# Product of determinants

Let $${{\Delta }_{1}}=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|$$ and $${{\Delta }_{2}}=\left| \begin{matrix}{{\alpha }_{1}} & {{\beta }_{1}} & {{\gamma }_{1}} \\{{\alpha }_{2}} & {{\beta }_{2}} & {{\gamma }_{2}} \\{{\alpha }_{3}} & {{\beta }_{3}} & {{\gamma }_{3}} \\\end{matrix} \right|$$ be two determinants. Then, the product Δ1Δ2 is defined as $${{\Delta }_{1}}{{\Delta}_{2}}=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|\left|\begin{matrix}{{\alpha }_{1}} & {{\beta }_{1}} & {{\gamma }_{1}} \\{{\alpha }_{2}} & {{\beta }_{2}} & {{\gamma }_{2}} \\{{\alpha }_{3}} & {{\beta }_{3}} & {{\gamma }_{3}} \\\end{matrix} \right|$$

= $$\left| \begin{matrix}{{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\alpha }_{2}}+{{c}_{1}}{{\alpha }_{3}} &{{a}_{1}}{{\beta }_{1}}+{{b}_{1}}{{\beta }_{2}}+{{c}_{1}}{{\beta}_{3}}&{{a}_{1}}{{\gamma}_{1}}+{{b}_{1}}{{\gamma}_{2}}+{{c}_{1}}{{\gamma}_{3}}\\{{a}_{2}}{{\alpha}_{1}}+{{b}_{2}}{{\alpha }_{2}}+{{c}_{2}}{{\alpha}_{3}}&{{a}_{2}}{{\beta}_{1}}+{{b}_{2}}{{\beta}_{2}}+{{c}_{2}}{{\beta}_{3}}&{{a}_{2}}{{\gamma}_{1}}+{{b}_{2}}{{\gamma}_{2}}+{{c}_{2}}{{\gamma}_{3}}\\{{a}_{3}}{{\alpha}_{1}}+{{b}_{3}}{{\alpha}_{2}}+{{c}_{3}}{{\alpha}_{3}}&{{a}_{3}}{{\beta}_{1}}+{{b}_{3}}{{\beta}_{2}}+{{c}_{3}}{{\beta}_{3}}&{{a}_{3}}{{\gamma}_{1}}+{{b}_{3}}{{\gamma }_{2}}+{{c}_{3}}{{\gamma }_{3}}\\\end{matrix} \right|$$

This is row-by-column multiplication value for finding the product of two determinants and it is same as the rule of multiplication of two matrices. Since the value of a determinant does not alter by interchanging the rows and columns. So, we can also follow the row-by-row or the column –by-row or column-by-column multiplication rule.

$${{\Delta }_{1}}{{\Delta }_{2}}=\left| \begin{matrix}{{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\{{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\{{a}_{3}} & {{b}_{3}} & {{c}_{3}} \\\end{matrix} \right|\left| \begin{matrix}{{\alpha }_{1}} & {{\beta }_{1}} & {{\gamma }_{1}} \\{{\alpha }_{2}} & {{\beta }_{2}} & {{\gamma }_{2}} \\{{\alpha }_{3}} & {{\beta }_{3}} & {{\gamma }_{3}} \\\end{matrix} \right|$$

= $$\left| \begin{matrix}{{a}_{1}}{{\alpha }_{1}}+{{b}_{1}}{{\beta }_{1}}+{{c}_{1}}{{\gamma }_{1}} & {{a}_{1}}{{\alpha }_{2}}+{{b}_{1}}{{\beta }_{2}}+{{c}_{1}}{{\gamma }_{2}} & {{a}_{1}}{{\alpha }_{3}}+{{b}_{1}}{{\beta }_{3}}+{{c}_{1}}{{\gamma }_{3}} \\{{a}_{2}}{{\alpha }_{1}}+{{b}_{2}}{{\beta }_{1}}+{{c}_{2}}{{\gamma }_{1}} & {{a}_{2}}{{\alpha}_{2}}+{{b}_{2}}{{\beta}_{2}}+{{c}_{2}}{{\gamma}_{2}}&{{a}_{2}}{{\alpha}_{3}}+{{b}_{2}}{{\beta}_{3}}+{{c}_{2}}{{\gamma }_{3}} \\{{a}_{3}}{{\alpha }_{1}}+{{b}_{3}}{{\beta }_{1}}+{{c}_{3}}{{\gamma }_{1}} & {{a}_{3}}{{\alpha}_{2}}+{{b}_{3}}{{\beta }_{2}}+{{c}_{3}}{{\gamma }_{2}} & {{a}_{3}}{{\alpha }_{3}}+{{b}_{3}}{{\beta}_{3}}+{{c}_{3}}{{\gamma }_{3}} \\\end{matrix} \right|$$

Ex: If Sr = αr + βr + γr. Show that $$\left| \begin{matrix}{{S}_{0}} & {{S}_{1}} & {{S}_{2}} \\{{S}_{1}} & {{S}_{2}} & {{S}_{3}} \\{{S}_{2}} & {{S}_{3}} & {{S}_{4}} \\\end{matrix} \right|$$ = (α – β)² (β – γ)² (γ – α)².

Sol: We have,

$$\left| \begin{matrix}{{S}_{0}} & {{S}_{1}} & {{S}_{2}} \\{{S}_{1}} & {{S}_{2}} & {{S}_{3}} \\{{S}_{2}} & {{S}_{3}} & {{S}_{4}} \\\end{matrix} \right|=\left| \begin{matrix}3 & \alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \\\alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha}^{3}}+{{\beta }^{3}}+{{\gamma }^{3}} \\{{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma}^{3}} & {{\alpha }^{4}}+{{\beta }^{4}}+{{\gamma }^{4}} \\\end{matrix} \right|$$

= $$\left| \begin{matrix}1 & 1 & 1 \\\alpha & \beta & \gamma \\{{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\\end{matrix} \right|\left| \begin{matrix}1 & 1 & 1 \\\alpha & \beta & \gamma \\{{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\\end{matrix} \right|$$

= $$\,{{\left| \begin{matrix}1 & 1 & 1 \\\alpha & \beta & \gamma \\{{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\\end{matrix} \right|}^{2}}$$

= [(α – β) (β – γ) (γ – α)]²

= (α – β)² (β – γ)² (γ – α)²