Question

Without expanding the determinant, prove that:
\left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&{{a^2}}&{{a^3}} \\\ 1&{{b^2}}&{{b^3}} \\\ 1&{{c^2}}&{{c^3}} \end{array}} \right|

Answer 1

We basically will make some row transformations and take our common factors as per the rule of the determinants and thus prove the required result. First, multiply each row to such a scalar quantity that you get abc in each row as at least one element and carry on.

Complete step-by-step answer:
First of all consider the left side matrix which is \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right|.
We know that if we make row transformations, we are not actually changing the value of the determinant.
Therefore, we will apply some row transformations.
Step 1: First of all, apply ${R_1} \to a{R_1},{R_2} \to b{R_2}$ and ${R_3} \to c{R_3}$ in the above matrix. Since, we are multiplying each of the row by some scalar, we will require to divide by each of the scalar as well.
$\therefore$ we will get the matrix \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| is equivalent to \dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}} {a \times a}&{a \times {a^2}}&{a \times bc} \\\ {b \times b}&{b \times {b^2}}&{b \times ca} \\\ {c \times c}&{c \times {c^2}}&{c \times ab} \end{array}} \right|.
Now, simplifying the inside of the matrix, we will then get:-
\Rightarrow \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv \dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}} {{a^2}}&{{a^3}}&{abc} \\\ {{b^2}}&{{b^3}}&{abc} \\\ {{c^2}}&{{c^3}}&{abc} \end{array}} \right|
Now, we can clearly observe that all of the rows have abc in it. Therefore, by taking it out of the determinant once, we have basically taken it out once from each of the row.
Step 2: Now, take out abc common.
We will now get the following equivalence:
\Rightarrow \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv abc\dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}} {{a^2}}&{{a^3}}&1 \\\ {{b^2}}&{{b^3}}&1 \\\ {{c^2}}&{{c^3}}&1 \end{array}} \right|
Simplifying it by dividing the common scalar, we will then get:-
\Rightarrow \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv \left| {\begin{array}{*{20}{c}} {{a^2}}&{{a^3}}&1 \\\ {{b^2}}&{{b^3}}&1 \\\ {{c^2}}&{{c^3}}&1 \end{array}} \right|
Step 3: Now, we will perform ${C_1} \leftrightarrow {C_2}$
Now, we know that, if we interchange row, we get a negative sign.
$\therefore$ we will get:

a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv - \left| {\begin{array}{*{20}{c}} {{a^3}}&{{a^2}}&1 \\\ {{b^3}}&{{b^2}}&1 \\\ {{c^3}}&{{c^2}}&1 \end{array}} \right|$$ Step 4: Now, we will perform ${C_1} \leftrightarrow {C_3}$ $\therefore $ we will get: $$ \Rightarrow \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv - \left( { - \left| {\begin{array}{*{20}{c}} 1&{{a^2}}&{{a^3}} \\\ 1&{{b^2}}&{{b^3}} \\\ 1&{{c^2}}&{{c^3}} \end{array}} \right|} \right)$$ Simplifying it by applying – ( - ) = + : $$ \Rightarrow \left| {\begin{array}{*{20}{c}} a&{{a^2}}&{bc} \\\ b&{{b^2}}&{ca} \\\ c&{{c^2}}&{ab} \end{array}} \right| \equiv \left| {\begin{array}{*{20}{c}} 1&{{a^2}}&{{a^3}} \\\ 1&{{b^2}}&{{b^3}} \\\ 1&{{c^2}}&{{c^3}} \end{array}} \right|$$ **Hence, proved.** **Note:** The students might make the mistake of forgetting that they require to multiply with a negative sign whenever they apply the column transformation. Because column transformation does not keep the determinant intact. Therefore, whatever change we make to determinant, we have to replace it back by doing the reverse of it. The students must note that they should not make any column transformation other than interchanging them because other transformations also engage a lot of change to determinant which will become chaotic if tried. So, always prefer to use row transformation.