If $$A$$ and $$B$$ are two matrices such that $$AB = B$$ and $$BA = A,$$ then $${A^2} + {B^2}$$ is equal to
A.
$$2AB$$
B.
$$2BA$$
C.
$$A + B$$
D.
$$AB$$
Answer :
$$A + B$$
Solution :
We have, $${A^2} + {B^2} = AA + BB = A\left( {BA} \right) + B\left( {AB} \right)\left( {\therefore AB = B{\text{ and }}BA + A} \right)$$
$$\eqalign{
& = \left( {AB} \right)A + \left( {BA} \right)B \cr
& = BA + AB = A + B\left( {\therefore AB = B{\text{ and }}BA = A} \right) \cr} $$
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