Question

Let $x<1$, then value of $\left|\begin{array}{ccc}{x}^2+2& 2x+1& 1\\ 2x+1& x+2& 1\\ 3& 3& 1\end{array}\right|$ is

• A. non-negative
• B. non-positive
• C. negative
• D. positve

Answer 1

The correct option is C negative

Let $\Delta =\left|\begin{array}{ccc}{x}^2+2& 2x+1& 1\\ 2x+1& x+2& 1\\ 3& 3& 1\end{array}\right|$
$R_1\to R_1-R_2,R_2\to R_2-R_3$
$=\left|\begin{array}{ccc}{(x-1)}^2& x-1& 0\\ 2(x-1)& x-1& 0\\ 3& 3& 1\end{array}\right|$
$={(x-1)}^2\left|\begin{array}{ccc}(x-1)& 1& 0\\ 2& 1& 0\\ 3& 3& 1\end{array}\right|$
$\Delta ={(x-1)}^2(x-3)$
Now, since x < 1
$\Rightarrow x-3<-2$
Also, ${(x-1)}^2>0$
So, $\Delta$ is negative