Question

Using properties of determinant, prove that: $\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=\text{ }abc\text{ }+\text{ }ab\text{ }+\text{ }ac\text{ }+\text{ }bc$

Answer 1

To solve this determinant, what we will do is firstly, we will take factor a common from ${{R}_{1}}$ , b common from ${{R}_{2}}$and c common from ${{R}_{3}}$, then we will use row and column elementary transformation to solve the determinant.

Complete step by step answer:
Now, before we start solving the questions, let us see how we calculate determinant -
$\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|={{a}_{11}}({{a}_{22}}{{a}_{33}}-{{a}_{32}}{{a}_{23}})-{{a}_{21}}({{a}_{12}}{{a}_{33}}-{{a}_{32}}{{a}_{13}})+{{a}_{31}}({{a}_{23}}{{a}_{12}}-{{a}_{22}}{{a}_{13}})$
Some of the properties of determinant are as follows,
( a ) Determinant evaluated across any row or column is the same.
( b ) If all the elements of a row or a column are zeros, then the value of the determinant is equal to zero.
( c ) If rows and columns are interchanged then the value of the determinant remains the same.
( d ) Determinant of an identity matrix is 1.
Now, let us move to question now, it is asked to prove that $\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=\text{ }abc\text{ }+\text{ }ab\text{ }+\text{ }ac\text{ }+\text{ }bc$
So, this can be proved by solving determinant across any row or column, as the value of the determinant will be the same.
Now, we can write $\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|$, as
$\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=abc\left| \begin{matrix} \dfrac{1}{a}+1 & \dfrac{1}{a} & \dfrac{1}{a} \\\ \dfrac{1}{b} & \dfrac{1}{b}+1 & \dfrac{1}{b} \\\ \dfrac{1}{c} & \dfrac{1}{c} & \dfrac{1}{c}+1 \\\ \end{matrix} \right|$, where we took a common from ${{R}_{1}}$ , b common from ${{R}_{2}}$and c common from ${{R}_{3}}$.
Now, using elementary row operation ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$ , we get
$=abc\left| \begin{matrix} \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 & \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \\\ \dfrac{1}{b} & \dfrac{1}{b}+1 & \dfrac{1}{b} \\\ \dfrac{1}{c} & \dfrac{1}{c} & \dfrac{1}{c}+1 \\\ \end{matrix} \right|$
Taking $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1$common from ${{R}_{1}}$, we get
$=abc\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left| \begin{matrix} 1 & 1 & 1 \\\ \dfrac{1}{b} & \dfrac{1}{b}+1 & \dfrac{1}{b} \\\ \dfrac{1}{c} & \dfrac{1}{c} & \dfrac{1}{c}+1 \\\ \end{matrix} \right|$
Using, ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}}$ and ${{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$, we get
$=abc\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left| \begin{matrix} 1 & 0 & 0 \\\ \dfrac{1}{b} & 1 & 0 \\\ \dfrac{1}{c} & 0 & 1 \\\ \end{matrix} \right|$
Expanding determinant along ${{R}_{1}}$, we get
$=abc\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left[ 1(1-0)-0\left( \dfrac{1}{b}-0\cdot \dfrac{1}{c} \right)+0\left( 0\cdot \dfrac{1}{b}-1\cdot \dfrac{1}{c} \right) \right]$
On simplifying, we get
$=abc\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left[ 1-0+0 \right]$
$=abc\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)$
On solving, we get
$=abc\text{ }+\text{ }ab\text{ }+\text{ }ac\text{ }+\text{ }bc$

Hence, determinant $\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=\text{ }abc\text{ }+\text{ }ab\text{ }+\text{ }ac\text{ }+\text{ }bc$

Note: It is very important to know how to solve determinant using it’s properties so knowledge of properties of determinant should be a priority. In determinant we can use both column and row elementary transformation. Calculation should be done carefully while solving determinant problems.