Without expanding, show that the value of each of the following determinants is zero:$\left| \begin{matrix} {{1}^{2}} & {{2}^{2}} & {{3}^{2}} & {{4}^{2}} \\ {{2}^{2}} & {{3}^{2}} & {{4}^{2}} & {{5}^{2}} \\ {{3}^{2}} & {{4}^{2}} & {{5}^{2}} & {{6}^{2}} \\ {{4}^{2}} & {{5}^{2}} & {{6}^{2}} & {{7}^{2}} \\ \end{matrix} \right|$

$\left| \begin{matrix}

{{1}^{2}} & {{2}^{2}} & {{3}^{2}} & {{4}^{2}}  \\

{{2}^{2}} & {{3}^{2}} & {{4}^{2}} & {{5}^{2}}  \\

{{3}^{2}} & {{4}^{2}} & {{5}^{2}} & {{6}^{2}}  \\

{{4}^{2}} & {{5}^{2}} & {{6}^{2}} & {{7}^{2}}  \\

\end{matrix} \right|$

Let $\vartriangle =\left| \begin{matrix}

{{1}^{2}} & {{2}^{2}} & {{3}^{2}} & {{4}^{2}}  \\

{{2}^{2}} & {{3}^{2}} & {{4}^{2}} & {{5}^{2}}  \\

{{3}^{2}} & {{4}^{2}} & {{5}^{2}} & {{6}^{2}}  \\

{{4}^{2}} & {{5}^{2}} & {{6}^{2}} & {{7}^{2}}  \\

\end{matrix} \right|$

Now we have to apply the column operation ${{C}_{3}}\to {{C}_{3}}-{{C}_{2}}$and ${{C}_{4}}\to {{C}_{4}}-{{C}_{1}}$, then we get,

$\vartriangle =\left| \begin{matrix}

{{1}^{2}} & {{2}^{2}} & {{3}^{2}}-{{2}^{2}} & {{4}^{2}}-{{1}^{2}}  \\

{{2}^{2}} & {{3}^{2}} & {{4}^{2}}-{{3}^{2}} & {{5}^{2}}-{{2}^{2}}  \\

{{3}^{2}} & {{4}^{2}} & {{5}^{2}}-{{4}^{2}} & {{6}^{2}}-{{3}^{2}}  \\

{{4}^{2}} & {{5}^{2}} & {{6}^{2}}-{{5}^{2}} & {{7}^{2}}-{{4}^{2}}  \\

\end{matrix} \right|$

$\vartriangle =\left| \begin{matrix}

{{1}^{2}} & {{2}^{2}} & 5 & 5  \\

{{2}^{2}} & {{3}^{2}} & 7 & 7  \\

{{3}^{2}} & {{4}^{2}} & 9 & 9  \\

{{4}^{2}} & {{5}^{2}} & 11 & 11  \\

\end{matrix} \right|$

As, ${{C}_{3}}={{C}_{4}}$so, the determinant is zero.