A) Non-negative
B) Non-positive
C) Negative
D) Positive
Correct Answer: C
Solution :
[c] \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}},\] \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}},\] (reduces the determinant to:) \[\left| \begin{matrix} {{x}^{2}}-2x+1 & x-1 & 0 \\ 2x-2 & x-1 & 0 \\ 3 & 3 & 1 \\ \end{matrix} \right|\] \[={{(x-1)}^{3}}-2{{(x-1)}^{2}}={{(x-1)}^{2}}(x-1-2)\] \[={{(x-1)}^{2}}(x-3),\] which is clearly negative for \[x<1\]You need to login to perform this action.
You will be redirected in
3 sec