1. ## Determinan

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 equation. Is there any other ways to solve this ?

thanks

2. There very well may be a sneaky, 'trick' answer to do this (like making use of the fact that det(AB)=det(A)det(B), but I don't see what two matrices would multiply to give you this)

If you're stuck using Sarrus, at the very least make the substitution x=n+2. That will balance out your terms a bit more, and at least give you some difference of squares while you're multiplying it all out.

3. ## use row and column operations

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
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

$\displaystyle 2 \cdot \begin{vmatrix} 2n+1 && 2n+3 \\ 2 && 2 \\ \end{vmatrix} = 2[4n+2-4n-6]=-8$