Question

using the properties of determinants , prove
| 1 x x+1 |
| 2x x(x-1) x(x+1) |
| 3x(1-x) x(x-1)(x-2) x(x+1)(x-1)|
= 6x​2(1-x2)

Answer 1

$\textbf{To prove:}$

$\left|\begin{array}{ccc}1&x&x+1\\2x&x(x-1)&x(x+1)\\3x(1-x)&x(x-1)(x-2)&x(x+1)(x-1)\end{array}\right|=6x^2(1-x^2)$

$\text{Consider}$

$\left|\begin{array}{ccc}1&x&x+1\\2x&x(x-1)&x(x+1)\\3x(1-x)&x(x-1)(x-2)&x(x+1)(x-1)\end{array}\right|$

$\text{Take x and x+1 common from C_2 and C_3 respectively}$

$=x(x+1)\left|\begin{array}{ccc}1&1&1\\2x&x-1&x\\3x(1-x)&(x-1)(x-2)&x(x-1)\end{array}\right|$

$\text{Take x-1 common from R_3 }$

$=x(x+1)(x-1)\left|\begin{array}{ccc}1&1&1\\2x&x-1&x\\-3x&x-2&x\end{array}\right|$

$=x(x+1)(x-1)\left|\begin{array}{ccc}1&0&0\\2x&-(x+1)&-x\\-3x&4x-2&4x\end{array}\right|$ $C_2\implies\,C_2-C_1\;\;C_3\implies\,C_3-C_1$

$\text{Expanding along R_1 we get}$

$=x(x+1)(x-1)[-4x(x+1)+x(4x-2)]$

$=x(x^2-1)[-4x^2-4x+4x^2-2x]$

$=x(x^2-1)[-6x]$

$=6x^2(1-x^2)$

$\therefore\bf\left|\begin{array}{ccc}1&x&x+1\\2x&x(x-1)&x(x+1)\\3x(1-x)&x(x-1)(x-2)&x(x+1)(x-1)\end{array}\right|=6x^2(1-x^2)$

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