A) 1
B) 0
C) \[(a+b+c)\]
D) 2
Correct Answer: B
Solution :
\[\left| \begin{matrix} 1/a & 1 & bc \\ 1/b & 1 & ca \\ 1/c & 1 & ab \\ \end{matrix} \right|\] \[=\left| \begin{matrix} 1/a & 1 & bc \\ 1/b-1/a & 0 & ca-bc \\ 1/c-1/a & 0 & ab-bc \\ \end{matrix} \right|\]\[\left[ \begin{align} & {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ & {{R}_{3}}\to {{R}_{3}}-{{R}_{1}} \\ \end{align} \right]\] \[=1\left[ (ca-bc)\left( \frac{1}{c}-\frac{1}{a} \right)-(ab-bc)\left( \frac{1}{b}-\frac{1}{a} \right) \right]\] \[=(ca-bc)\frac{(a-c)}{ac}-(ab-bc)\left( \frac{a-b}{ab} \right)\] \[=\frac{(a-b)(a-c)}{a}-\frac{(a-c)(a-b)}{a}=0\]
You need to login to perform this action.
You will be redirected in
3 sec
