Question

Without expanding the determinant, prove that $\left. \left| \begin{matrix} a & {{a}^{2}} & bc \\\ b & {{b}^{2}} & ca \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|=\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$.

Answer 1

Hint: First, start with the LHS. Use the property that any number which is common in any row or column can be taken outside of the determinant. Multiply and divide $R_1$​ by a , $R_2$​ by b and $R_3$​ by c. Then interchange column $C_1$ and column $C_3$. Again interchange column $C_2$ and column $C_3$. The expression obtained is equal to the RHS.

Complete step-by-step answer:
In this question, we need to prove that $\left. \left| \begin{matrix} a & {{a}^{2}} & bc \\\ b & {{b}^{2}} & ca \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|=\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$. But we have to do this without actually expanding the determinants.
So, if we cannot expand the determinants, that means that we have to solve the question using row and column operations only.
To solve the question, we will take the LHS, modify it using row and column operations only and arrive at the RHS.
So, let us start by simplifying the LHS.
We are given that:
LHS = $\left. \left| \begin{matrix} a & {{a}^{2}} & bc \\\ b & {{b}^{2}} & ca \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|$
Multiplying and dividing $R_1$​ by a, we will get the following:

{{a}^{2}} & {{a}^{3}} & abc \\\ b & {{b}^{2}} & ca \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|$$ We know that any number which is common in any row or column can be taken outside of the determinant. Using this property, we have taken out a in the denominator. Multiplying and dividing $R_2$​ by b and taking the b in the denominator outside the matrix, we will get the following: $$\dfrac{1}{ab}\left. \left| \begin{matrix} {{a}^{2}} & {{a}^{3}} & abc \\\ {{b}^{2}} & {{b}^{3}} & abc \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|$$ Similarly, multiplying and dividing $R_3$​ by c and taking the c in the denominator outside the matrix, we will get the following: $$\dfrac{1}{abc}\left. \left| \begin{matrix} {{a}^{2}} & {{a}^{3}} & abc \\\ {{b}^{2}} & {{b}^{3}} & abc \\\ {{c}^{2}} & {{c}^{3}} & abc \\\ \end{matrix} \right. \right|$$ Now, we see that the term abc is common in the column $C_3$ . So, we will take abc common from the column $C_3$ outside the determinant. $$\dfrac{abc}{abc}\left. \left| \begin{matrix} {{a}^{2}} & {{a}^{3}} & 1 \\\ {{b}^{2}} & {{b}^{3}} & 1 \\\ {{c}^{2}} & {{c}^{3}} & 1 \\\ \end{matrix} \right. \right|$$ $$\left. \left| \begin{matrix} {{a}^{2}} & {{a}^{3}} & 1 \\\ {{b}^{2}} & {{b}^{3}} & 1 \\\ {{c}^{2}} & {{c}^{3}} & 1 \\\ \end{matrix} \right. \right|$$ Now, we know the fact that: When two rows or columns of a determinant are interchanged, then the value of the determinant differs by a negative sign. Using this, we interchange the column $C_1$ and column $C_3$. $$-\left. \left| \begin{matrix} 1 & {{a}^{3}} & {{a}^{2}} \\\ 1 & {{b}^{3}} & {{b}^{2}} \\\ 1 & {{c}^{3}} & {{c}^{2}} \\\ \end{matrix} \right. \right|$$ Again, we will interchange the column $C_2$ and column $C_3$. $$\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$$ So, LHS = $$\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$$ RHS = $$\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$$ Hence, LHS = RHS $$\left. \left| \begin{matrix} a & {{a}^{2}} & bc \\\ b & {{b}^{2}} & ca \\\ c & {{c}^{2}} & ab \\\ \end{matrix} \right. \right|=\left. \left| \begin{matrix} 1 & {{a}^{2}} & {{a}^{3}} \\\ 1 & {{b}^{2}} & {{b}^{3}} \\\ 1 & {{c}^{2}} & {{c}^{3}} \\\ \end{matrix} \right. \right|$$ Hence proved. Note: In this question, it is important to know the following facts about row and column operations of determinants: any number which is common in any row or column can be taken outside of the determinant and that when two rows or columns of a determinant are interchanged, then the value of determinant differs by a negative sign.