The number of values of $$k$$ for which the system of equations $$\left( {k + 1} \right)x + 8y = 4k;kx + \left( {k + 3} \right)y = 3k - 1$$ has infinitely many solutions is
A.
0
B.
1
C.
2
D.
infinite
Answer :
1
Solution :
For infinitely many solutions the two equations become identical
$$\eqalign{
& \Rightarrow \frac{{k + 1}}{k} = \frac{8}{{k + 3}} = \frac{{4k}}{{3k - 1}} \cr
& \Rightarrow k = 1. \cr} $$
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