Question

Using the properties of determinants, prove that $\left| \begin{matrix} b+c & c+a & a+b \\\ q+r & r+p & p+q \\\ y+z & z+x & x+y \\\ \end{matrix} \right|=2\left| \begin{matrix} a & b & c \\\ p & q & r \\\ x & y & z \\\ \end{matrix} \right|$

Answer 1

To solve this question, we will first consider the left hand side. We will apply column transformations and operations to prove that the left hand side is equal to the right hand side. The operations we can perform are addition of one column to another, subtraction of one column from another, multiplication of a column with any real number and taking a multiple common from a column.

Complete step by step answer:
The left hand side is given to us as $\left| \begin{matrix} b+c & c+a & a+b \\\ q+r & r+p & p+q \\\ y+z & z+x & x+y \\\ \end{matrix} \right|$
We will change the column 3 as the addition of column 1, column 2 and column 3.
The changed determinant after ${{C}_{3}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}$ will be as follows: $\begin{aligned} & \Rightarrow LHS=\left| \begin{matrix} b+c & c+a & a+b+b+c+c+a \\\ q+r & r+p & p+q+q+r+r+p \\\ y+z & z+x & x+y+y+z+z+x \\\ \end{matrix} \right| \\\ & \Rightarrow LHS=\left| \begin{matrix} b+c & c+a & 2\left( a+b+c \right) \\\ q+r & r+p & 2\left( p+q+r \right) \\\ y+z & z+x & 2\left( x+y+z \right) \\\ \end{matrix} \right| \\\ \end{aligned}$
We will take 2 common from the third row. The 2 will come outside the determinant.

b+c & c+a & \left( a+b+c \right) \\\ q+r & r+p & \left( p+q+r \right) \\\ y+z & z+x & \left( x+y+z \right) \\\ \end{matrix} \right|$$ We will now transform column 1 as difference of column 1 and column 3. Thus, after the transformation ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$, the left hand side will be as follows: $$\Rightarrow LHS=2\left| \begin{matrix} -a & c+a & \left( a+b+c \right) \\\ -p & r+p & \left( p+q+r \right) \\\ -x & z+x & \left( x+y+z \right) \\\ \end{matrix} \right|$$ Similar transformation will be done on column 2 and column 2 will change as difference of column 2 and column 3. Thus, after the transformation ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$, the left hand side is changed as follows: $$\Rightarrow LHS=2\left| \begin{matrix} -a & -b & \left( a+b+c \right) \\\ -p & -q & \left( p+q+r \right) \\\ -x & -y & \left( x+y+z \right) \\\ \end{matrix} \right|$$ Now, the third column will be transformed as the sum of column 1, column 2 and column 3. After the transformation ${{C}_{3}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}$, the left hand side is changed as follows: $$\Rightarrow LHS=2\left| \begin{matrix} -a & -b & c \\\ -p & -q & r \\\ -x & -y & z \\\ \end{matrix} \right|$$ Now, we will take ─1 from column 1 and ─1 from column 2. $$\Rightarrow LHS=2\left( -1 \right)\left( -1 \right)\left| \begin{matrix} a & b & c \\\ p & q & r \\\ x & y & z \\\ \end{matrix} \right|$$ The product of ─1 and ─1 is 1. $$\begin{aligned} & \Rightarrow LHS=2\left| \begin{matrix} a & b & c \\\ p & q & r \\\ x & y & z \\\ \end{matrix} \right| \\\ & \Rightarrow LHS=RHS \\\ \end{aligned}$$ **Hence, we proved that left hand side is equal to right hand side, i.e. $$\left| \begin{matrix} b+c & c+a & a+b \\\ q+r & r+p & p+q \\\ y+z & z+x & x+y \\\ \end{matrix} \right|=2\left| \begin{matrix} a & b & c \\\ p & q & r \\\ x & y & z \\\ \end{matrix} \right|$$** **Note:** Students have to be careful while row and column transformation. The operations like ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$ and ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ cannot happen simultaneously. Hence, it is a good practice to carry operations with only one reference column, like ${{C}_{1}}\to {{C}_{1}}-{{C}_{3}}$ and ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$.