If the least number of zeroes in a lower triangular matrix is 10, then what is the order of the matrix ?
A.
$$3 \times 3$$
B.
$$4 \times 4$$
C.
$$5 \times 5$$
D.
$$10 \times 10$$
Answer :
$$4 \times 4$$
Solution :
Number of zeroes in a lower triangular matrix of order $$n \times n$$ is
$$1 + 2 + 3 + ..... + n = \frac{{n\left( {n + 1} \right)}}{2}$$
Number of zeros = 10
$$\eqalign{
& \Rightarrow \frac{{n\left( {n + 1} \right)}}{2} = 10 \cr
& \Rightarrow {n^2} + n - 20 = 0 \cr
& \Rightarrow \left( {n + 5} \right)\left( {n - 4} \right) = 0 \cr
& \Rightarrow n = 4\,\,{\text{or }} - 5\left( { - 5{\text{ is meaningless}}} \right) \cr
& \Rightarrow n = 4. \cr} $$
⇒ order of the matrix is $$4 \times 4$$
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