Question

Number of real roots of the equation $\left|\begin{array}{ccc}1& x& x\\ x& 1& x\\ x& x& 1\end{array}\right|+\left|\begin{array}{ccc}1-x& 1& 1\\ 1& 1-x& 1\\ 1& 1& 1-x\end{array}\right|=0$ is

• A. $0$
• B. $-1$
• C. $-2$
• D. $-3$

Answer 1

The correct option is B $-1$

Now,

$\left|\begin{array}{ccc}1& x& x\\ x& 1& x\\ x& x& 1\end{array}\right|+\left|\begin{array}{ccc}1-x& 1& 1\\ 1& 1-x& 1\\ 1& 1& 1-x\end{array}\right|=0$

or, $\left\{1\right(1-x^2)-x(x-x^2)+x(x^2-x\left)\right\}+\left\{\right(1-x\left)\right[(1-x{)}^2-1]-1(1-x-1)+1(1-1+x)\}=0$

or, $(1-x^2)+2x(x^2-x)+(1-x)(x^2-2x)+2x=0$

or, $1-3x^2+2x^3+x^2-x^3+2x^2=0$

or, $1+x^3=0$

or, $(x+1)(x^2-x+1)=0$.

or, $x+1=0$ and $x^2-x+1=0$

or, $x=-1$.

The second root will give imaginary root.

So only one real root.