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A set of vectors $$\left \{ \left ( a_{1}, a_{2}, a_{3} \right ), \left ( b_{1}, b_{2}, b_{3} \right ), \left ( c_{1}, c_{2}, c_{3} \right ) \right \}$$ is said to be linearly independent if and only if
$$\begin{vmatrix}
a_{1} & a_{2} & a_{3}\\ 
b_{1} & b_{2} & b_{3}\\ 
c_{1} & c_{2} & c_{3}
\end{vmatrix}\neq 0$$
otherwise the set is said to be linearly dependent. A similar result holds for $$\left \{ \left ( a_{1}, a_{2} \right ), \left ( b_{1}, b_{2} \right ) \right \}$$.
If $$\left ( a_{1}, a_{2}, a_{3} \right )$$, $$\left ( b_{1}, b_{2}, b_{3} \right )$$ and $$\left ( c_{1}, c_{2}, c_{3} \right )$$ are linearly independent and $$x, y, z\epsilon R$$, then


A
there exist α,β,γϵR such that (x,y,z)+α(a1,a2,a3)+β(b1,b2,b3)+γ(c1,c2,c3)=(0,0,0)
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B
if x(a1,a2,a3)+y(b1,b2,b3)+z(c1,c2,c3)=(0,0,0)x+y+z0.
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C
α(a1,a2,a3)+β(b1,b2,b3)+γ(c1,c2,c3)=(x,y,z) cannot hold for any values of α,β,γ
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D
none of these
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Solution

The correct option is A there exist $$\alpha , \beta , \gamma \epsilon R$$ such that $$\left ( x, y, z \right )+\alpha \left ( a_{1}, a_{2}, a_{3} \right )+\beta \left ( b_{1}, b_{2}, b_{3} \right )+\gamma \left ( c_{1}, c_{2}, c_{3} \right )=\left ( 0, 0, 0 \right )$$
$$\left( a_{ 1 },a_{ 2 },a_{ 3 } \right) ,\left( b_{ 1 },b_{ 2 },b_{ 3 } \right) $$ and 

$$\left( c_{ 1 },c_{ 2 },c_{ 3 } \right) $$ are linearly independent when
$$a_{ 1 }\alpha +b_{ 1 }\beta +c_{ 1 }\gamma =-x\\ a_{ 2 }\alpha +b_{ 2 }\beta +c_{ 2 }\gamma =-y\\ a_{ 3 }\alpha +b_{ 3 }\beta +c_{ 3 }\gamma =-z$$
From Cramer's rule
$$\alpha \left( a_{ 1 },a_{ 2 },a_{ 3 } \right) +\beta \left( b_{ 1 },b_{ 2 },b_{ 3 } \right) +\gamma \left( c_{ 1 },c_{ 2 },c_{ 3 } \right) =-\left( x,y,z \right) \\$$

$$ \Rightarrow \left( x,y,z \right) +\alpha \left( a_{ 1 },a_{ 2 },a_{ 3 } \right) +\beta \left( b_{ 1 },b_{ 2 },b_{ 3 } \right) +\gamma \left( c_{ 1 },c_{ 2 },c_{ 3 } \right) =\left( 0,0,0 \right) $$

Mathematics

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