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જો $f(x) = \left| {\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x - 1)}&{(x + 1)x}\\{3x(x - 1)}&{x(x - 1)(x - 2)}&{(x + 1)x(x - 1)}\end{array}} \right|$ તો $f(100)$ મેળવો.
$0$
$1$
$100$
$-100$
Solution
(a) Given determinant
$ = \left| {\,\begin{array}{*{20}{c}}1&x&{x + 1}\\{2x}&{x(x – 1)}&{(x + 1)\,x}\\{3x(x – 1)}&{x(x – 1)\,(x – 2)}&{(x + 1)\,x(x – 1)}\end{array}\,} \right|$
$ = x(x + 1)\,\left| {\,\begin{array}{*{20}{c}}1&x&1\\{2x}&{x – 1}&{\,x}\\{3x(x – 1)}&{(x – 1)\,(x – 2)}&{x(x – 1)}\end{array}\,} \right|$
= $x(x + 1)\,(x – 1)\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\{2x}&{x – 1}&x\\{3x}&{x – 2}&x\end{array}\,} \right|$
Applying ${C_1} – {C_3}$ and ${C_2} – {C_3}$
$x(x + 1)\,(x – 1)\,\left| {\,\begin{array}{*{20}{c}}0&0&1\\x&{ – 1}&x\\{2x}&{ – 2}&x\end{array}\,} \right|$=$x(x + 1)(x – 1)\,[ – 2x + 2x] = 0$
$\therefore $ $f(x) = 0 \Rightarrow f(100) = 0$.