Question

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

Answer 1

To solve this question, we should perform row and column operations on the determinants. For the determinant of order $n\times n$, there will be n rows and n columns. We can perform linear operations on rows. An example of it is ${{R}_{3}}\to {{R}_{3}}-2{{R}_{1}}-3{{R}_{2}}$ . Similarly, we can perform linear operations on columns. In the question, performing row operation ${{R}_{3}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$, we can take 2 common from the ${{R}_{3}}$of the determinant. Now, we should do the row operations ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\text{ and }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ . Now, we should do the row operation ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}+{{R}_{2}}$ to get the required R.H.S. The key note in performing row operations is that we cannot perform row operations on all the rows at a time. For example, let us consider ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\text{, }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$, ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}+{{R}_{2}}$. The three operations should not be done in the same step. Instead, we should perform the first two operations in one step and we should use the modified ${{R}_{1}}$, ${{R}_{2}}$ from the first two operations in the third step ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}+{{R}_{2}}$.

Complete step by step answer:
Let us consider the given determinant
$\left| \begin{matrix} b+c & q+r & y+z \\\ c+a & r+p & z+x \\\ a+b & p+q & x+y \\\ \end{matrix} \right|$
Let us perform the row operation ${{R}_{3}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$, we get
$\begin{aligned} & \left| \begin{matrix} b+c & q+r & y+z \\\ c+a & r+p & z+x \\\ a+b+b+c+c+a & p+q+q+r+r+p & x+y+y+z+z+x \\\ \end{matrix} \right| \\\ & \Rightarrow \left| \begin{matrix} b+c & q+r & y+z \\\ c+a & r+p & z+x \\\ 2\left( a+b+c \right) & 2\left( p+q+r \right) & 2\left( x+y+z \right) \\\ \end{matrix} \right| \\\ \end{aligned}$
We know the property that
$\left| \begin{matrix} a & b & c \\\ d & e & f \\\ kg & kh & ki \\\ \end{matrix} \right|=k\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|$
Using this and taking 2 outside the determinant, we get
$L.H.S=2\left| \begin{matrix} b+c & q+r & y+z \\\ c+a & r+p & z+x \\\ a+b+c & p+q+r & x+y+z \\\ \end{matrix} \right|$
Performing the row operations ${{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\text{ and }{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}$ step by step, we get
$\begin{aligned} & 2\left| \begin{matrix} b+c-a-b-c & q+r-p-q-r & y+z-x-y-z \\\ c+a & r+p & z+x \\\ a+b+c & p+q+r & x+y+z \\\ \end{matrix} \right| \\\ & \Rightarrow 2\left| \begin{matrix} -a & -p & -x \\\ c+a-a-b-c & r+p-p-q-r & z+x-x-y-z \\\ a+b+c & p+q+r & x+y+z \\\ \end{matrix} \right| \\\ & \Rightarrow 2\left| \begin{matrix} -a & -p & -x \\\ -b & -q & -y \\\ a+b+c & p+q+r & x+y+z \\\ \end{matrix} \right| \\\ \end{aligned}$
Performing the operation ${{R}_{3}}\to {{R}_{3}}+{{R}_{1}}+{{R}_{2}}$, we get
$2\left| \begin{matrix} -a & -p & -x \\\ -b & -q & -y \\\ a+b+c-a-b & p+q+r-p-q & x+y+z-x-y \\\ \end{matrix} \right|=2\left| \begin{matrix} -a & -p & -x \\\ -b & -q & -y \\\ c & r & z \\\ \end{matrix} \right|$
Taking -1 outside in both numerator and denominator, we get
$2\left| \begin{matrix} -a & -p & -x \\\ -b & -q & -y \\\ c & r & z \\\ \end{matrix} \right|=2\left( -1 \right)\left( -1 \right)\left| \begin{matrix} a & p & x \\\ b & q & y \\\ c & r & z \\\ \end{matrix} \right|=2\left| \begin{matrix} a & p & x \\\ b & q & y \\\ c & r & z \\\ \end{matrix} \right|$
So we got the result as
$\left| \begin{matrix} b+c & q+r & y+z \\\ c+a & r+p & z+x \\\ a+b & p+q & x+y \\\ \end{matrix} \right|=2\left| \begin{matrix} a & p & x \\\ b & q & y \\\ c & r & z \\\ \end{matrix} \right|$
Hence, we have proved the required equation.

Note:
Students can make mistakes by applying many row operations in one step. This leads to confusion and in some cases, it will go wrong if we use a previous configuration of an altered row. To avoid these mistakes, it is good practice to perform the operations step by step. Students can start using the property
$\left| \begin{matrix} a+x & b & c \\\ d+y & e & f \\\ g+z & h & i \\\ \end{matrix} \right|=\left| \begin{matrix} a & b & c \\\ d & e & f \\\ g & h & i \\\ \end{matrix} \right|+\left| \begin{matrix} x & b & c \\\ y & e & f \\\ z & h & i \\\ \end{matrix} \right|$ on the L.H.S which will give us two determinants to work with. We can still prove that L.H.S=R.H.S but it will be lengthy. So, it is good to remember the procedure that suits the given determinant.