If $$a \ne b \ne c$$ are all positive, then the value of the determinant \[\left| {\begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array}} \right|\] is
A.
non-negative
B.
non-positive
C.
negative
D.
positive
Answer :
negative
Solution :
\[\left| {\begin{array}{*{20}{c}}
a&b&c\\
b&c&a\\
c&a&b
\end{array}} \right| = \left| {\begin{array}{*{20}{c}}
{a + b + c}&b&c\\
{a + b + c}&c&a\\
{a + b + c}&a&b
\end{array}} \right|\]
$$\left( {\because {C_1} \to {C_1} + {C_2} + {C_3}} \right)$$
\[ = \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}}
1&b&c\\
1&c&a\\
1&a&b
\end{array}} \right|\]
[ on taking $$(a + b + c)$$ common from $$C_1$$ ]
$$\eqalign{
& = \left( {a + b + c} \right)\left[ {1\left( {bc - {a^2}} \right) - b\left( {b - a} \right) + c\left( {a - c} \right)} \right] \cr
& = \left( {a + b + c} \right)\left[ {bc - {a^2} - {b^2} + ab + ac - {c^2}} \right] \cr
& = \left( {a + b + c} \right)\left[ { - \left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right] \cr
& = - \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr} $$
Hence, the determinant is negative value.
Releted MCQ Question on Algebra >> Matrices and Determinants
Releted Question 1
Consider the set $$A$$ of all determinants of order 3 with entries 0 or 1 only. Let $$B$$ be the subset of $$A$$ consisting of all determinants with value 1. Let $$C$$ be the subset of $$A$$ consisting of all determinants with value $$- 1.$$ Then
A.
$$C$$ is empty
B.
$$B$$ has as many elements as $$C$$
C.
$$A = B \cup C$$
D.
$$B$$ has twice as many elements as elements as $$C$$
Let $$a, b, c$$ be the real numbers. Then following system of equations in $$x, y$$ and $$z$$
$$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1,$$ $$ - \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} + \frac{{{z^2}}}{{{c^2}}} = 1$$ has