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Show that the determinant of the matrix \begin{bmatrix} 1& a& a^3\\ 1& b& b^3\\ 1& c& c^3\end{bmatrix}

is $(a-b)(b-c)(c-a)(a+b+c)$ without expanding.

I was able to get out $(a-b)(b-c)(c-a)$ but couldn't complete.

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Even though your determinant is perfectly clear, there will probably be some complaints because you didn't format it using Latex commands. Personally, I'm much more concerned about the lack of periods and upper-case letters. :-) –  bubba Mar 21 at 1:06
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It is the first time that I use this website so I didn't know how to do it right –  Ahmad Amr Ebeid Mar 21 at 1:10
    
You don't know how to type a period?? :-) I'm not complaining about the math formatting, but I expect someone will. Or, some nice person might even fix it for you. –  bubba Mar 21 at 1:12
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Anyway, back to the mathematics. After you factored out $(a-b)(b-c)(c-a)$, what determinant did you have left? –  bubba Mar 21 at 1:14

3 Answers 3

up vote 7 down vote accepted

Use row operations to simplify. In the process, the factorization drops right out. \begin{align} \det \begin{bmatrix} 1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \end{bmatrix} &= \det \begin{bmatrix} 1 & a & a^3 \\ 0 & b-a & b^3-a^3 \\ 0 & c-a & c^3-a^3 \end{bmatrix} \\ &= \det \begin{bmatrix} 1 & a & a^3 \\ 0 & b-a & (b-a)(b^2+ab+b^2) \\ 0 & c-a & (c-a)(c^2+ac+a^2) \end{bmatrix} \\ &= (b-a)(c-a)\det \begin{bmatrix} 1 & a & a^3 \\ 0 & 1 & b^2+ab+a^2 \\ 0 & 1 & c^2+ac+a^2 \end{bmatrix} \\ &= (b-a)(c-a)\det \begin{bmatrix} 1 & a & a^3 \\ 0 & 1 & b^2+ab+a^2 \\ 0 & 0 & c^2-b^2+ac-ab \end{bmatrix} \\ &= (b-a)(c-a)(c^2-b^2 + ac-ab)\det \begin{bmatrix} 1 & a & a^3 \\ 0 & 1 & b^2+ab+a^2 \\ 0 & 0 & 1 \end{bmatrix} \\ &= (b-a)(c-a)(c-b)(c+b+a) \\ &= (a-b)(b-c)(c-a)(a+b+c). \end{align}

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Hint: reduce the matrix to upper triangular form and then read off the determinant as the product of the diagonals.

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\begin{bmatrix} 1& a& a^2&a^3\\ 1& b& b^2 &b^3\\ 1& c& c^2 &c^3\\ 1& X& X^2 &X^3 \end{bmatrix}is the well known Vandermonde determinant. When expend with respect to the last line, this is a polynomial whose $X^2$ coefficient is the opposite of the result.

Now, expending $$ (c-a)(b-a)(c-b)(X-c)(X-b)(X-a) \\= (c-a)(b-a)(c-b)(X^3 - (a+b+c)X+\cdots) $$gives the result $$ (a+b+c)(c-a)(b-a)(c-b) $$

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I'd imagine if the question is asking to work out the determinant specifically without row expansion, Vandermonde determinants won't have been covered yet (row reducing to an upper triangular matrix is usually covered in introductory linear algebra courses). –  ah11950 Mar 21 at 1:21
    
that's ok. Take my solution as an alternative for more experimented people in this case. The chosen one is very much similar to the vandermonde. –  mookid Mar 21 at 1:25
    
Oh of course. I certainly prefer it over the more brute force methods, but it might not enlighten the OP! –  ah11950 Mar 21 at 1:26

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