First slide

Introduction to Determinants

Question

If $f (x) = |\begin{array}{ccc} a & - 1 & 0 \\ ax & a & - 1 \\ {ax}^{2} & ax & a \end{array}|$ , then $f (2 x) - f (x)$ is divisible by

Moderate

A

x
B

a
C

2a + 3x
D

x²

Solution

Applying $R_{3} \to R_{3} - {xR}_{2} and R_{2} \to R_{2} - {xR}_{1}$ , we get
$f (x) = |\begin{array}{ccc} a & - 1 & 0 \\ 0 & a + x & - 1 \\ 0 & 0 & a + x \end{array}| = a (a + x)^{2}$
Hence,
$\begin{matrix} f (2 x) - f (x) = a [(a + 2 x)^{2} - (a + x)^{2}] \\ = a (a + 2 x - a - x) (a + 2 x + a + x) \\ = ax (2 a + 3 x) \end{matrix}$

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