If - 1-10 4 3 17 7 4 4 -5 7 - - 2-4 x-5 3 7 x-12 4 -5 x-1 7such that - 1- 2-0 then

The correct option is C x=5 Given, #160; Δ _1=10 amp;4 #160; amp;3 17 amp;7 #160; amp;4 4 amp; 5 #160; amp;7 ...

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Properties of Determinants

If Δ 1=10 ...

Question

If $Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 10 & 4 & 3 17 & 7 & 4 4 & - 5 & 7 \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 4 & x + 5 & 3 7 & x + 12 & 4 - 5 & x - 1 & 7 \end{matrix} ∣ ∣ ∣ ∣$
such that $Δ_{1} + Δ_{2} = 0$ then

x = 5

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x

has no real value

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x = 0

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none of these

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Solution

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Similar questions

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} 7 & x & 2 - 5 & x + 1 & 3 4 & x & 7 \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & 2 & 7 x + 1 & 3 & - 5 x & 7 & 4 \end{matrix} ∣ ∣ ∣ ∣

Δ_{1} - Δ_{2} = 0

∆_{1} = |\begin{matrix} 1 & 1 & 1 \\ a & b & c \\ a^{2} & b^{2} & c^{2} \end{matrix}|, ∆_{2} = |\begin{matrix} 1 & b c & a \\ 1 & c a & b \\ 1 & a b & c \end{matrix}|, then

∆_{1} + ∆_{2} = 0

∆_{1} + 2 ∆_{2} = 0

∆_{1} = ∆_{2}

Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin θ & cos θ - sin θ & - x & 1 cos θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} x & sin 2 θ & cos 2 θ - sin 2 θ & - x & 1 cos 2 θ & 1 & x \end{matrix} ∣ ∣ ∣ ∣

x \neq 0

θ \in (0, \frac{π}{2})

Δ_{1}, Δ_{2}

3, 4

5

\frac{Δ_{1}}{Δ_{2}} =

△

A M N, B N L

C L M

Δ_{1}, Δ_{2}

Δ_{3}

Δ_{1} + Δ_{2} + Δ_{3}

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