Question

If $a,b,c$ are non-zero real numbers then $\left| {\begin{array}{*{20}{c}}
  {bc}&{ca}&{ab} \\
  {ca}&{ab}&{bc} \\
  {ab}&{bc}&{ca}
\end{array}} \right|$ vanishes, when
A. $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0$
B. $\dfrac{1}{a} - \dfrac{1}{b} - \dfrac{1}{c} = 0$
C. $\dfrac{1}{b} + \dfrac{1}{c} - \dfrac{1}{a} = 0$
D. $\dfrac{1}{b} - \dfrac{1}{c} - \dfrac{1}{a} = 0$

Answer 1

Hint: If the value of determinant is zero then we can say that determinant vanishes. So, in this problem first we will expand the given determinant along any row and any column. Then, we will equate this value of determinant with zero. Also we will use the factorization of ${a^3} + {b^3} + {c^3} - 3abc$.

Complete step-by-step solution:
In this problem, it is given that $\left| {\begin{array}{*{20}{c}}
  {bc}&{ca}&{ab} \\
  {ca}&{ab}&{bc} \\
  {ab}&{bc}&{ca}
\end{array}} \right|$ vanishes. So, we can write $\left| {\begin{array}{*{20}{c}}
  {bc}&{ca}&{ab} \\
  {ca}&{ab}&{bc} \\
  {ab}&{bc}&{ca}
\end{array}} \right| = 0$.
Now we know that $\left| {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right| = {a_{11}}\left| {\begin{array}{*{20}{c}}
  {{a_{22}}}&{{a_{23}}} \\
  {{a_{32}}}&{{a_{33}}}
\end{array}} \right| - {a_{12}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{33}}}
\end{array}} \right| + {a_{13}}\left| {\begin{array}{*{20}{c}}
  {{a_{21}}}&{{a_{22}}} \\
  {{a_{31}}}&{{a_{32}}}
\end{array}} \right|$. By using this information, let us expand $\left| {\begin{array}{*{20}{c}}
  {bc}&{ca}&{ab} \\
  {ca}&{ab}&{bc} \\
  {ab}&{bc}&{ca}
\end{array}} \right| = 0$ along the first row. Therefore, we can write
$bc\left| {\begin{array}{*{20}{c}}
  {ab}&{bc} \\
  {bc}&{ca}
\end{array}} \right| - ca\left| {\begin{array}{*{20}{c}}
  {ca}&{bc} \\
  {ab}&{ca}
\end{array}} \right| + ab\left| {\begin{array}{*{20}{c}}
  {ca}&{ab} \\
  {ab}&{bc}
\end{array}} \right| = 0$
Also we know that $\left| {\begin{array}{*{20}{c}}
  a&b \\
  c&d
\end{array}} \right| = ad - bc$.
 By using this information, now we can write
$bc\left[ {\left( {ab} \right)\left( {ca} \right) - \left( {bc} \right)\left( {bc} \right)} \right] - ca\left[ {\left( {ca} \right)\left( {ca} \right) - \left( {ab} \right)\left( {bc} \right)} \right] + ab\left[ {\left( {ca} \right)\left( {bc} \right) - \left( {ab} \right)\left( {ab} \right)} \right] = 0$
Let us simplify the above equation. Therefore, we get
$
   \Rightarrow bc\left[ {{a^2}bc - {b^2}{c^2}} \right] - ca\left[ {{c^2}{a^2} - a{b^2}c} \right] + ab\left[ {ab{c^2} - {a^2}{b^2}} \right] = 0 \\
   \Rightarrow {a^2}{b^2}{c^2} - {b^3}{c^3} - {c^3}{a^3} + {a^2}{b^2}{c^2} + {a^2}{b^2}{c^2} - {a^3}{b^3} = 0 \\
 $
Change the sign of each term and add all equal terms in above equation, we get
${a^3}{b^3} + {b^3}{c^3} + {c^3}{a^3} - 3{a^2}{b^2}{c^2} = 0$
Rewrite the above equation, we get
$
  {\left( {ab} \right)^3} + {\left( {bc} \right)^3} + {\left( {ca} \right)^3} - 3\left( {aa} \right)\left( {bb} \right)\left( {cc} \right) = 0 \\
   \Rightarrow {\left( {ab} \right)^3} + {\left( {bc} \right)^3} + {\left( {ca} \right)^3} - 3\left( {ab} \right)\left( {bc} \right)\left( {ca} \right) = 0 \cdots \cdots \left( 1 \right) \\
 $
Now we are going to use the factorization of ${a^3} + {b^3} + {c^3} - 3abc$ which is given by ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + ab + bc + ca} \right)$. Let us find the factorization of the LHS of equation $\left( 1 \right)$. Therefore, we get
$\left( {ab + bc + ca} \right)\left( {{a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + a{b^2}c + b{c^2}a + {a^2}bc} \right) = 0 \cdots \cdots \left( 2 \right)$
From equation $\left( 2 \right)$, we can write either $ab + bc + ca = 0 \cdots \cdots \left( 3 \right)$ or ${a^2}{b^2} + {b^2}{c^2} + {c^2}{a^2} + a{b^2}c + b{c^2}a + {a^2}bc = 0 \cdots \cdots \left( 4 \right)$.
Note that here $a,b,c$ are non-zero real numbers. Let us divide by $abc$ on both sides of equation $\left( 3 \right)$. Therefore, we get
$
  \dfrac{{ab + bc + ca}}{{abc}} = 0 \\
   \Rightarrow \dfrac{{ab}}{{abc}} + \dfrac{{bc}}{{abc}} + \dfrac{{ca}}{{abc}} = 0 \\
   \Rightarrow \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{1}{b} = 0 \\
   \Rightarrow \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0 \\
 $
Therefore, if $a,b,c$ are non-zero real numbers then $\left| {\begin{array}{*{20}{c}}
  {bc}&{ca}&{ab} \\
  {ca}&{ab}&{bc} \\
  {ab}&{bc}&{ca}
\end{array}} \right|$ vanishes, when $\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = 0$.
Therefore, option A is correct.

Note: If ${a^3} + {b^3} + {c^3} - 3abc = 0$ then we can say that either $a + b + c = 0$ or $a = b = c$. The converse of this result is also true. That is, if $a + b + c = 0$ then we can prove that ${a^3} + {b^3} + {c^3} - 3abc = 0$. If $a = b = c$ then it is obvious that ${a^3} + {b^3} + {c^3} - 3abc = 0$. We can expand the determinant along any row or any column.