Question

Show that $\left| \begin{matrix} b+c & c+a & a+b \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ .

Answer 1

Hint : We first use the row and column operations to simplify the determinant values. We take $a+b+c$ as the common component from the same row. Then we expand the determinant to find the final value. We use formula of $\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$.

Complete step-by-step answer :
We need to show that the determinant value will be equal to ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$ .
We can apply row operations on the determinant value without changing the initial form.
So, we take the form of ${{R}_{1}}^{'}={{R}_{1}}+{{R}_{3}}$ .
We get $\left| \begin{matrix} b+c & c+a & a+b \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|=\left| \begin{matrix} a+b+c & b+c+a & a+b+c \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|$ .
We take common the term $a+b+c$ from the first row.
$\left| \begin{matrix} b+c & c+a & a+b \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|=\left( a+b+c \right)\left| \begin{matrix} 1 & 1 & 1 \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|$
Now we simplify further by taking the operations like ${{C}_{2}}^{'}={{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}^{'}={{C}_{3}}-{{C}_{1}}$ .
We get $\left( a+b+c \right)\left| \begin{matrix} 1 & 1 & 1 \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|=\left( a+b+c \right)\left| \begin{matrix} 1 & 0 & 0 \\\ a+b & c-a & c-b \\\ a & b-a & c-a \\\ \end{matrix} \right|$
Now we expand the determinant value through the first row.
So, $\left| \begin{matrix} b+c & c+a & a+b \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|=\left( a+b+c \right)\left[ {{\left( c-a \right)}^{2}}-\left( c-b \right)\left( b-a \right) \right]$ .
We simplify the expression and get

& \left( a+b+c \right)\left[ {{\left( c-a \right)}^{2}}-\left( c-b \right)\left( b-a \right) \right] \\\ & =\left( a+b+c \right)\left( {{c}^{2}}+{{a}^{2}}-2ac-bc+ac+{{b}^{2}}-ab \right) \\\ & =\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right) \\\ \end{aligned}$$ We also know that $$\left( a+b+c \right)\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac \right)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc$$. Therefore, $ \left| \begin{matrix} b+c & c+a & a+b \\\ a+b & b+c & c+a \\\ a & b & c \\\ \end{matrix} \right|={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc $ . **Note** : There are certain operations which we can apply for the problems. We can switch two rows or columns which causes the determinant to switch sign. We can add a multiple of one row to another which causes the determinant to remain the same. We can multiply a row as a constant result in the determinant scaling by that constant.