Determinants Matrices Question 71
Question: Let $ S _{k}={{\alpha }^{k}}+{{\beta }^{k}}+{{\gamma }^{k}}, $ then $ \Delta = \begin{vmatrix} S_0 & S_1 & S_2 \\ S_1 & S_2 & S_3 \\ S_2 & S_3 & S_4 \\ \end{vmatrix} $ is equal to
Options:
A) $ S_6 $
B) $ S_5-S_3 $
C) $ S_6-S_4 $
D) None
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ \Delta \begin{vmatrix} S_0 & S_1 & S_2 \\ S_1 & S_2 & S_3 \\ S_2 & S_3 & S_4 \\ \end{vmatrix} $
$ = \begin{vmatrix} 1+1+1 & \alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} \\ \alpha +\beta +\gamma & {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}} \\ {{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}} & {{\alpha }^{3}}+{{\beta }^{3}}+{{\gamma }^{3}} & {{\alpha }^{4}}+{{\beta }^{4}}+{{\gamma }^{4}} \\ \end{vmatrix} $ The above determinant can be expressed as product of two determinants. Thus, $ \Delta = \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ {{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\ \end{vmatrix} \begin{vmatrix} 1 & 1 & 1 \\ \alpha & \beta & \gamma \\ {{\alpha }^{2}} & {{\beta }^{2}} & {{\gamma }^{2}} \\ \end{vmatrix} $
$ ={{[(\beta -\alpha )(\gamma -\alpha )(\gamma -\beta )]}^{2}} $
BETA
