Find the determinant without row expansion

Question

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.

Answer 1

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}

Answer 2

\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 $$ (a+b+c)(c-a)(b-a)(c-b) $$

Answer 3

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