Question

Prove \[\left| \begin{matrix}
   {{x}^{3}} & 3{{x}^{2}} & 3x & 1 \\
   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|={{\left( x-1 \right)}^{6}}\]

Answer 1

Hint: To solve the question, we have to apply properties of determinants which state that the value of determinant is unchanged when the row and columns are subtracted or added. After simplifying the given determinant, apply the formula for 4x4 matrix to calculate the answer.

Complete step-by-step answer:

The LHS of the given expression is \[\left| \begin{matrix}
   {{x}^{3}} & 3{{x}^{2}} & 3x & 1 \\
   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

We know that when the \[{{R}_{1}}={{R}_{1}}-{{R}_{2}}\], the determinant is unchanged.

Here \[{{R}_{i}}\] is the \[{{i}^{th}}\] row of the determinant.

Thus, we get

\[\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 3{{x}^{2}}-\left( {{x}^{2}}+2x \right) & 3x-\left( 2x+1 \right) & 1-1 \\
   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 3{{x}^{2}}-{{x}^{2}}-2x & 3x-2x-1 & 0 \\
   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

We know that when the \[{{R}_{2}}={{R}_{2}}-{{R}_{3}}\], the determinant is unchanged. Thus, we get


\[\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}+2x-(2x+1) & 2x+1-(x+2) & 1-1 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}+2x-2x-1 & 2x+1-x-2 & 0 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 & 0 \\
   x & 2x+1 & x+2 & 1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

We know that when the \[{{R}_{3}}={{R}_{3}}-{{R}_{4}}\], the determinant is unchanged. Thus, we get

\[\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 & 0 \\
   x-1 & 2x+1-3 & x+2-3 & 1-1 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 & 0 \\
   x-1 & 2x-2 & x-1 & 0 \\
   1 & 3 & 3 & 1 \\
\end{matrix} \right|\]

We know that the formula for determinant of 4x4 matrix is given by

\[\left| \begin{matrix}
   {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} & {{a}_{1,4}} \\
   {{a}_{2,1}} & {{a}_{2,2}} & {{a}_{2,3}} & {{a}_{2,4}} \\
   {{a}_{3,1}} & {{a}_{3,2}} & {{a}_{3,3}} & {{a}_{3,4}} \\
   {{a}_{4,1}} & {{a}_{4,2}} & {{a}_{4,3}} & {{a}_{4,4}} \\
\end{matrix} \right|=\sum\limits_{i=1}^{4}{{{a}_{i,4}}{{(-1)}^{i+4}}{{\Delta
}_{i,4}}}={{a}_{1,4}}{{(-1)}^{1+4}}{{\Delta }_{1,4}}+{{a}_{2,4}}{{(-1)}^{2+4}}{{\Delta }_{2,4}}+{{a}_{3,4}}{{(-1)}^{3+4}}{{\Delta }_{3,4}}+{{a}_{4,4}}{{(-1)}^{4+4}}{{\Delta }_{4,4}}\]
\[=-{{a}_{1,4}}{{\Delta }_{1,4}}+{{a}_{2,4}}{{\Delta }_{2,4}}-{{a}_{3,4}}{{\Delta }_{3,4}}+{{a}_{4,4}}{{\Delta }_{4,4}}\]

By comparing the given matrix with the general 4x4 matrix we get

\[{{a}_{1,4}}=0,{{a}_{2,4}}=0,{{a}_{3,4}}=0,{{a}_{4,4}}=1\]

By substituting the above values in the formula, we get

\[=0+0+0+(1){{\Delta }_{4,4}}\]

\[={{\Delta }_{4,4}}\]

We know the formula for \[{{\Delta }_{4,4}}=\left| \begin{matrix}
   {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} \\
   {{a}_{1,2}} & {{a}_{2,2}} & {{a}_{2,3}} \\
   {{a}_{1,3}} & {{a}_{3,2}} & {{a}_{3,3}} \\
\end{matrix} \right|\]

By substituting the above values in the formula, we get

\[{{\Delta }_{4,4}}=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}} & 2{{x}^{2}}-2x & x-1 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

We know that when the \[{{R}_{1}}={{R}_{1}}-{{R}_{2}}\], the determinant is unchanged. Thus, we get

\[\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}}-\left( {{x}^{2}}-x \right) & 2{{x}^{2}}-2x-\left( {{x}^{2}}-1 \right) & x-1-\left( x-1 \right) \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-{{x}^{2}}-{{x}^{2}}+x & 2{{x}^{2}}-2x-{{x}^{2}}+1 & x-1-x+1 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-2{{x}^{2}}+x & {{x}^{2}}-2x+1 & 0 \\
   {{x}^{2}}-x & {{x}^{2}}-1 & x-1 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

We know that when the \[{{R}_{2}}={{R}_{2}}-{{R}_{3}}\], the determinant is unchanged. Thus, we get

\[\left| \begin{matrix}
   {{x}^{3}}-2{{x}^{2}}+x & {{x}^{2}}-2x+1 & 0 \\
   {{x}^{2}}-x-(x-1) & {{x}^{2}}-1-(2x-2) & x-1-(x-1) \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-2{{x}^{2}}+x & {{x}^{2}}-2x+1 & 0 \\
   {{x}^{2}}-x-x+1 & {{x}^{2}}-1-2x+2 & x-1-x-1 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   {{x}^{3}}-2{{x}^{2}}+x & {{x}^{2}}-2x+1 & 0 \\
   {{x}^{2}}-2x+1 & {{x}^{2}}+1-2x & 0 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

\[=\left| \begin{matrix}
   x\left( {{x}^{2}}-2x+1 \right) & {{x}^{2}}-2x+1 & 0 \\
   {{x}^{2}}-2x+1 & {{x}^{2}}-2x+1 & 0 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

We know the formula \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]. Thus, by applying the formula we get

\[=\left| \begin{matrix}
   x\left( {{(x-1)}^{2}} \right) & {{(x-1)}^{2}} & 0 \\
   {{(x-1)}^{2}} & {{(x-1)}^{2}} & 0 \\
   x-1 & 2x-2 & x-1 \\
\end{matrix} \right|\]

We know that the formula for determinant of 3x3 matrix is given by
\[\left| \begin{matrix}

   {{a}_{1,1}} & {{a}_{1,2}} & {{a}_{1,3}} \\
   {{a}_{2,1}} & {{a}_{2,2}} & {{a}_{2,3}} \\
   {{a}_{3,1}} & {{a}_{3,2}} & {{a}_{3,3}} \\
\end{matrix} \right|={{a}_{1,3}}\left( {{a}_{2,1}}{{a}_{3,2}}-{{a}_{1,3}}{{a}_{2,2}}

\right)-{{a}_{2,3}}\left( {{a}_{1,1}}{{a}_{3,2}}-{{a}_{1,3}}{{a}_{1,2}} \right)+{{a}_{3,3}}\left( {{a}_{2,2}}{{a}_{1,1}}-{{a}_{2,1}}{{a}_{1,2}} \right)\]

By comparing the given matrix with the general 4x4 matrix we get

\[{{a}_{1,3}}=0,{{a}_{2,3}}=0,{{a}_{3,3}}=x-1\]

By substituting the above values in the formula, we get

\[=0+0+\left( x-1 \right)\left( x{{(x-1)}^{2}}{{(x-1)}^{2}}-{{(x-1)}^{2}}{{(x-1)}^{2}} \right)\]

We know that \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\]

\[=\left( x-1 \right)\left( x{{(x-1)}^{2+2}}-{{(x-1)}^{2+2}} \right)\]

\[=\left( x-1 \right)\left( x{{(x-1)}^{4}}-{{(x-1)}^{4}} \right)\]

\[=(x-1){{(x-1)}^{4}}\left( x-1 \right)\]

\[={{\left( x-1 \right)}^{1+1+4}}\]

\[={{(x-1)}^{6}}\]

Thus, \[\left| \begin{matrix}

   {{x}^{3}} & 3{{x}^{2}} & 3x & 1 \\

   {{x}^{2}} & {{x}^{2}}+2x & 2x+1 & 1 \\

   x & 2x+1 & x+2 & 1 \\

   1 & 3 & 3 & 1 \\

\end{matrix} \right|={{\left( x-1 \right)}^{6}}\]

Hence proved.

Note: The possibility of mistake can be not applying appropriate formulae for calculations. The other possibility of mistake can be applying formula directly without rearranging the determinant which will lead to more calculation instead of easing the procedure of solving.