Question

What is the expansion of \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right|\]

Answer 1

Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we need to know how to expand the determinant term to solve this question. Also, we need to know the formula to expand the matrix or determinant term to make the easy calculation.

Complete step by step solution:
The given question is shown below,
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = ?\] \[ \to \left( 1 \right)\]
The formula which is used to expand the determinant of \[3 \times 3\] a matrix is shown below,
If,
 \[A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right) \to \left( 2 \right)\]
Then,
 \[\left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{22}}{a_{31}}} \right)\] \[ \to \left( 3 \right)\]
Let’s compare the equation \[\left( 1 \right)\] and \[\left( 2 \right)\] , we get
 \[\left( 1 \right) \to \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = ?\]
 \[\left( 2 \right) \to A = \left( {\begin{array}{*{20}{c}}
  {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\
  {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\
  {{a_{31}}}&{{a_{32}}}&{{a_{33}}}
\end{array}} \right)\]
By comparing the equation \[\left( 1 \right)\] and \[\left( 2 \right)\] , we get
 \[{a_{11}} = a,{a_{12}} = b,{a_{13}} = c,{a_{21}} = b,{a_{22}} = c,{a_{23}} = a,\] \[{a_{31}} = c,{a_{32}} = a,{a_{33}} = b\]
Let’s substitute the above values in the equation \[\left( 3 \right)\] , we get
 \[\left( 3 \right) \to \left| A \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{12}}\left( {{a_{21}}{a_{33}} - {a_{23}}{a_{31}}} \right) + {a_{13}}\left( {{a_{21}}{a_{32}} - {a_{22}}{a_{31}}} \right)\]
(Here let’s take \[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right|\] )
 \[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = a\left( {cb - aa} \right) - b\left( {bb - ac} \right) + c\left( {ba - cc} \right)\]
We know that \[n \cdot n = {n^2}\] , so the above equation can also be written as,
 \[\left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = a\left( {cb - {a^2}} \right) - b\left( {{b^2} - ac} \right) + c\left( {ba - {c^2}} \right) \to \left( 4 \right)\]
Let’s solve the above equation, we get
 \[a\left( {cb - {a^2}} \right) = abc - {a^3}\]
 \[b\left( {{b^2} - ac} \right) = {b^3} - abc\]
 \[c\left( {ba - {c^2}} \right) = abc - {c^3}\]
Let’s substitute the above three simplifications in the equation \[\left( 4 \right)\] , we get
 \[\left( 4 \right) \to \left| A \right| = \left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = a\left( {cb - {a^2}} \right) - b\left( {{b^2} - ac} \right) + c\left( {ba - {c^2}} \right)\]
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = abc - {a^3} - \left( {{b^3} - abc} \right) + abc - {c^3}\]
By solving the above equation we get,
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = abc - {a^3} - {b^3} + abc + abc - {c^3}\]
We have three \[abc\] in the above equation. So, we can simplify the above equation as given below,
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = 3abc - {a^3} - {b^3} - {c^3}\]
Let’s take \[ - \] outside from RHS part to make the simplified form of the above equation
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = - \left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\]
So, the final answer is,
 \[\left| {\begin{array}{*{20}{c}}
  a&b&c \\
  b&c&a \\
  c&a&b
\end{array}} \right| = - \left( {{a^3} + {b^3} + {c^3} - 3abc} \right)\]
So, the correct answer is “$ - \left( {{a^3} + {b^3} + {c^3} - 3abc} \right) $ ”.

Note: In this question, we would involve the arithmetic operations like addition/ subtraction/ multiplication/ division. Also, we would remember the formula to expand the determinant of the matrix with three columns and three rows. To find the simplified form of expression we would separate the cubic terms, square terms and constant terms separately. Also, note that the final answer would be the most simplified form of the given question.