Question

If we have $x\ne 0,y\ne 0,z\ne 0$ and $\left| \begin{matrix} 1+x & 1 & 1 \\\ 1+y & 1+2y & 1 \\\ 1+z & 1+z & 1+3z \\\ \end{matrix} \right|=0$ then ${{x}^{-1}}+{{y}^{-1}}+{{z}^{-1}}$ is equal to
(a) -1
(b) -2
(c) -3
(d) $\dfrac{1}{3}$

Answer 1

In this type of question we have to use the concept of determinants. We will apply row and column transformations on the given determinant to simplify the calculation of finding determinant. And then we expand the determinant along the first row and then equate it to 0 so that we get an equation. We simplify this equation to obtain the value of ${{x}^{-1}}+{{y}^{-1}}+{{z}^{-1}}$.

Complete step-by-step solution:
Now, we have to find the value of ${{x}^{-1}}+{{y}^{-1}}+{{z}^{-1}}$ if $x\ne 0,y\ne 0,z\ne 0$ and $\left| \begin{matrix} 1+x & 1 & 1 \\\ 1+y & 1+2y & 1 \\\ 1+z & 1+z & 1+3z \\\ \end{matrix} \right|=0$.
We have given that,

1+x & 1 & 1 \\\ 1+y & 1+2y & 1 \\\ 1+z & 1+z & 1+3z \\\ \end{matrix} \right|=0$$ Let us apply row transformation to simplify the determinant, By dividing $${{\text{R}}_{1}},{{\text{R}}_{2}}\text{ }\\!\\!\And\\!\\!\text{ }{{\text{R}}_{3}}$$ by $$x,y\And z$$ respectively we can write, $$\Rightarrow xyz\left| \begin{matrix} 1+\dfrac{1}{x} & \dfrac{1}{x} & \dfrac{1}{x} \\\ 1+\dfrac{1}{y} & 2+\dfrac{1}{y} & \dfrac{1}{y} \\\ 1+\dfrac{1}{z} & 1+\dfrac{1}{z} & 3+\dfrac{1}{z} \\\ \end{matrix} \right|=0$$ Now by applying $${{\text{R}}_{1}}\to {{\text{R}}_{1}}\text{+ }{{\text{R}}_{2}}\text{+ }{{\text{R}}_{3}}$$, we will have, $$\Rightarrow xyz\left| \begin{matrix} 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} & 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} & 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \\\ 1+\dfrac{1}{y} & 2+\dfrac{1}{y} & \dfrac{1}{y} \\\ 1+\dfrac{1}{z} & 1+\dfrac{1}{z} & 3+\dfrac{1}{z} \\\ \end{matrix} \right|=0$$ By taking out common $$\left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)$$ from $${{\text{R}}_{1}}$$ we can write $$\Rightarrow \left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)xyz\left| \begin{matrix} 1 & 1 & 1 \\\ 1+\dfrac{1}{y} & 2+\dfrac{1}{y} & \dfrac{1}{y} \\\ 1+\dfrac{1}{z} & 1+\dfrac{1}{z} & 3+\dfrac{1}{z} \\\ \end{matrix} \right|=0$$ Now, let us apply column transformation to simplify it further. By applying $${{\text{C}}_{2}}\to {{\text{C}}_{2}}-{{\text{C}}_{1}},\text{ }{{\text{C}}_{3}}\to {{\text{C}}_{3}}-{{\text{C}}_{1}}$$ we will have, $$\Rightarrow \left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)xyz\left| \begin{matrix} 1 & 0 & 0 \\\ 1+\dfrac{1}{y} & 1 & -1 \\\ 1+\dfrac{1}{z} & 0 & 2 \\\ \end{matrix} \right|=0$$ Now we will expand the determinant along the first row, $$\begin{aligned} & \Rightarrow \left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)xyz\left[ 2-0 \right]=0 \\\ & \Rightarrow 2xyz\left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)=0 \\\ & \Rightarrow \left( 3+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)=0 \\\ & \Rightarrow \left( \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z} \right)=-3 \\\ & \Rightarrow {{x}^{-1}}+{{y}^{-1}}+{{z}^{-1}}=-3 \\\ \end{aligned}$$ **Hence, option (c) is the correct.** **Note:** In this type of question students have to note that we can expand the determinant along any row or any column. Also students have to take into consideration that we can apply row and column transformations simultaneously to simplify the calculations of finding the value of the determinants. Also students have to remember that if the value of the determinant is equal to zero, then the corresponding matrix is known as a singular matrix.