Prove that if - - - 0- then - - - a 1 b 1 a 1 b 2 a 1 b 3 a 2 b 1 - -a 2 b 2 a 2 b 3 a 3 b 1 a 3 b 2 - - a 3 b 3 - - 1- a 1 b 1 - - a 2 b 2 - - a 3 b 3 -

Take α ,β ,γ from R _ 1 , R _ 2 and R _ 3 Δ =αβγ 1+ a _ 1 b _ 1 α amp; a _ 1 b _ 2 α amp; a _ 1 b _ 3 α 0+ a _ 2 b _ 1 β amp; ...

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Discriminant

Prove that if...

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Prove that if $α, β, γ \neq 0,$ then
$∣ ∣ ∣ ∣ \begin{matrix} α + a_{1} b_{1} & a_{1} b_{2} & a_{1} b_{3} a_{2} b_{1} & β + a_{2} b_{2} & a_{2} b_{3} a_{3} b_{1} & a_{3} b_{2} & γ + a_{3} b_{3} \end{matrix} ∣ ∣ ∣ ∣$
$= α β γ [1 + \frac{a_{1} b_{1}}{α} + \frac{a_{2} b_{2}}{β} + \frac{a_{3} b_{3}}{γ}]$

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∣ ∣ ∣ ∣ \begin{matrix} 1 + a_{1} + b_{1} & a_{1} + b_{2} & a_{1} + b_{3} a_{2} + b_{1} & 1 + a_{2} + b_{2} & a_{2} + b_{3} a_{3} + b_{1} & a_{3} + b_{2} & 1 + a_{3} + b_{3} \end{matrix} ∣ ∣ ∣ ∣

2 y - x = 8; 5 y - x = 14, y - 2 x = 1

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

(a_{1} + a_{2} + a_{3} + b_{1} + b_{2} + b_{3})

(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 x^{2} + b x + c = 0

a, b, c \in R

a \neq 0

α, β

α + β = - \frac{b}{a}, α β = \frac{c}{a}

a x^{2} + b x + c = a (x - α) (x - β)

a_{1}, a_{2}, a_{3}, a_{4}

a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = . . . \neq 0

b_{1}, b_{2}, b_{3}, b_{4}

\frac{b_{2}}{b_{1}} = \frac{b_{3}}{b_{2}} = \frac{b_{4}}{b_{3}} =

\neq 1

c_{1}

c_{2}

c_{3}

c_{4}

\frac{1}{c_{2}} - \frac{1}{c_{1}} = \frac{1}{c_{3}} - \frac{1}{c_{2}} = \frac{1}{c_{4}} - \frac{1}{c_{3}} =

\neq 0

a (b - c) x^{2} + b (c - a) x + c (a - b) = 0

a, b, c

| 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

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