Question

If $\left|\begin{array}{ccc}x+1& 3& 5\\ 2& x+2& 5\\ 2& 3& x+4\end{array}\right|=0,$ then x =

• A. 1,9
• B. -1,9
• C. -1,-9
• D. 1,-9

Answer 1

The correct option is D 1,-9

By $C_1\to C_1+C_2+C_3,$
we have $(9+x)\left|\begin{array}{ccc}x+1& 3& 5\\ 2& x+2& 5\\ 2& 3& x+4\end{array}\right|=0\Rightarrow (x+9)\left|\begin{array}{ccc}0& 1-x& 0\\ 0& -(1-x)& 1-x\\ 1& 3& x+4\end{array}\right|=0\Rightarrow (x+9)(1-x{)}^2\left|\begin{array}{ccc}0& 1& 0\\ 0& -1& 1\\ 1& 3& x+4\end{array}\right|=0\Rightarrow x=1,1,-9,$ (Since the determinant = 1).