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

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,

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{aligned} & \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{aligned}$$ 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.