Prove the following by appropriate manipulations using Facts 1 to 4; do not just evaluate the determinants.

$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|\mathbf{=}\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|\mathbf{=}\left(c-a\right)\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|\mathbf{=}\left(c-a\right)\left(b-a\right)\left(c-b\right)$

Given determinant is$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|$

Facts used about determinants:

1. If each element of a row is multiplied by a number m, the value of the determinant becomes m times.
2. The determinant will change sign, if two rows (or two columns) are interchanged.
3. The determinant remains the same if each element of one row is added by m times the corresponding element of another row.

Consider: $\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|$.

Multiply the first row with ‘a’, second row with ‘b’ and third row with ‘c’ using Fact 1.

So,$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\frac{1}{abc}\left|\begin{array}{ccc}a& a^2& abc\\ b& b^2& abc\\ c& c^2& abc\end{array}\right|$

Multiply third column with ‘$\frac{1}{abc}$’ using Fact 1.

So,

$$\begin{gathered} \left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\frac{abc}{abc}\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right| \\ =\left|\begin{array}{ccc}a& a^2& 1\\ b& b^2& 1\\ c& c^2& 1\end{array}\right| \end{gathered}$$

Interchange column 1 and 3 using Fact 2

So,$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=-\left|\begin{array}{ccc}1& a^2& a\\ 1& b^2& b\\ 1& c^2& c\end{array}\right|$

Finally, interchange column 2 and 3.

role="math" localid="1664361081021" $$\begin{gathered} \left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\left(-\right)\left(-\right)\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right| \\ =\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right| \end{gathered}$$

Finally,$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

Consider, $\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|$

Using Fact 3, perform, $$\begin{gathered} R_2\to R_2-R_1 \\ {R}_3\to R_3-R_1 \end{gathered}$$as:

$\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|=\left|\begin{array}{ccc}1& a& a^2\\ 0& b-a& b^2-a^2\\ 0& c-a& c^2-a^2\end{array}\right|$

Multiply Row 2 by $\frac{1}{b-a}$and Row 3 by $\frac{1}{c-a}$Using Fact 1

So,$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\left(b-a\right)\left(c-a\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 1& c+a\end{array}\right|$

Using fact 3, perform$R_3\to R_3-R_1$

So,

$$\begin{gathered} \left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& c-a\end{array}\right| \\ =\left(c-a\right)\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right| \end{gathered}$$

Hence,$\left|\begin{array}{ccc}1& a& bc\\ 1& b& ac\\ 1& c& ab\end{array}\right|=\left(c-a\right)\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|$

Consider$\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|$

Expand the determinant along column 1

$\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|=1\left(1-0\right)=1$

Thus,$\left(c-a\right)\left(b-a\right)\left(c-b\right)\left|\begin{array}{ccc}1& a& a^2\\ 0& 1& b+a\\ 0& 0& 1\end{array}\right|=\left(c-a\right)\left(b-a\right)\left(c-b\right)$