Question

If $x,y,z$ are all different and if $\left( {\begin{array}{*{20}{c}}
  x&{{x^2}}&{1 + {x^3}} \\
  y&{{y^2}}&{1 + {y^3}} \\
  z&{{z^2}}&{1 + {z^3}}
\end{array}} \right) = 0$ then $1 + xyz = $
A) -1
B) 0
C) 1
D) 2

Answer 1

Hint: Use the properties of determinants to simplify the given equation into factors of the given variables. Then use the given condition of different values of variables to evaluate the given expression.

Complete step-by-step answer:
It is given in the problem

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

  x&{{x^2}}&{1 + {x^3}} \\

  y&{{y^2}}&{1 + {y^3}} \\

  z&{{z^2}}&{1 + {z^3}}

\end{array}} \right| = 0{\text{ (1)}}\]

Where $x,y,z$ are all different.

We need to find the value of $1 + xyz$

In problems like above, the properties of determinants are to be used in order to reduce the determinant into the desired expression.

We know that if two determinants differ by just one column, we can add them together by just adding up these two columns. Also, the reverse is also true.

Hence using the above rule in equation $(1)$, we split column 3

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

  x&{{x^2}}&1 \\

  y&{{y^2}}&1 \\

  z&{{z^2}}&1

\end{array}} \right| + \left| {\begin{array}{*{20}{c}}

  x&{{x^2}}&{{x^3}} \\

  y&{{y^2}}&{{y^3}} \\

  z&{{z^2}}&{{z^3}}

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

Taking $x,y,z$ common from row ${R_1}$,${R_2}$ and ${R_3}$ respectively in the second determinant, we get

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

  x&{{x^2}}&1 \\

  y&{{y^2}}&1 \\

  z&{{z^2}}&1

\end{array}} \right| + xyz\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

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

We know that interchanging the columns of determinant odd times leads to change in the sign of determinant.

Replacing ${C_3}$ by ${C_2}$ in the first determinant in the above equation, we get
$ \Rightarrow \left( { - 1} \right)\left| {\begin{array}{*{20}{c}}

  x&1&{{x^2}} \\

  y&1&{{y^2}} \\

  z&1&{{z^2}}

\end{array}} \right| + xyz\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

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

Now replacing ${C_1}$ by ${C_2}$ in the first determinant in the above equation, we get
$ \Rightarrow \left( { - 1} \right)\left( { - 1} \right)\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

\end{array}} \right| + xyz\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

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

$ \Rightarrow \left( { + 1} \right)\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

\end{array}} \right| + xyz\left| {\begin{array}{*{20}{c}}

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

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

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

  1&x&{{x^2}} \\

  1&y&{{y^2}} \\

  1&z&{{z^2}}

\end{array}} \right|\left( {1 + xyz} \right) = 0$

We know that by elementary row operations that multiplying a row (or column) by a non-zero scalar and adding the result to another row (or column) does not alter the determinant.

Hence performing elementary row operation ${R_2} \to {R_2} - {R_1}$and ${R_3} \to {R_3} - {R_1}$ on the determinant in the above equation

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

  1&x&{{x^2}} \\

  {1 - 1}&{y - x}&{{y^2} - {x^2}} \\

  {1 - 1}&{z - x}&{{z^2} - {x^2}}

\end{array}} \right|\left( {1 + xyz} \right) = 0$

Evaluating the determinant in the above equation, we get

\[ \Rightarrow \left\{ {1 \times \left\{ {\left( {y - x} \right)\left( {{z^2} - {x^2}} \right) - \left( {z - x} \right)\left( {{y^2} - {x^2}} \right)} \right\} + 0 + 0} \right\}\left( {1 + xyz} \right) = 0\]

Using the property ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$ to factorise, the

above equation can be simplified as

\[

   \Rightarrow \left\{ {\left( {y - x} \right)\left( {z - x} \right)\left( {z + x} \right) - \left( {z - x}

\right)\left( {y - x} \right)\left( {y + x} \right)} \right\}\left( {1 + xyz} \right) = 0 \\

   \Rightarrow \left\{ {\left( {y - x} \right)\left( {z - x} \right)\left( {z + x - y - x} \right)}

\right\}\left( {1 + xyz} \right) = 0 \\

   \Rightarrow \left( {y - x} \right)\left( {z - x} \right)\left( {z - y} \right)\left( {1 + xyz} \right) =

0{\text{ (2)}} \\

\]

Since it is given in the problem that $x,y,z$ are all different, \[\left( {y - x} \right),\left( {z - x}

\right),\left( {z - y} \right)\] are non-zero in equation $(2)$.

Hence for equation $(2)$to be valid the following condition should be true

\[ \Rightarrow \left( {1 + xyz} \right) = 0\]

Hence option (B). 0 is the correct answer.

Note: It is important to keep all the properties of determinants in mind while solving problems like above. Effort should be made to simplify the determinant in order to have a maximum number of elements as zero as it is easy to evaluate then. Scalar multiplied to a determinant is multiplied to a particular row or column unlike in a matrix where it is multiplied to all the elements. Hence a clear distinction between properties of matrix and determinants should be developed in order to avoid confusion in problems like above.