If $$C$$ is skew-symmetric matrix of order $$n$$ and $$X$$ is $$n \times 1$$ column matrix, then $$X'CX$$ is a
A.
scalar matrix
B.
unit matrix
C.
null matrix
D.
None of these
Answer :
null matrix
Solution :
Here, $$X$$ is $$n \times 1,C$$ is $$n \times n$$ and $$X'$$ is $$1 \times n$$
Hence, $$X'CX$$ is $$1 \times 1$$ matrix. Let $$X'CX = k .$$
Then, $$\left( {X'CX} \right)' = X'C'X' = X'C'X' = X'\left( { - C} \right)X = - X'CX = - k$$
$$\eqalign{
& \Rightarrow k = - k \cr
& \Rightarrow k = 0 \cr
& \Rightarrow X'CX{\text{ is null matrix}}{\text{.}} \cr} $$
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