If f(x) = a & -1 & 0 ax & a & -1 ax^ 2 & ax & a , using propertie… - Vidyadip

Question

$\text{If f(x)} = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^{2} & ax & a \end{vmatrix} , $ using properties of determinants find the value of $\text{f}(2x) - \text{f}(x).$

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Answer

$\text{ f(x)} = \begin{vmatrix} \text{a} & -1 & 0 \\ \text{ax} & \text{a} & -1 \\ \text{ax}^{2} & \text{ax} & \text{a} \end{vmatrix} , $
$\text{R}_{2} \rightarrow\text{R}_{2} - \text{x R}_{1}$ and $\text{R}_{3} \rightarrow\text{R}_{3} - \text{x}^{2} \text{ R}_{1}$
$\text{f (x)} = \begin{bmatrix} \text{a} & -1 & 0 \\ 0 & \text{a + x} & -1 \\ 0 & \text{ax} + \text{x}^{2} & \text{a} \end{bmatrix} \text{(For bringing 2 zeroes in any row/column}) $
$\therefore \text{f(x)} = \text{a (a}^{2} + \text{2ax + x}^{2}) = \text{a(x + a)}^{2}$
$\therefore\text{f (2x) - f(x)} = \text{a[2x + a]}^{2} - \text{a(x + a})^{2} $
$= \text{a x (3x + 2a)} $

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