Question

Using properties of determinants, prove the following :$\left(\begin{array}{ccc}1& x& x+1\\ 2x& x(x-1)& x(x+1)\\ 3x(1-x)& x(x-1)(x-2)& x(x+1)(x-1)\end{array}\right)=6x^2(1-x^2)$

Answer 1

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

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

$=1\cdot x(x-1)x(x+1)(x-1)-x(x+1)x(x-1)(x-2)-x\cdot 2xx(x+1)(x-1)-x(x+1)\cdot 3x(1-x)+(1+x)\cdot 2xx(x-1)(x-2)-x(x-1)\cdot 3x(1-x)$

$=1\cdot (x^4-x^2)-x(5x^4-5x^2)+(x+1)(5x^4-12x^3+7x^2)$

$=x^4-x^2-x(5x^4-5x^2)+(x+1)(5x^4-12x^3+7x^2)-(x^4-x^2-5x^5+5x^3+(x+1)(5x^4-12x^3+7x^2))+x^4-x^2-5x^5+5x^3+5x^5-7x^4-5x^3+7x^2$

$=-6x^4+6x^2$

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