Question

Using the properties determinants show that \[\left| \left. \begin{matrix}
   b+c & a & a \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.=4abc\]

Answer 1

Hint: Use row operations or column operation of determinants to eliminate (conversion to zero) as many as elements of a row or a column respectively of the given determinant and then calculate the value of the determinant using the conventional multiplication by row or a column.

Complete step by step answer:
We know that from the property of determinants that if multiple of a row is subtracted from another row then the values of the determinant does not change. Mathematically, if the determinant constitutes has $\text{m}^{th}$ and $n^{th}$row denoted as ${{R}_{m}}\text{ and }{{R}_{n}}$ respectively, then by row operation, ${{R}_{m}}\text{ }\leftarrow {{R}_{m}}-k{{R}_{n}}$ where $k$ is an integer.\[\]
Similarly if multiple of a column is subtracted from another column then the values of the determinant does not change. Mathematically, if the determinant constitutes has $\text{m}^{th}$ and $n^{th}$ row denoted as ${{C}_{m}}\text{ and }{{C}_{n}}$ respectively, then by row operation, ${{C}_{m}}\text{ }\leftarrow {{C}_{m}}-k{{C}_{n}}$ where $k$ is an integer.
We are only going to use the row operation here.
Let the given determinant denoted by $\Delta $ which consists of three rows${{R}_{1}},{{R}_{2}},{{R}_{3}}$. \[\]
At the Left Hand Side,
\[\Delta =\left| \left. \begin{matrix}
   b+c & a & a \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.\]
Applying row operation ${{R}_{1}}={{R}_{1}}+(-1){{R}_{2}}$,\[\]
\[\Delta =\left| \left. \begin{matrix}
   b+c & a & a \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.=\left| \left. \begin{matrix}
   b+c-b & a-c-a & a-b \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.=\left| \left. \begin{matrix}
   c & -c & a-b \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.\]

 Again applying row operation, ${{R}_{1}}={{R}_{1}}+(-1){{R}_{3}}$, \[\Delta =\left| \left. \begin{matrix}
   c & -c & a-b \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.=\left| \left. \begin{matrix}
   c-c & -c-c & a-b-a-b \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.=\left| \left. \begin{matrix}
   0 & -2c & -2b \\
   b & c+a & b \\
   c & c & a+b \\
\end{matrix} \right| \right.\]
We can observe that only one element can only be eliminated.
Now expanding along ${{R}_{1}}$
\[\begin{align}
  & \Delta =0\left| \left. \begin{matrix}
   c+a & b \\
   c & a+b \\
\end{matrix} \right| \right.-\left( -2c \right)\left| \left. \begin{matrix}
   b & b \\
   c & a+b \\
\end{matrix} \right| \right.+\left( -2b \right)\left| \left. \begin{matrix}
   b & c+a \\
   c & c \\
\end{matrix} \right| \right. \\
 & =2c\left( ab+{{b}^{2}}-bc \right)-2b\left( bc-{{c}^{2}}-ac \right) \\
 & =2abc+2{{b}^{2}}c-2b{{c}^{2}}-2{{b}^{2}}c+2b{{c}^{2}}+2abc \\
 & =4abc \\
\end{align}\]
This is equal to the value at the right hand side.

Note: The calculation of value of the determinant may require a little more time than usual when we cannot eliminate at least two elements of a row or a column. At that time we should be patient or keep the problem to solve at the end after we finish solving other problems.