Matrices And Determinants Ques 67
Let $a, b, c$ be positive and not all equal. Show that the value of the determinant $\begin{vmatrix} a & b & c\\ b & c & a\\ c & a & b \end{vmatrix}$ is negative.
$(1981,4 M)$
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Solution:
Formula:
- Let $\Delta=\begin{vmatrix}a & b & c \\ b & c & a \\ c & a & b\end{vmatrix} $
Applying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$
$ \Delta=\begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix}=(a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} $
Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we get
$ \begin{aligned} & =(a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} \\ & =(a+b+c)\left[-(c-b)^{2}-(a-b)(a-c)\right] \\ & =-(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right) \\ & =-\frac{1}{2}(a+b+c)\left(2 a^{2}+2 b^{2}+2 c^{2}-2 a b-2 b c-2 c a\right) \\ & =-\frac{1}{2}(a+b+c)\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] \end{aligned} $
which is always negative.
BETA
