$∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ =$

(1 - a)^{3}

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{(a - 1)}^{2}

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{(a - 1)}^{3}

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{(a + 1)}^{2}

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Solution

The correct option is D ${(a - 1)}^{3}$
$A = ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$

By row trasformation on A,

$r_{1} \to r_{1} - r_{2}$ and $r_{2} \to r_{2} - r_{3}$

$A = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} - 1 & a - 1 & 0 2 (a - 1) & (a - 1) & 0 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

So, $A = (a - 1)^{2} ∣ ∣ ∣ ∣ \begin{matrix} a + 1 & 1 & 0 2 & 1 & 0 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣$

$= (a - 1)^{2} [1 (a + 1 - 2)]$

$\Rightarrow A = (a - 1)^{3}$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = {(a - 1)}^{3}

∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣

∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = {(a - 1)}^{2}

|\begin{matrix} 1 + a & 1 & 1 \\ 1 & 1 + a & 1 \\ 1 & 1 & 1 + a \end{matrix}| = a^{3} + 3 a^{2}

|\begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 \\ 2 a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{matrix}| = {(a - 1)}^{3}

∣ ∣ ∣ ∣ \begin{matrix} a^{2} + 2 a & 2 a + 1 & 1 2 a + 1 & a + 2 & 1 3 & 3 & 1 \end{matrix} ∣ ∣ ∣ ∣ = (a - 1)^{3}

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