If A- - a1 a2 a3- b1 b2 b3- c1 c2 c3 - and rank A - 3- then a-a1- a2- a3-b-b1- b2- b3 - c-c1- c2- c3 are

The correct option is B coplanarRank is lt;3 implies, determinant of the matrix is zero which means the vectors are coplanar ...

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Scalar Matrix

If A= [ a1 ...

Question

If A= $⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦$
and rank (A) < 3, then $¯ ¯ ¯ a = (a_{1}, a_{2}, a_{3}); ¯ ¯ b = (b_{1}, b_{2}, b_{3})$ ;
$¯ ¯ c = (c_{1}, c_{2}, c_{3})$ are

non-coplanar

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coplanar

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collinear

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cannot be said

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Solution

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Similar questions

{(a_{1}, a_{2}, a_{3}), (b_{1}, b_{2}, b_{3}), (c_{1}, c_{2}, c_{3})}

{(a_{1}, a_{2}), (b_{1}, b_{2})}

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

x, y, z ϵ R

(a_{1}, a_{2}, a_{3})

(b_{1}, b_{2}, b_{3})

(c_{1}, c_{2}, c_{3})

x (a_{1}, a_{2}, a_{3}) + y (b_{1}, b_{2}, b_{3}) + z (c_{1}, c_{2}, c_{3}) = 0

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

a_{3} x^{2} + b_{3} x + c_{3} = 0

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

| a_{1} | > | a_{2} | + | a_{3} |, | b_{2} | > | b_{1} | + | b_{3} |

| c_{3} | > | c_{1} | + | c_{2} |

∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣ = 0

a_{1}, b_{1}, c_{1}

a_{2} = g c d (b_{1}, c_{1}), b_{2} = g c d (c_{1}, a_{1}), c_{2} = g c d (a_{1}, b_{1}),

a_{3} = l c m (b_{2}, c_{2}), b_{3} = l c m (c_{2}, a_{2}), c_{3} = l c m (a_{2}, b_{2}) .

g c d (b_{3}, c_{3}) = a_{2}

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