Question

Using the properties of determinants, prove the following:

1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=ab+bc+ca+abc$$

Answer 1

Let us assume that determinant is $\Delta =\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|$. We need to solve the given determinant to prove that the determinant is equal to $\Delta =abc+ab+bc+ac$. To solve the determinant, we can expand the whole determinant along its first row and get the final result.

Complete step by step answer:
Since we need to prove that:

1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|=ab+bc+ca+abc$$ So, we can start by taking LHS = $$\Delta =\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|$$ We need to find the value of the determinant, so we can expand it along the first row. While expanding a determinant of order $3\times 3$, say $\Delta =\left| \begin{matrix} {{a}_{11}} & {{a}_{12}} & {{a}_{13}} \\\ {{a}_{21}} & {{a}_{22}} & {{a}_{23}} \\\ {{a}_{31}} & {{a}_{32}} & {{a}_{33}} \\\ \end{matrix} \right|$, we can write: $$\begin{aligned} & \Delta =+{{a}_{11}}\left\\{ \left( {{a}_{22}}\times {{a}_{33}} \right)-\left( {{a}_{32}}\times {{a}_{23}} \right) \right\\} \\\ & -{{a}_{12}}\left\\{ \left( {{a}_{21}}\times {{a}_{33}} \right)-\left( {{a}_{31}}\times {{a}_{23}} \right) \right\\} \\\ & +{{a}_{13}}\left\\{ \left( {{a}_{21}}\times {{a}_{32}} \right)-\left( {{a}_{31}}\times {{a}_{22}} \right) \right\\} \end{aligned}$$ So, we can write the expansion of the determinant $$\Delta =\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|$$ as:$\begin{aligned} & \Delta =\left( 1+a \right)\left\\{ \left( 1+b \right)\left( 1+c \right)-1 \right\\} \\\ & -1\left\\{ \left( 1+c \right)-1 \right\\} \\\ & +1\left\\{ 1-\left( 1+b \right) \right\\}......(1) \end{aligned}$ By solving the equation (1), we get: $\begin{aligned} & \Delta =\left[ \left( 1+a \right)\left( c+b+bc \right)-c-b \right] \\\ & =\left[ c+b+bc+ac+ab+abc-c-b \right] \\\ & =\left[ bc+ac+ab+abc \right]......(2) \end{aligned}$ Since it was mentioned in the question that $RHS=\left[ ab+bc+ca+abc \right]$ So, comparing the obtained result of LHS with RHS, we get LHS = RHS Hence proved that, $\Delta =\left[ ab+bc+ca+abc \right]$. **Note:** Another way to solve the given determinant $$\Delta =\left| \begin{matrix} 1+a & 1 & 1 \\\ 1 & 1+b & 1 \\\ 1 & 1 & 1+c \\\ \end{matrix} \right|$$ is to get a, b, and c common from each row, i.e.: $$\Delta =\left( abc \right)\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|$$ Now, apply ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$, we have: $$\Delta =\left( abc \right)\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|$$ Now, take $$\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)$$ common from ${{R}_{1}}$, we get: $$\Delta =\left( abc \right)\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|$$ Now, apply ${{C}_{2}}\to {{C}_{2}}-{{C}_{1}};{{C}_{3}}\to {{C}_{3}}-{{C}_{1}}$, we get: $$\Delta =\left( abc \right)\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|$$ Now, we can expand the determinant to get the final result: $$\begin{aligned} & \Delta =\left( abc \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left( 1\left( 1-0 \right)-0\left( \dfrac{1}{b}-0 \right)+0\left( 0-\dfrac{1}{c} \right) \right) \\\ & =\left( abc \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right)\left( 1-0-0 \right) \\\ & =\left( abc \right)\left( \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+1 \right) \\\ & =bc+ac+ab+abc \\\ & =RHS \end{aligned}$$ Hence proved that, $\Delta =\left[ ab+bc+ca+abc \right]$.