Question

Using the properties of the determinants prove the following expression.
$\left| \begin{matrix}
   1 & x & {{x}^{2}} \\
   {{x}^{2}} & 1 & x \\
   x & {{x}^{2}} & 1 \\
\end{matrix} \right|={{\left( 1-{{x}^{3}} \right)}^{2}}.$

Answer 1

Hint: To prove the following expression we will take the LHS of the expression and apply properties of the determinant to prove it equal to the RHS of the expression. First, we will apply ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$. After that we will apply ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$ and then ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$ and after that we will expand the matrix to solve it.

Complete step by step answer:
We can observe that if we add values of row 2 and row 3 to row 1 we will be able to take $1+{{x}^{2}}+x$ common out of row 1, so applying ${{R}_{1}}\to {{R}_{1}}+{{R}_{2}}+{{R}_{3}}$, we get
$\begin{align}
  & \left| \begin{matrix}
   1 & x & {{x}^{2}} \\
   {{x}^{2}} & 1 & x \\
   x & {{x}^{2}} & 1 \\
\end{matrix} \right| \\
 & =\left| \begin{matrix}
   1+{{x}^{2}}+x & x+1+{{x}^{2}} & {{x}^{2}}+x+1 \\
   {{x}^{2}} & 1 & x \\
   x & {{x}^{2}} & 1 \\
\end{matrix} \right| \\
\end{align}$
Now taking $1+{{x}^{2}}+x$ common from row 1, we get
$=\left( 1+x+{{x}^{2}} \right)\left| \begin{matrix}
   1 & 1 & 1 \\
   {{x}^{2}} & 1 & x \\
   x & {{x}^{2}} & 1 \\
\end{matrix} \right|$
Now, applying ${{C}_{1}}\to {{C}_{1}}-{{C}_{2}}$, we get
$\begin{align}
  & =\left( 1+x+{{x}^{2}} \right)\left| \begin{matrix}
   1-1 & 1 & 1 \\
   {{x}^{2}}-1 & 1 & x \\
   x-{{x}^{2}} & {{x}^{2}} & 1 \\
\end{matrix} \right| \\
 & =\left( 1+x+{{x}^{2}} \right)\left| \begin{matrix}
   0 & 1 & 1 \\
   {{x}^{2}}-1 & 1 & x \\
   x\left( 1-x \right) & {{x}^{2}} & 1 \\
\end{matrix} \right| \\
\end{align}$
Using the property ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in second row and 1st column, we get
$=\left( 1+x+{{x}^{2}} \right)\left| \begin{matrix}
   0 & 1 & 1 \\
   \left( x-1 \right)\left( x+1 \right) & 1 & x \\
   -x\left( x-1 \right) & {{x}^{2}} & 1 \\
\end{matrix} \right|$
Taking $x-1$ common from column 1, we get
$=\left( 1+x+{{x}^{2}} \right)\left( x-1 \right)\left| \begin{matrix}
   0 & 1 & 1 \\
   \left( x+1 \right) & 1 & x \\
   -x & {{x}^{2}} & 1 \\
\end{matrix} \right|$
Now applying ${{C}_{2}}\to {{C}_{2}}-{{C}_{3}}$, and also using the property ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in third row and 2nd column, we get

$\begin{align}
  & =\left( 1+x+{{x}^{2}} \right)\left( x-1 \right)\left| \begin{matrix}
   0 & 1-1 & 1 \\
   \left( x+1 \right) & 1-x & x \\
   -x & {{x}^{2}}-1 & 1 \\
\end{matrix} \right| \\
 & =\left( 1+x+{{x}^{2}} \right)\left( x-1 \right)\left| \begin{matrix}
   0 & 0 & 1 \\
   \left( x+1 \right) & -\left( x-1 \right) & x \\
   -x & \left( x+1 \right)\left( x-1 \right) & 1 \\
\end{matrix} \right| \\
\end{align}$
Now, we will take $x-1$ common from column 2,
$=\left( 1+x+{{x}^{2}} \right){{\left( x-1 \right)}^{2}}\left| \begin{matrix}
   0 & 0 & 1 \\
   \left( x+1 \right) & -1 & x \\
   -x & \left( x+1 \right) & 1 \\
\end{matrix} \right|$
Now we will expand the determinant along row 1, and we will get,
\[\begin{align}
  & =\left( 1+x+{{x}^{2}} \right){{\left( x-1 \right)}^{2}}\times \left[ 0+0+1\times \left( {{\left( x+1 \right)}^{2}}-x \right) \right] \\
 & =\left( 1+x+{{x}^{2}} \right){{\left( x-1 \right)}^{2}}\times \left( {{\left( x+1 \right)}^{2}}-x \right) \\
 & =\left( 1+x+{{x}^{2}} \right){{\left( x-1 \right)}^{2}}\times \left( \left( {{x}^{2}}+1+2x \right)-x \right) \\
 & =\left( 1+x+{{x}^{2}} \right){{\left( x-1 \right)}^{2}}\times \left( 1+x+{{x}^{2}} \right) \\
 & ={{\left( 1+x+{{x}^{2}} \right)}^{2}}{{\left( x-1 \right)}^{2}} \\
 & ={{\left( \left( 1+x+{{x}^{2}} \right)\left( x-1 \right) \right)}^{2}} \\
 & ={{\left( -\left( 1-x \right)\left( 1+x+{{x}^{2}} \right) \right)}^{2}} \\
\end{align}\]
As we know square of negative symbol is positive only so we get,
\[={{\left( \left( 1-x \right)\left( 1+x+{{x}^{2}} \right) \right)}^{2}}\]
And we know that ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+{{b}^{2}}+ab \right)$, in the above expression taking a = 1, and b = $x$, we get,
$\begin{align}
  & ={{\left( {{1}^{3}}-{{x}^{3}} \right)}^{2}} \\
 & ={{\left( 1-{{x}^{3}} \right)}^{2}} \\
\end{align}$
= R. H. S
Hence we proved LHS is equal to the RHS by using the properties of the determinants.

Note: You need to note that this question can be solved by the number of ways and the above shown method is one of the ways. Whenever you are trying to solve questions related to determinant try to analyse what result will we get after applying that property if the result is useful then only apply that property otherwise don’t.