Find the determinant using colum or row operations

Question

I find problem in simplification. When I tried to simplify I ended up doing the regular process of finding the determinant value. The matrix is $\begin{pmatrix} 1 & 1 & 1 \\ a & b & c \\ ab & bc & ca \end{pmatrix}$.

Answer 1

$$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}&\frac{1}{b}&\frac{1}{c}\\1&1&1\\b&c&a\end {array}\right)=abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\0&1&0\\b-c&c&a-c\end {array}\right)=$$ $$abc\det\left(\begin{array}{ccc}\frac{1}{a}-\frac{1}{b}&\frac{1}{c}-\frac{1}{b}\\b-c&a-c\end {array}\right)=c(b-a)(a-c)-a(b-c)^{2}$$

Answer 2

Subtract the first column from the other two, to get $\begin{vmatrix} 1 & 0 & 0 \\ a & b-a & c-a \\ ab & bc - ab & ca - ab \end{vmatrix}$. Now compute the determinant, developing it along the first row, and obtain $\begin{vmatrix} b-a & c-a \\ bc-ab & ca-ab \end{vmatrix}$. This can further be calculated (with the usual formula for $2\times2$ determinants) to get $a(b-a)(c-b) - b(c-a)^2$, which you can't simplify anymore.

Answer 3

What about computing the determinant directly based on the first row?

$$\det\left(\begin{array}{ccc}1&1&1\\a&b&c\\ab&bc&ca\end {array}\right)=(bca-bc^2)-(ca^2-abc)+(abc-ab^2)=3abc-bc^2-ca^2-ab^2$$

Answer 4

Maybe it's unlikely that you would have learned of this method, but via Dodgson condensation, you can reduce this to a problem of computing the determinants of several $2\times2$ matrices, which you might find easier to compute strictly via row operations. $$\begin{vmatrix}1&1&1\\a&b&c\\ab&bc&ca\end{vmatrix}=\begin{vmatrix}\dfrac{\begin{vmatrix}1&1\\a&b\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}1&1\\b&c\end{vmatrix}}{b}\\[1ex] \dfrac{\begin{vmatrix}a&b\\ab&bc\end{vmatrix}}{b}&\dfrac{\begin{vmatrix}b&c\\bc&ca\end{vmatrix}}{b} \end{vmatrix}$$ The first determinant (in the $(1,1)$ position), for instance, would be $$\begin{vmatrix}1&1\\a&b\end{vmatrix}=\begin{vmatrix}1&1\\0&b-a\end{vmatrix}=\begin{vmatrix}1&0\\0&b-a\end{vmatrix}=(b-a)\begin{vmatrix}1&0\\0&1\end{vmatrix}=b-a$$