# Find all scalars c_{1} , c_{2}, c_{3} such that c_{1}(1 , -1, 0) + c_{2}(4, 5, 1) + c_{3}(0, 1, 5) = (3, 2, -19)

Find all scalars ${c}_{1},{c}_{2},{c}_{3}$ such that ${c}_{1}\left(1,-1,0\right)+{c}_{2}\left(4,5,1\right)+{c}_{3}\left(0,1,5\right)=\left(3,2,-19\right)$
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Nathanael Webber
The equation ${c}_{1}\left(1,-1,0\right)+{c}_{2}\left(4,5,1\right)+{c}_{3}\left(0,1,5\right)=\left(3,2,-19\right)$
becomes $\left({c}_{1}-{c}_{1},0\right)+\left(4{c}_{2},5{c}_{2},{c}_{2}\right)+\left(0,{c}_{3}5{c}_{3}\right)=\left(3,2,-19\right)$
which can be written as $\left({c}_{1}+4{c}_{2},-c2+5{c}_{2}+{c}_{3},{c}_{2}+5{c}_{3}\right)=\left(3,2,-19\right)$
This gives us the the system ${c}_{1}+4{c}_{2}=3$
$-{c}_{1}+5{c}_{2}={c}_{3}=2$
${c}_{2}+5{c}_{3}=-19$
From the first equation we get ${c}_{1}=3-4{c}_{2}$
From the third equation we get ${c}_{2}=-19-5{c}_{3}$
Thus, ${c}_{1}=3-4{c}_{2}=3-4\left(-19-5{c}_{3}\right)+{c}_{3}=2⇒-44{c}_{3}-174=2$
This yields $-44{c}_{3}=176⇒{c}_{3}=-4$
Thus, ${c}_{2}=-19-5{c}_{3}=-19-5\cdot \left(-4\right)⇒{c}_{2}=1$
and ${c}_{2}=79+20{c}_{3}=79+20\cdot \left(-4\right)⇒{c}_{1}=-1$
Therefore, to conclusude, ${c}_{1}=-1,{c}_{2}=1,{c}_{3}=-4$