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Math Help - Determinant with Very Hard Algebra

  1. #1
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    Determinant with Very Hard Algebra

    The problem is
    <br />
\det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ]<br />
    Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
    <br />
abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)<br />
    There are many ways to factor out terms from this.
    <br />
abc[bc(c - b) + ac(a - c) + ab(b - a)] =<br />
    <br />
abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =<br />
    <br />
abc[c^2(b - a) + b^2(a - c) + a^2(c - b)]<br />
    There are even more ways of course. None of these ways lead me to the simple solution:
    <br />
abc(b - a)(c - a)(c - b)<br />
    Can someone explain how to factor this? Thanks.
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  2. #2
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    Quote Originally Posted by binarybob0001 View Post
    The problem is
    <br />
\det \left [ \begin{matrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{matrix} \right ]<br />
    Sorry I'm new. That's the best I could write the problem. It's easy to find the determinant with expansion by minors but to get the simplified solution involves some sort of clever trick that I don't know of. Here's what I can get to.
    <br />
abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)<br />
    There are many ways to factor out terms from this.
    <br />
abc[bc(c - b) + ac(a - c) + ab(b - a)] =<br />
    <br />
abc[b(c + a)(c - a) + c(a + b)(a - b) + a(b + c)(b - c)] =<br />
    <br />
abc[c^2(b - a) + b^2(a - c) + a^2(c - b)]<br />
    There are even more ways of course. None of these ways lead me to the simple solution:
    <br />
abc(b - a)(c - a)(c - b)<br />
    Can someone explain how to factor this? Thanks.
    Multiply out

    \begin{aligned}(b - a)(c - a)(c - b) &= (bc -ac -ab +a^2)(c-b) \\<br />
&= bc^2 - ac^2 -abc + a^2c -b^2c +abc + ab^2 - a^2b \\<br />
&= bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b \end{aligned}<br />

    as you wanted.
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  3. #3
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    You can only do that if you already know the solution. What if you didn't know the solution? I would have just stopped thinking I was done. Is there a systematic way of arriving at that result?
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  4. #4
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    Lexington, MA (USA)
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    Hello, binarybob0001!

    The problem is: . <br />
\begin{bmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{bmatrix}

    Sorry I'm new. . . . . Welcome aboard!
    That's the best I could write the problem. . . . . Good job!

    It's easy to find the determinant with expansion by minors
    but to get the simplified solution involves some sort of clever trick that I don't know of.

    Here's what I can get to:
    . . abc(bc^2 - b^2c - ac^2 + a^2c + ab^2 - a^2b)

    There are many ways to factor out terms from this.
    None of these ways lead me to the simple solution:
    . . abc(b - a)(c - a)(c - b)
    Can someone explain how to factor this?
    Here's an approach I "discovered" years ago . . .


    Collect terms with respect to one of the variables.

    Let's use c
    . . Collect terms with c^2, and with c, and 'constants' (no c's).

    We have: . abc\bigg[bc^2 - ac^2 - b^2c + a^2c + ab^2 - a^2b\bigg]

    Factor "by grouping": . abc\bigg[c^2(b-a) - c(b^2-a^2) + ab(b-a)\bigg]
    . . = \;abc\bigg[c^2(b-a) - c(b-a)(b+a) + ab(b-a)\bigg]

    Factor out (b-a)\!:\;\;abc(b-a)\bigg[c^2 - c(b+a) + ab\bigg]
    . . = \;abc(b-a)\bigg[c^2 - bc - ac + ab\bigg]

    Factor "by grouping": . abc(b-a)\bigg[c(c-b) - a(c-b)\bigg]

    Factor out (c-b)\!:\;\;abc(b-a)(c-b)(c-a) . . . . ta-DAA!

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  5. #5
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    Aug 2007
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    ahah, so I just need to chose a letter, stupid me. Thank you very much. Hey, are there more problems this hard to practice on?
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