Question

Without expanding the determinant, prove that:

1&a&{bc} \\\ 1&b&{ca} \\\ 1&c&{ab} \end{array}} \right| = \left| {\begin{array}{*{20}{c}} 1&a&{{a^2}} \\\ 1&b&{{b^2}} \\\ 1&c&{{c^2}} \end{array}} \right|$$

Answer 1

To obtain the required RHS we multiply and divide the determinant in LHS by ‘$abc$’. Multiply ‘a’ to the first row, ‘b’ to the second row and ‘c’ to the third row. Interchange the required columns and form the determinant in RHS of the equation.

• If we multiply a term ‘k’ to the determinant then the value ‘k’ is multiplied to a complete row i.e.

m&n&o \\\ p&q&r \\\ s&t&u \end{array}} \right| = \left| {\begin{array}{*{20}{c}} {km}&{kn}&{ko} \\\ p&q&r \\\ s&t&u \end{array}} \right|$$ **Complete step-by-step answer:** We are given the determinant on LHS of the equation$$\left| {\begin{array}{*{20}{c}} 1&a&{bc} \\\ 1&b&{ca} \\\ 1&c&{ab} \end{array}} \right|$$ ……….… (1) Since we know if we multiply and divide a value by same number then there will be no change in the original value. So, we multiply the determinant in equation (1) by $$abc$$ So, the equation (1) becomes $$ \Rightarrow \dfrac{{abc}}{{abc}}\left| {\begin{array}{*{20}{c}} 1&a&{bc} \\\ 1&b&{ca} \\\ 1&c&{ab} \end{array}} \right|$$ Now we use the property that if any determinant multiplied by a number, then that number is multiplied to each component of a row. Since, the determinant in RHS has $${a^2},{b^2},{c^2}$$in the first, second and third row respectively so we multiply the determinant in LHS as first row by ‘a’, second row by ‘b’ and third row by ‘c’

\Rightarrow \dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}}
{1 \times a}&{a \times a}&{bc \times a} \\
{1 \times b}&{b \times b}&{ca \times b} \\
{1 \times c}&{c \times c}&{ab \times c}
\end{array}} \right|$$
Calculate the product of each element in the determinant

a&{{a^2}}&{abc} \\\ b&{{b^2}}&{abc} \\\ c&{{c^2}}&{abc} \end{array}} \right|$$ Now interchange the required columns. $${C_1} \leftrightarrow {C_3}$$ $$ \Rightarrow \dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}} {abc}&{{a^2}}&a \\\ {abc}&{{b^2}}&b \\\ {abc}&{{c^2}}&c \end{array}} \right|$$ $${C_2} \leftrightarrow {C_3}$$ $$ \Rightarrow \dfrac{1}{{abc}}\left| {\begin{array}{*{20}{c}} {abc}&a&{{a^2}} \\\ {abc}&b&{{b^2}} \\\ {abc}&c&{{c^2}} \end{array}} \right|$$ Take ‘$$abc$$’ common from the first column $$ \Rightarrow \dfrac{1}{{abc}} \times abc\left| {\begin{array}{*{20}{c}} 1&a&{{a^2}} \\\ 1&b&{{b^2}} \\\ 1&c&{{c^2}} \end{array}} \right|$$ Cancel same terms that are in the multiplication with the determinant $$ \Rightarrow \left| {\begin{array}{*{20}{c}} 1&a&{{a^2}} \\\ 1&b&{{b^2}} \\\ 1&c&{{c^2}} \end{array}} \right|$$ This is equal to the RHS of the equation. $$ \Rightarrow $$LHS$$ = $$RHS **Hence Proved.** **Note:** Students many times get confused and try to form the elements forcefully by applying row transformations, this will make the solution more and more complex and it will become difficult to take common factors from rows and elements.