If $$A$$ and $$B$$ are symmetric matrices of the same order and $$X = AB + BA$$ and $$Y = AB - BA,$$ then $${\left( {XY} \right)^T}$$ is equal to
A.
$$XY$$
B.
$$YX$$
C.
$$- YX$$
D.
None of these
Answer :
$$- YX$$
Solution :
Given that, $$X = AB + BA$$
$$\eqalign{
& \Rightarrow X = {X^T}{\text{ and }}Y = AB - BA \cr
& \Rightarrow Y = - {Y^T}. \cr
& {\text{Now, }}{\left( {XY} \right)^T} = {Y^T}{X^T} = - YX \cr} $$
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