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If $\mathrm{a, b, c}$ are positive and unequal, show that value of the determinant $\Delta=\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ is negative.
Solution
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$ to the given determinant, we get
$\Delta = \left| {\begin{array}{*{20}{l}}
{a + b + c}&b&c \\
{a + b + c}&c&a \\
{a + b + c}&a&b
\end{array}} \right| = (a + b + c)\left| {\begin{array}{*{20}{c}}
1&b&c \\
1&c&a \\
1&a&b
\end{array}} \right|$
${ = (a + b + c)\left| {\begin{array}{*{20}{c}}
1&b&c \\
0&{c – b}&{a – c} \\
0&{a – b}&{b – c}
\end{array}} \right|}$ ${{\text{(Applying }}{{\text{R}}_2} \to {{\text{R}}_2} – {{\text{R}}_1},{\text{ and }}{{\text{R}}_3} \to {{\text{R}}_3} – {{\text{R}}_1})}$
$ = (a + b + c)[(c – b)(b – c) – (a – c)(a – b)]$ ${\text{(Expanding along }}{{\text{C}}_1}{\text{ ) }}$
$ = (a + b + c)\left( { – {a^2} – {b^2} – {c^2} + ab + bc + ca} \right)$
which is negative (since $\left.a+b+c>0 \text { and }(a-b)^{2}+(b-c)^{2}+(c-a)^{2}>0\right)$
