Question

Determinant $ \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix} \right| $ is equal to

• A. $ a+b+c $
• B. $ {{a}^{2}}+{{b}^{2}}+{{c}^{2}} $
• C. 0
• D. 1

Answer 1

Answer: (C)

$ \left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 1 & b & {{b}^{2}}-ac \\ 1 & c & {{c}^{2}}-ab \\ \end{matrix} \right| $
$ =\left| \begin{matrix} 1 & a & {{a}^{2}}-bc \\ 0 & b-a & {{b}^{2}}-ac-{{a}^{2}}+bc \\ 0 & c-a & {{c}^{2}}-ab-{{a}^{2}}+bc \\ \end{matrix} \right| $
$ [{{R}_{2}}\to {{R}_{2}}-{{R}_{1}},{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}] $
$ =1[(b-a)({{c}^{2}}-ab-{{a}^{2}}+bc)-(c-a) $ $ ({{b}^{2}}-ac-{{a}^{2}}+bc)] $
$ =b{{c}^{2}}-a{{b}^{2}}-{{a}^{2}}b+{{b}^{2}}c-a{{c}^{2}}+{{a}^{2}}b $
$ +{{a}^{3}}-abc-{{b}^{2}}c+a{{c}^{2}}+{{a}^{2}}c-b{{c}^{2}}+a{{b}^{2}} $
$ -{{a}^{2}}c-{{a}^{3}}+abc $ $ =0 $