Answer

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Hint- Here, after applying some row operations we will expand the determinant through first column as $\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( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{21}}\left( {{a_{12}}{a_{33}} - {a_{13}}{a_{32}}} \right) + {a_{31}}\left( {{a_{12}}{a_{23}} - {a_{13}}{a_{22}}} \right)$.

Complete step-by-step answer:

Let us suppose D is the value of the given determinant.

i.e., ${\text{D = }}\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

1&b&{ca} \\

1&c&{ab}

\end{array}} \right|$

We can simplify the above given determinant by using some row operations.

Here, we will take the first row as the reference row and with the help of this row, we will make the second row and third row elements which are common with the first column elements as zero.

The row operations to be applied on the given determinant are ${{\text{R}}_2} \to {{\text{R}}_2} - {{\text{R}}_1}$ and ${{\text{R}}_3} \to {{\text{R}}_3} - {{\text{R}}_1}$

After applying above row operations, the determinant simplifies to

${\text{D}} = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

{\left( {1 - 1} \right)}&{\left( {b - a} \right)}&{\left( {ca - bc} \right)} \\

{\left( {1 - 1} \right)}&{\left( {c - a} \right)}&{\left( {ab - bc} \right)}

\end{array}} \right| = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{\left( {ca - bc} \right)} \\

0&{\left( {c - a} \right)}&{\left( {ab - bc} \right)}

\end{array}} \right| = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{ - c\left( {b - a} \right)} \\

0&{\left( {c - a} \right)}&{ - b\left( {c - a} \right)}

\end{array}} \right|$

By taking $\left( {b - a} \right)$ and common from second and third rows respectively, we get

${\text{D}} = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{ - c\left( {b - a} \right)} \\

0&{\left( {c - a} \right)}&{ - b\left( {c - a} \right)}

\end{array}} \right| = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&1&{ - c} \\

0&1&{ - b}

\end{array}} \right|$

Now, expanding the determinant through the first column, we get

\[ {\text{D}} = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&1&{ - c} \\

0&1&{ - b}

\end{array}} \right| = \left( {b - a} \right)\left( {c - a} \right) \times \left[ {1\left( { - b - \left( { - c} \right)} \right)} \right] = \left( {b - a} \right)\left( {c - a} \right)\left( { - b + c} \right) \\

\Rightarrow {\text{D}} = \left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right) \\

\]

Therefore, the value of the given determinant is \[\left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right)\].

Note- In this particular problem, we will simplify the given determinant in such a way that there will exist only one non-zero element in the first column and then we will expand the determinant through the first column in order to determine the value of the determinant.

{{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\

{{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\

{{a_{31}}}&{{a_{32}}}&{{a_{33}}}

\end{array}} \right| = {a_{11}}\left( {{a_{22}}{a_{33}} - {a_{23}}{a_{32}}} \right) - {a_{21}}\left( {{a_{12}}{a_{33}} - {a_{13}}{a_{32}}} \right) + {a_{31}}\left( {{a_{12}}{a_{23}} - {a_{13}}{a_{22}}} \right)$.

Complete step-by-step answer:

Let us suppose D is the value of the given determinant.

i.e., ${\text{D = }}\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

1&b&{ca} \\

1&c&{ab}

\end{array}} \right|$

We can simplify the above given determinant by using some row operations.

Here, we will take the first row as the reference row and with the help of this row, we will make the second row and third row elements which are common with the first column elements as zero.

The row operations to be applied on the given determinant are ${{\text{R}}_2} \to {{\text{R}}_2} - {{\text{R}}_1}$ and ${{\text{R}}_3} \to {{\text{R}}_3} - {{\text{R}}_1}$

After applying above row operations, the determinant simplifies to

${\text{D}} = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

{\left( {1 - 1} \right)}&{\left( {b - a} \right)}&{\left( {ca - bc} \right)} \\

{\left( {1 - 1} \right)}&{\left( {c - a} \right)}&{\left( {ab - bc} \right)}

\end{array}} \right| = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{\left( {ca - bc} \right)} \\

0&{\left( {c - a} \right)}&{\left( {ab - bc} \right)}

\end{array}} \right| = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{ - c\left( {b - a} \right)} \\

0&{\left( {c - a} \right)}&{ - b\left( {c - a} \right)}

\end{array}} \right|$

By taking $\left( {b - a} \right)$ and common from second and third rows respectively, we get

${\text{D}} = \left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&{\left( {b - a} \right)}&{ - c\left( {b - a} \right)} \\

0&{\left( {c - a} \right)}&{ - b\left( {c - a} \right)}

\end{array}} \right| = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&1&{ - c} \\

0&1&{ - b}

\end{array}} \right|$

Now, expanding the determinant through the first column, we get

\[ {\text{D}} = \left( {b - a} \right)\left( {c - a} \right)\left| {\begin{array}{*{20}{c}}

1&a&{bc} \\

0&1&{ - c} \\

0&1&{ - b}

\end{array}} \right| = \left( {b - a} \right)\left( {c - a} \right) \times \left[ {1\left( { - b - \left( { - c} \right)} \right)} \right] = \left( {b - a} \right)\left( {c - a} \right)\left( { - b + c} \right) \\

\Rightarrow {\text{D}} = \left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right) \\

\]

Therefore, the value of the given determinant is \[\left( {b - a} \right)\left( {c - a} \right)\left( {c - b} \right)\].

Note- In this particular problem, we will simplify the given determinant in such a way that there will exist only one non-zero element in the first column and then we will expand the determinant through the first column in order to determine the value of the determinant.

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