\end{matrix}\right)+s\left(\begin{matrix}4\9\1\0\end{matrix}\right)+t\left(\begin{matrix}-3\6\0\1\end{matrix}\right)]
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lofty nebula
solely because x_3 is a free variable
just wait until you get to bases
for solving a system into vector parametric form, reduced row echelon isn't necessary, but it is helpful
vector parametric is just the equation in the TeXit thing there
basically a smart-looking way of writing the solutions for all x_n
remember that reduced row echelon is just row echelon but with 0s above each pivot
.close
.... nice bot :T
tame needleBOT
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