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Math Help - Determinan

  1. #1
    Junior Member Singular's Avatar
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    Determinan

    Show that
    <br />
\left| {\begin{array}{*{20}c}<br />
   {n^2 } & {(n + 1)^2 } & {(n + 2)^2 }  \\<br />
   {(n + 1)^2 } & {(n + 2)^2 } & {(n + 3)^2 }  \\<br />
   {(n + 2)^2 } & {(n + 3)^2 } & {(n + 4)^2 }  \\<br />
\end{array}} \right| =  - 8<br />

    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
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  2. #2
    Member Henderson's Avatar
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    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.
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  3. #3
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Thumbs up use row and column operations

    Quote Originally Posted by Singular View Post
    Show that
    <br />
\left| {\begin{array}{*{20}c}<br />
   {n^2 } & {(n + 1)^2 } & {(n + 2)^2 }  \\<br />
   {(n + 1)^2 } & {(n + 2)^2 } & {(n + 3)^2 }  \\<br />
   {(n + 2)^2 } & {(n + 3)^2 } & {(n + 4)^2 }  \\<br />
\end{array}} \right| =  - 8<br />

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

    \begin{vmatrix}<br />
n^2 && (n+1)^2 && (n+2)^2 \\<br />
2n+1 && 2n+3 && 2n+ 5 \\<br />
2n+3 && 2n+5 && 2n+7 \\<br /> <br /> <br />
\end{vmatrix}

    Now do Column3 - Column2 and colum2n-column1
    and we get

    \begin{vmatrix}<br />
n^2 && 2n+1 && 2n+3 \\<br />
2n+1 && 2 && 2 \\<br />
2n+3 && 2 && 2 \\<br />
\end{vmatrix}

    Row3 -Row2 gives

    \begin{vmatrix}<br />
n^2 && 2n+1 && 2n+3 \\<br />
2n+1 && 2 && 2 \\<br />
2 && 0 && 0 \\<br />
\end{vmatrix}

    expanding along the bottom row

    2 \cdot \begin{vmatrix}<br />
2n+1 && 2n+3 \\<br />
2 && 2 \\<br />
\end{vmatrix} = 2[4n+2-4n-6]=-8
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