Question

Show that the determinant \[\left| \begin{matrix}
   a+b+2c & a & b \\
   c & b+c+2a & b \\
   c & a & a+c+2b \\
\end{matrix} \right|=2{{\left( a+b+c \right)}^{3}}\].

Answer 1

Hint:In this question, in order to find the determinant of the given matrix. We will first use the property of the determinant of a matrix that performing any row or column operation in the given matrix will not change the determinant of the matrix. Thus we will use a column operation where \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\]. We will then get a matrix with elements of the first column that are all equal to \[2\left( a+b+c \right)\]. Then using the property of determinant of a matrix \[A\] that if in any row or a column all the elements are equal, say \[x\] . Then we can take the value \[x\] out of the determinant of the matrix \[A\]leaving the entries of that row or column equal to 1. Using this will take the common factor \[2\left( a+b+c \right)\] from the first column of the resultant determinant leaving all the entries of the first column equal to 1. Then we will perform two row operations \[{{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\] and \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] and then evaluate the determinant in order to get the desired result.

Complete step by step answer:
Let us suppose that the matrix \[A=\left[ \begin{matrix}
   a+b+2c & a & b \\
   c & b+c+2a & b \\
   c & a & a+c+2b \\
\end{matrix} \right]\].
We have to evaluate the determinant of the matrix \[A\].
Now we will be using the property of the determinant of a matrix that performing any row or column operation in the given matrix will not change the determinant of the matrix.
We also have \[\left| A \right|=\left| \begin{matrix}
   a+b+2c & a & b \\
   c & b+c+2a & b \\
   c & a & a+c+2b \\
\end{matrix} \right|\].
Now we will use the column operation \[{{C}_{1}}\to {{C}_{1}}+{{C}_{2}}+{{C}_{3}}\] in matrix \[A\] and we know that the determinant of the matrix will not change.
Thus we have
\[\left| A \right|=\left| \begin{matrix}
   2a+2b+2c & a & b \\
   2a+2b+2c & b+c+2a & b \\
   2a+2b+2c & a & a+c+2b \\
\end{matrix} \right|\]
We now have that elements of the first column are all equal to \[2\left( a+b+c \right)\].
Then using the property of determinant of a matrix \[A\] that if in any row or a column all the elements are equal, say \[x\] . Then we can take the value \[x\] out of the determinant of the matrix \[A\]leaving the entries of that row or column equal to 1.
Using this will take the common factor \[2\left( a+b+c \right)\] from the first column of the resultant determinant leaving all the entries of the first column equal to 1.
Then we have
\[\begin{align}
  & \left| A \right|=\left( 2a+2b+2c \right)\left| \begin{matrix}
   1 & a & b \\
   1 & b+c+2a & b \\
   1 & a & a+c+2b \\
\end{matrix} \right| \\
 & =2\left( a+b+c \right)\left| \begin{matrix}
   1 & a & b \\
   1 & b+c+2a & b \\
   1 & a & a+c+2b \\
\end{matrix} \right|
\end{align}\]
Now we will use the row operation \[{{R}_{1}}\to {{R}_{1}}-{{R}_{3}}\] in the above determinant.
Then we get
\[\left| A \right|=2\left( a+b+c \right)\left| \begin{matrix}
   0 & 0 & -a-b-c \\
   1 & b+c+2a & b \\
   1 & a & a+c+2b \\
\end{matrix} \right|\]
Again we will use the row operation \[{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] in the above determinant.
We now have
\[\left| A \right|=2\left( a+b+c \right)\left| \begin{matrix}
   0 & 0 & -a-b-c \\
   0 & b+c+a & -a-b-c \\
   1 & a & a+c+2b \\
\end{matrix} \right|\]
Now on evaluating the above determinant we get
\[\begin{align}
  & \left| A \right|=2\left( a+b+c \right)\left( 0-0+1\left( 0+{{\left( a+b+c \right)}^{2}} \right) \right) \\
 & =2\left( a+b+c \right){{\left( a+b+c \right)}^{2}} \\
 & =2{{\left( a+b+c \right)}^{3}}
\end{align}\]
Thus we have shown that
\[\left| \begin{matrix}
   a+b+2c & a & b \\
   c & b+c+2a & b \\
   c & a & a+c+2b \\
\end{matrix} \right|=2{{\left( a+b+c \right)}^{3}}\].

Note:
In this problem, please do not evaluate the determinant of the given matrix directly using the definition of determinants otherwise it will be difficult to get the desired answer. Instead we can use properties of determinants of matrices in order to simplify the problem.