If the system of equations
$$\eqalign{
& ax + by + c = 0 \cr
& bx + cy + a = 0 \cr
& cx + ay + b = 0 \cr} $$
has a solution then the system of equations
$$\eqalign{
& \left( {b + c} \right)x + \left( {c + a} \right)y + \left( {a + b} \right)z = 0 \cr
& \left( {c + a} \right)x + \left( {a + b} \right)y + \left( {b + c} \right)z = 0 \cr
& \left( {a + b} \right)x + \left( {b + c} \right)y + \left( {c + a} \right)z = 0 \cr} $$
has
A.
only one solution
B.
no solution
C.
infinite number of solutions
D.
None of these
Answer :
infinite number of solutions
Solution :
For existence of a solution of the first system,
\[\left| {\begin{array}{*{20}{c}}
a&b&c \\
b&c&a \\
c&a&b
\end{array}} \right| = 0.\]
The second system will have a nontrivial solution if we can prove that
\[\left| {\begin{array}{*{20}{c}}
{b + c}&{c + a}&{a + b} \\
{c + a}&{a + b}&{b + c} \\
{a + b}&{b + c}&{c + a}
\end{array}} \right| = 0.\]
Establish \[\left| {\begin{array}{*{20}{c}}
{b + c}&{c + a}&{a + b} \\
{c + a}&{a + b}&{b + c} \\
{a + b}&{b + c}&{c + a}
\end{array}} \right| = 2\left| {\begin{array}{*{20}{c}}
a&b&c \\
b&c&a \\
c&a&b
\end{array}} \right| = 0.\]
Remember that the existence of one nontrivial solution implies existence of infinite number of non-trivial solutions
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