Answer

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**Hint:**This clear that this question belongs to determinants. We’ll try to make $ - \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ in the determinant by using some elementary properties of determinant, so that, in the end by expansion we can easily get its value.

__Complete step-by-step answer__:Here we have given a condition in the form of determinant. Better to say, we have given an equation in the form of determinant because determinant is nothing but an arrangement of numbers or algebraic expressions.

We’ll use some elementary row and column operations to make the term in the determinant whose value we need to find. The steps are as follows:

Given,

$\left| {\begin{array}{*{20}{c}}

{1 + x}&1&1 \\

{1 + y}&{1 + 2y}&1 \\

{1 + z}&{1 + z}&{1 + 3z}

\end{array}} \right| = 0$

Step 1: ${C_2} \to {C_2} - {C_1}$ and ${C_3} \to {C_3} - {C_1}$

It means, we’ll replace the second column by its difference from 1 column and similarly, the third column is replaced by its difference from the first column.

\[

\left| {\begin{array}{*{20}{c}}

{1 + x}&1&1 \\

{1 + y}&{1 + 2y}&1 \\

{1 + z}&{1 + z}&{1 + 3z}

\end{array}} \right| = 0 \\

\sim \left| {\begin{array}{*{20}{c}}

{1 + x}&{ - x}&{ - x} \\

{1 + y}&y&{ - y} \\

{1 + z}&0&{2z}

\end{array}} \right| = 0 \\

\]

Step 2: Now, we’ll take common x from the first row, y from the second row, and z from the third row.

\[

\left| {\begin{array}{*{20}{c}}

{1 + x}&{ - x}&{ - x} \\

{1 + y}&y&{ - y} \\

{1 + z}&0&{2z}

\end{array}} \right| = 0 \\

\sim \left( {xyz} \right)\left| {\begin{array}{*{20}{c}}

{\dfrac{1}{x} + 1}&{ - 1}&{ - 1} \\

{\dfrac{1}{y} + 1}&1&{ - 1} \\

{\dfrac{1}{z} + 1}&0&2

\end{array}} \right| = 0 \\

\]

Step 3: we wanted to make $ - \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ in the determinant. Now observe that by adding all the rows we’ll get it without negative sign. But it won’t be problematic for us. So, on adding all the rows that is on ${R_3} \to {R_3} + {R_2} + {R_1}$ we get,

\[

\left( {xyz} \right)\left| {\begin{array}{*{20}{c}}

{\dfrac{1}{x} + 1}&{ - 1}&{ - 1} \\

{\dfrac{1}{y} + 1}&1&{ - 1} \\

{\dfrac{1}{z} + 1}&0&2

\end{array}} \right| = 0 \\

\sim \left( {xyz} \right)\left| {\begin{array}{*{20}{c}}

{\dfrac{1}{x} + 1}&{ - 1}&{ - 1} \\

{\dfrac{1}{y} + 1}&1&{ - 1} \\

{\dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y} + 3}&0&0

\end{array}} \right| = 0 \\

\]

Now we can easily expand the determinant, which will give us,

$\left( {xyz} \right)\left( {3 + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)\left( {1 + 1} \right) = 0$

Now, observe that, we need to find the value of $ - \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$. That means none of x, y and z can be 0. Hence, we’ll discard the case $\left( {xyz} \right) = 0$. Also 1+1 is 2 which is obviously not equal to 0. So, for this expression to be zero, $\left( {3 + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ has to be equal to zero. On considering the same,

$

\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 \\

$

**Hence the required value of $ - \left( {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}} \right)$ is equal to 3.**

**Note**: If the given determinant is $\left| {\begin{array}{*{20}{c}}

{{x_1}}&{{x_2}}&{{x_3}} \\

{{y_1}}&{{y_2}}&{{y_3}} \\

{{z_1}}&{{z_2}}&{{z_3}}

\end{array}} \right|$ then it’s expansion along first row will be

${x_1}\left( {{y_2}{z_3} - {z_2}{y_3}} \right) - {x_2}\left( {{y_1}{z_3} - {z_1}{y_3}} \right) + {x_3}\left( {{y_1}{z_2} - {z_1}{y_2}} \right)$. Also, observe that, if ${x_2} = 0$ and ${x_3} = 0$ then it’s easier to expand. Its value will become ${x_1}\left( {{y_2}{z_3} - {z_2}{y_3}} \right)$. This was the same case in our solution. That’s why it’s always recommended to make zeros in determinant or even in matrices.

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