beginvmatrix b - 2- ab - b - c - bc - aclla b-a- 2 - a-b - b- 2-a b lbc-ac - c - a - ab - a - 2 e n d v m a t r i x -1 A- abc a - b - c B- 3 a2 b2 c2C- 0D- None of these

The correct option is C 0Δ =(b a)(b a). b amp; b c amp; c a amp; a b amp; b c amp; c a amp; a =(a b)^2 b amp; b amp; c a amp; a ...

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Byju's Answer

Standard XII

Mathematics

Properties of Determinants

beginvmatrix ...

Question

$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

abc(a+b+c)

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3 a^{2} b^{2} c^{2}

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0.0

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None of these

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Solution

The correct option is C 0.0
$Δ = (b - a) (b - a) . ∣ ∣ ∣ ∣ \begin{matrix} b & b - c & c a & a - b & b c & c - a & a \end{matrix} ∣ ∣ ∣ ∣ = (a - b)^{2} ∣ ∣ ∣ ∣ \begin{matrix} b & b & c a & a & b c & c & a \end{matrix} ∣ ∣ ∣ ∣ = 0, [b y C_{2} \to C_{2} + C_{3}]$ .

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Similar questions

$∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 1 & a b & a c a b & b^{2} + 1 & b c c a & c b & c^{2} + 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 1 + a^{2} + b^{2} + c^{2}$

The determinant
$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣$ equals to:

(a) abc(b-c)(c-a)(a-b)
(b) (b-c)(c-a)(a-b)
(c) (a+b+c)(b-c)(c-a)(a-b)
(d) None of these

$Find x : \frac{x - b - c}{a} + \frac{x - c - a}{b} + \frac{x - a - b}{c} = 3, if \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \neq 0$

Q. If the vectors

(- b c, b^{2} + b c, c^{2} + b c), (a^{2} + a c, - a c, c^{2} + a c)

and

(a^{2} + a b, b^{2} + a b, - a b)

are coplanar (where none of a, b or c is zero). Then?

∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =

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∣∣ ∣ ∣∣b2−abb−cbc−acab−a2a−bb2−abbc−acc−aab−a2∣∣ ∣ ∣∣=

The correct option is C 0.0Δ=(b−a)(b−a).∣∣ ∣∣bb−ccaa−bbcc−aa∣∣ ∣∣=(a−b)2∣∣ ∣∣bbcaabcca∣∣ ∣∣=0,[by C2→C2+C3].

$∣ ∣ ∣ ∣ ∣ \begin{matrix} b^{2} - a b & b - c & b c - a c a b - a^{2} & a - b & b^{2} - a b b c - a c & c - a & a b - a^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ =$

The correct option is C 0.0
$Δ = (b - a) (b - a) . ∣ ∣ ∣ ∣ \begin{matrix} b & b - c & c a & a - b & b c & c - a & a \end{matrix} ∣ ∣ ∣ ∣ = (a - b)^{2} ∣ ∣ ∣ ∣ \begin{matrix} b & b & c a & a & b c & c & a \end{matrix} ∣ ∣ ∣ ∣ = 0, [b y C_{2} \to C_{2} + C_{3}]$ .