If - - x-3 2x-1 2 3x-2 x-2 1 5x-1 5x-4 5 then - is

The correct option is C 15 Now, #160; Δ = x 3 amp; 2x+1 amp; 2 3x+2 amp; x+2 amp; 1 5x+1 amp; 5x+4 amp; 5 [ R_1'=R ...

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Properties of Determinants

If Δ = x-3 ...

Question

If $Δ = ∣ ∣ ∣ ∣ \begin{matrix} x - 3 & 2 x + 1 & 2 3 x + 2 & x + 2 & 1 5 x + 1 & 5 x + 4 & 5 \end{matrix} ∣ ∣ ∣ ∣$ then $Δ$ is

multiple of

x^{2}

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15

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a multiple of

x

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- 15

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Solution

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Δ

Δ = ∣ ∣ ∣ ∣ \begin{matrix} - x & a & b b & - x & a a & b & - x \end{matrix} ∣ ∣ ∣ ∣

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

x_{i}, y_{i}, z_{i} \in R (i = 1, 2, 3), x \in R

x

Δ = 0

x = 1, - 1

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

Δ = 0

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

Δ

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