A) \[2ac\]
B) \[-ac\]
C) \[ac\]
D) \[-2ac\]
E) \[a\]
Correct Answer: A
Solution :
Given linear equations are \[x+4ay+az=0\] ... (i) \[x+3by+bz=0\] ...(ii) and \[x+2cy+cz=0\] ...(iii) For non-trivial solution \[\left| \begin{matrix} 1 & 4a & a \\ 1 & 3b & b \\ 1 & 2c & c \\ \end{matrix} \right|=0\] Applying\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[\Rightarrow \] \[\left| \begin{matrix} 1 & 4a & a \\ 0 & 3b-4a & b-a \\ 0 & 2c-4a & c-a \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[1[(3b-4a)(c-a)-2(b-a)(c-2a)]=0\] \[\Rightarrow \]\[3bc-3ab-4ac+4{{a}^{2}}\] \[-2(bc-2ab-ac+2{{a}^{2}})=0\] \[\Rightarrow \] \[bc+ab-2ac=0\] \[\Rightarrow \] \[ab+bc=2ac\]
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