Browse Questions

Using properties of determinants, prove that: $\begin{vmatrix} \alpha & \alpha^2 & \beta+\gamma\\ \beta & \beta^2 & \alpha+\gamma\\ \gamma & \gamma^2 & \alpha+\beta \end{vmatrix} = (\beta-\gamma) (\gamma - \alpha) (\alpha-\beta) (\alpha+\beta+\gamma)$

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 column of a given determinant.
• Elementary transformations can be done by
• 1. Interchanging any two rows or columns.
• 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.
Let $\Delta = \begin{vmatrix} \alpha & \alpha^2 & \beta+\gamma\\\beta-\alpha & \beta^2-\alpha^2 & \alpha-\beta\\\gamma-\alpha& \gamma^2-\alpha^2& \alpha-\gamma\end{vmatrix}$

Apply $R_2\rightarrow R_2-R_1$ and $R_3\rightarrow R_3-R_1$ we have,

$\Delta = (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha^2 & \beta+\gamma\\1& \beta+\alpha & -1\\1& \gamma+\alpha& 1\end{vmatrix}$

Apply $R_3\rightarrow R_3-R_2$ we get,

$\Delta = (\beta-\alpha)(\gamma-\alpha)\begin{vmatrix} \alpha & \alpha^2 & \beta+\gamma\\1& \beta+\alpha & -1\\0& \gamma-\beta& 0\end{vmatrix}$

Now expanding along $R_3$ we get,

$\Delta=(\beta-\alpha)(\gamma-\alpha)[-(\gamma-\beta)(-\alpha-\beta-\gamma)]$

$\;\;\;=(\beta-\alpha)(\gamma-\alpha)[(\gamma-\beta)(\alpha+\beta+\gamma)]$

$\Delta=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)$

edited Mar 5, 2013