Matrices and Determinants 2 Question 25
28. The value of the determinant $\begin{vmatrix}1 & a & a^{2}-b c \\ 1 & b & b^{2}-c a \\ 1 & c & c^{2}-a b\end{vmatrix}$ is ….
$(1988,2 \mathrm{M})$
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Answer:
Correct Answer: 27. (0)
Solution:
- $\quad\begin{vmatrix}1 & a & a^{2}-b c \\ 1 & b & b^{2}-c a \\ 1 & c & c^{2}-a b\end{vmatrix}=\begin{vmatrix}1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2}\end{vmatrix}-\begin{vmatrix}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{vmatrix}$
Now, $\quad\begin{vmatrix}1 & a & b c \\ 1 & b & c a \\ 1 & c & a b\end{vmatrix}=\frac{1}{a b c}\begin{vmatrix}a & a^{2} & a b c \\ b & b^{2} & a b c \\ c & c^{2} & a b c\end{vmatrix}$
Applying $R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, R_{3} \rightarrow c R_{3}$
$ \begin{aligned} & =\frac{1}{a b c} \cdot a b c\begin{vmatrix} a & a^{2} & 1 \\ b & b^{2} & 1 \\ c & c^{2} & 1 \end{vmatrix}=\begin{vmatrix} 1 & a & a^{2} \\ 1 & b & b^{2} \\ 1 & c & c^{2} \end{vmatrix} \\ \therefore\begin{vmatrix} 1 & a & a^{2}-b c \\ 1 & b & b^{2}-c a \\ 1 & c & c^{2}-a b \end{vmatrix} & =0 \end{aligned} $
