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Using the property of determinants and without expanding, prove that $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a}&{a - b}&{b - c} \end{array}} \right| = 0$.

Last updated date: 14th Jun 2024
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Hint: Here, we will first apply the rule of rows addition ${R_1} \to {R_1} + {R_2}$in the given determinant and then make the two rows identical by taking common any constant $k$ from one row, then the value of determinant gets multiplied by $k$. Using the property of a determinant is when any two rows of a determinant are identical, and then the value of the determinant is zero.

Complete step by step solution:
We are given that the determinant $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a}&{a - b}&{b - c} \end{array}} \right| = 0$.

First, applying the rows addition ${R_1} \to {R_1} + {R_2}$in the given determinant, we get

$\Rightarrow \left| {\begin{array}{*{20}{c}} {a - c}&{b - a}&{c - b} \\ {b - c}&{c - a}&{a - b} \\ {c - a}&{a - b}&{b - c} \end{array}} \right| = 0$

Taking the negative sign common from the third row, we get

$\Rightarrow \left| {\begin{array}{*{20}{c}} {a - c}&{b - a}&{c - b} \\ {b - c}&{c - a}&{a - b} \\ { - \left( {a - c} \right)}&{ - \left( {b - a} \right)}&{ - \left( {c - b} \right)} \end{array}} \right| = 0$

We know that the property of determinant tells us when each element of a row (or a column) of a determinant is multiplied by any constant $k$, then its value gets multiplied by $k$.

Since the third row of the above determinant is multiplied by a constant $- 1$, then we will multiply this value by $- 1$, we get
$\Rightarrow - \left| {\begin{array}{*{20}{c}} {a - c}&{b - a}&{c - b} \\ {b - c}&{c - a}&{a - b} \\ {a - c}&{b - a}&{c - b} \end{array}} \right| = 0$

We also know that the property of a determinant is when any two rows of a determinant are identical, and then the value of the determinant is zero.

Since we have two rows ${R_1}$ and ${R_2}$ identical, we can say that the value of the determinant is 0.

Therefore, we have $\left| {\begin{array}{*{20}{c}} {a - b}&{b - c}&{c - a} \\ {b - c}&{c - a}&{a - b} \\ {c - a}&{a - b}&{b - c} \end{array}} \right| = 0$.

Hence, proved.

Note: Calculation is the important side of the question. In questions like these, our main goal is to create two rows or columns identical, then we can easily use the property of determinant when any two rows of a determinant are identical, and then the value of the determinant is zero. We can also solve this question by applying ${C_1} \to {C_1} + {C_2} + {C_3}$ in the columns of the given determinant, we get

$\Rightarrow \left| {\begin{array}{*{20}{c}} 0&{b - c}&{c - a} \\ 0&{c - a}&{a - b} \\ 0&{a - b}&{b - c} \end{array}} \right|$

Using the property of determinant of when a column (or a row) is 0, then its value is zero.

$\Rightarrow \left| {\begin{array}{*{20}{c}} 0&{b - c}&{c - a} \\ 0&{c - a}&{a - b} \\ 0&{a - b}&{b - c} \end{array}} \right| = 0$

Hence, proved.