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Properties of Determinants

Solve: x 2 ...

Question

Solve: $∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} - 1 & x^{2} + 2 x + 1 & 2 x^{2} + 3 x + 1 2 x^{2} + x - 1 & 2 x^{2} + 5 x - 3 & 4 x^{2} + 4 x - 3 6 x^{2} - x - 2 & 6 x^{2} - 7 x + 2 & 12 x^{2} - 5 x - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ . Let sum of all values of x be k. Find -2k.

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Solution

$∣ ∣ ∣ ∣ ∣ \begin{matrix} x^{2} - 1 & x^{2} + 2 x + 1 & 2 x^{2} + 3 x + 1 2 x^{2} + x - 1 & 2 x^{2} + 5 x - 3 & 4 x^{2} + 4 x - 3 6 x^{2} - x - 2 & 6 x^{2} - 7 x + 2 & 12 x^{2} - 5 x - 2 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ \begin{matrix} (x - 1) (x + 1) & (x + 1) (x + 1) & (x + 1) (2 x + 1) (2 x - 1) (x + 1) & (2 x - 1) (x + 3) & (2 x - 1) (2 x + 3) (3 x - 2) (2 x + 1) & (3 x - 2) (2 x - 1) & (3 x - 2) (4 x + 1) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

$\Rightarrow (x + 1) (2 x - 1) (3 x - 2) ∣ ∣ ∣ ∣ \begin{matrix} x - 1 & x + 1 & 2 x + 1 x + 1 & x + 3 & 2 x + 3 2 x + 1 & 2 x - 1 & 4 x + 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

applying $R_{1} \to R_{1} - R_{2}$ and taking $- 2$ common gives
$\Rightarrow (x + 1) (2 x - 1) (3 x - 2) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x + 1 & x + 3 & 2 x + 3 2 x + 1 & 2 x - 1 & 4 x + 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

applying $C_{3} - C_{1}$ and $C_{2} - C_{1}$ gives

$\Rightarrow (x + 1) (2 x - 1) (3 x - 2) ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 x + 1 & 2 & x + 2 2 x + 1 & - 2 & 2 x \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow (x + 1) (2 x - 1) (3 x - 2) (3 x + 2) = 0$

Sum of all values of $x$ ,

$k = - 1 + \frac{1}{2} + \frac{2}{3} - \frac{2}{3}$

$\Rightarrow k = \frac{- 1}{2}$

$∴ - 2 k = 1$

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\frac{6 x^{2} + 13 x - 4}{2 x + 5} = \frac{12 x^{2} + 5 x - 2}{4 x + 3}

Q. Solve the folowing equation:

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