Question

If $\left|\begin{array}{ccc}x& 3& 6\\ 3& 4& x\\ 6& x& 3\end{array}\right|=\left|\begin{array}{ccc}2& x& 7\\ x& 7& 2\\ 7& 2& x\end{array}\right|=\left|\begin{array}{ccc}4& 5& x\\ 5& x& 4\\ x& 4& 5\end{array}\right|=0$, then $x$ is equal to

• A. $0$
• B. $-9$
• C. $3$
• D. none of these

Answer 1

The correct option is B $-9$

1) Given $\left|\begin{array}{ccc}x& 3& 6\\ 3& 4& x\\ 6& x& 3\end{array}\right|=0$

$\therefore x\left|\begin{array}{cc}4& x\\ x& 3\end{array}\right|-3\left|\begin{array}{cc}3& x\\ 6& 3\end{array}\right|+6\left|\begin{array}{cc}3& 4\\ 6& x\end{array}\right|=0$

$\therefore x(12-x^2)-3(9-6x)+6(3x-24)=0$

$\therefore 12x-x^3-27+18x+18x-144=0$

$\therefore -x^3+48x-171=0$

$\therefore x^3-48x+171=0$ (1)

2) $\left|\begin{array}{ccc}2& x& 7\\ x& 7& 2\\ 7& 2& x\end{array}\right|=0$

$\therefore 2\left|\begin{array}{cc}7& 2\\ 2& x\end{array}\right|-x\left|\begin{array}{cc}x& 2\\ 7& x\end{array}\right|+7\left|\begin{array}{cc}x& 7\\ 7& 2\end{array}\right|=0$

$\therefore 2(7x-4)-x(x^2-14)+7(2x-49)=0$

$\therefore 14x-8-x^3+14x+14x-343=0$

$\therefore -x^3+42x-351=0$

$\therefore x^3-42x+351=0$ (2)

3) $\left|\begin{array}{ccc}4& 5& x\\ 5& x& 4\\ x& 4& 5\end{array}\right|=0$

$\therefore 4\left|\begin{array}{cc}x& 4\\ 4& 5\end{array}\right|-5\left|\begin{array}{cc}5& 4\\ x& 5\end{array}\right|+x\left|\begin{array}{cc}5& x\\ x& 4\end{array}\right|=0$

$\therefore 4(5x-16)-5(25-4x)+x(20-x^2)=0$

$\therefore 20x-64-125+20x+20x-x^3=0$

$\therefore -x^3+60x-189=0$

$\therefore x^3-60x+189=0$ (3)

Equate equation (1) and (2), we get,

$x^3-48x+171=x^3-42x+351$

$\therefore 6x+180=0$

$\therefore 6x=-180$

$\therefore x=-30$

Equate equation (2) and (3), we get,

$x^3-42x+351=x^3-60x+189$

$\therefore 18x+162=0$

$\therefore 18x=-162$

$\therefore x=-9$

Equate equations (1) and (3), we get,

$x^3-48x+171=x^3-60x+189$

$\therefore 12x-18=0$

$\therefore x=\frac{3}{2}$