Using properties of determinants, prove that a^ 2 + 2a & 2a + 1 &… - Vidyadip

Question

Using properties of determinants, prove that
$\begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix} = \text{(a - 1)}^{3}$

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Answer

$\Delta = \begin{vmatrix} \text{a}^{2} + \text{2a} & \text{2a + 1} & 1 \\ \text{2a + 1} & \text{a + 2} & 1 \\ 3 & 3 & 1 \end{vmatrix}$ $\text{R}_{1} \rightarrow \text{R}_{1} - \text{R}_{2} \text{ and } \text{R}_{2} \rightarrow \text{R}_{2} - \text{R}_{3}$$\Delta = \begin{vmatrix} \text{a}^{2} - 1 & \text{a - 1} & 0 \\ 2\text{(a - 1)} & \text{a - 1} & 0 \\ 3 & 3 & 1 \end{vmatrix}$
$ = \text{(a - 1)}^{2}\begin{vmatrix} \text{a + 1} & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix}$ Expanding $\text{(a - 1)}^{2}. \text{(a - 1)} = \text{(a - 1)}^{3}.$

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