# Prove that $\begin{vmatrix} a^2&bc&ac+c^2\\ a^2+ab& b^2& ac\\ ab&b^2+bc&c^2 \end{vmatrix}=4\,a^2\;b^2\;c^2.$

Toolbox:
• If each element of a row (or a column) of a determinant is multiplied by a constant k,then its value gets multiplied by k.
• By this property we can take out any common factor from any one row or any one column.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns. rows.
• 2. Mutiplication of the elements of any row or column by a non-zero number
• 3. The addition of any row or column , the corresponding elemnets of any other row or column multiplied by any non zero number.
$\Delta=\begin{vmatrix} a^2&bc&ac+c^2\\ a^2+ab& b^2& ac\\ ab&b^2+bc&c^2 \end{vmatrix}=4\,a^2\;b^2\;c^2.$
We can take common factor a from $C_1$,b from $C_2$ and c from $C_2$
Therefore $\Delta=abc\begin{vmatrix}a & c & a+c\\a+b & b & a\\b & b+c & c\end{vmatrix}$
Apply $R_2\rightarrow R_2-R_1$ and $R_3-R_1$
$\Delta=abc\begin{vmatrix}a & c & a+c\\b & b-c & -c\\b-a & b & -a\end{vmatrix}$
Now let us apply $R_2\rightarrow R_2+R_1$
$\Delta=abc\begin{vmatrix}a & c & a+c\\a+b & b & a\\b-a & b & -a\end{vmatrix}$
Now let us apply $R_3\rightarrow R_3+R_2$
$\Delta=abc\begin{vmatrix}a & c & a+c\\a+b & b & a\\2b & 2b & 0\end{vmatrix}$
Let us take 2b as the common factor from $R_3$
$\Delta=2ab^2c\begin{vmatrix}a & c & a+c\\a+b & b & a\\1 & 1 & 0\end{vmatrix}$
Apply $C_2\rightarrow C_1-C_2$
$\Delta=2ab^2c\begin{vmatrix}a & c-a & a+c\\a+b & -a & a\\1 & 0 & 0\end{vmatrix}$
Now let us expand along $R_3$ we get,
$\Delta=2ab^2c[1[a(c-a)+a(a+c)]+a(a+c)]+0+0]$
$\;\;\;=2ab^2c[ac-a^2+a^2+ac]$
$\;\;\;=2ab^2c[ac-a^2+a^2+ac]$
$\;\;\;=2ab^2c\times 2ac.$
$\Delta=4a^2b^2c^2$
Hence proved
edited Mar 2, 2013