use row and column operations

Quote:

Originally Posted by

**Singular** Show that

$\displaystyle

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

{n^2 } & {(n + 1)^2 } & {(n + 2)^2 } \\

{(n + 1)^2 } & {(n + 2)^2 } & {(n + 3)^2 } \\

{(n + 2)^2 } & {(n + 3)^2 } & {(n + 4)^2 } \\

\end{array}} \right| = - 8

$

I know that I can use **Sarrus** but I think I will get a complicated. Is there any other ways to solve this ?

thanks (Bow)

try Row3 - Row2 and Row2 - Row 1

and note the pattern with the difference of squares.

+5

$\displaystyle (n+2)^2-(n+1)^2=2n+3$ and $\displaystyle (n+3)^2-(n+2)^2=2n+5$

$\displaystyle \begin{vmatrix}

n^2 && (n+1)^2 && (n+2)^2 \\

2n+1 && 2n+3 && 2n+ 5 \\

2n+3 && 2n+5 && 2n+7 \\

\end{vmatrix}$

Now do Column3 - Column2 and colum2n-column1

and we get

$\displaystyle \begin{vmatrix}

n^2 && 2n+1 && 2n+3 \\

2n+1 && 2 && 2 \\

2n+3 && 2 && 2 \\

\end{vmatrix}$

Row3 -Row2 gives

$\displaystyle \begin{vmatrix}

n^2 && 2n+1 && 2n+3 \\

2n+1 && 2 && 2 \\

2 && 0 && 0 \\

\end{vmatrix}$

expanding along the bottom row

$\displaystyle 2 \cdot \begin{vmatrix}

2n+1 && 2n+3 \\

2 && 2 \\

\end{vmatrix} = 2[4n+2-4n-6]=-8$