The equations $$x + y + z = 6, x + 2y + 3z = 10, x + 2y + mz = n$$ give infinite number of values of the triplet $$(x, y, z)$$ if
A.
$$m = 3,n \in R$$
B.
$$m = 3,n \ne 10$$
C.
$$m = 3,n = 10$$
D.
None of these
Answer :
$$m = 3,n = 10$$
Solution :
Each of the first three options contains $$m = 3.$$
When $$m = 3,$$ the last two equations become $$x + 2y + 3z = 10$$ and $$x + 2y + 3z = n.$$
Obviously, when $$n = 10$$ these equations become the same. So, we are left with only two independent equations to find the values of the three unknowns.
Consequently, there will be infinite 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