Question

Find the value of the following determinant using column and row operations on it and choose the correct alternative$\left| \begin{matrix}
   1 & 1 & 1 \\
   bc & ca & ab \\
   b+c & c+a & a+b \\
\end{matrix} \right|$.
a)1
b)0
c)$\left( a-b \right)\left( b-c \right)\left( c-a \right)$
d)$\left( a+b \right)\left( b+c \right)\left( c+a \right)$

Answer 1

Hint: We will first simplify the given determinant by applying various row and column operations as ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and then ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$ and finally expand the obtained determinant along ${{R}_{1}}$ to get the value of the given determinant.

Complete step-by-step answer:
It is given in the question that we have to find the value of determinant $\left| \begin{matrix}
   1 & 1 & 1 \\
   bc & ca & ab \\
   b+c & c+a & a+b \\
\end{matrix} \right|$. We will first simplify the given determinant by performing the following operations ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and then ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$. Now first applying ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$, we get the determinant modified as \[\left| \begin{matrix}
   1 & 1-1 & 1 \\
   bc & ca-bc & ab \\
   b+c & \left( c+a \right)-\left( b+c \right) & a+b \\
\end{matrix} \right|\], that is, \[\left| \begin{matrix}
   1 & 0 & 1 \\
   bc & c\left( a-b \right) & ab \\
   b+c & \left( a-b \right) & a+b \\
\end{matrix} \right|\].
Now, taking $\left( a-b \right)$ common from ${{C}_{2}}$, we get $\left( a-b \right)\left| \begin{matrix}
   1 & 0 & 1 \\
   bc & c & ab \\
   b+c & 1 & a+b \\
\end{matrix} \right|$. Again we apply the column operation as ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$, we get $\left( a-b \right)\left| \begin{matrix}
   1-1 & 0 & 1 \\
   bc-ab & c & ab \\
   \left( b+c \right)-\left( a+b \right) & 1 & a+b \\
\end{matrix} \right|$, further solving and simplifying we get $\left( a-b \right)\left| \begin{matrix}
   1-1 & 0 & 1 \\
   b\left( c-a \right) & c & ab \\
   \left( c-a \right) & 1 & a+b \\
\end{matrix} \right|$. Now, taking $\left( c-a \right)$ common from ${{C}_{1}}$ we get $\left( a-b \right)\left( c-a \right)\left| \begin{matrix}
   0 & 0 & 1 \\
   b & c & ab \\
   1 & 1 & a+b \\
\end{matrix} \right|$.
Now expanding the determinant along row ${{R}_{1}}$ we get, $\left( a-b \right)\left( c-a \right)\left[ 1\times \left( b-c \right) \right]$ or simplified as \[\left( a-b \right)\left( b-c \right)\left( c-a \right)\]. Therefore we expanded the determinant to get an expression \[\left( a-b \right)\left( b-c \right)\left( c-a \right)\], therefore option c) is the correct answer.

Note: Usually students directly expand the given determinant directly which is not the correct method to find the value of any determinant. Before expanding any determinant we have to simplify it and try to make our expression as simple as it is possible. Also, while taking common in any row and column we have to keep our eyes on the sign of each term individually. In general, always try to make two zeroes in any of the row or the column using row and column operations, it makes it easy to expand the determinant along the row or column containing maximum number of zeroes.