Question

If $$A$$ and $$B$$ are two matrices such that $$AB = A$$   and $$BA = B,$$   then which one of the following is correct ?

A. $${\left( {{A^T}} \right)^2} = {A^T}$$  
B. $${\left( {{A^T}} \right)^2} = {B^T}$$
C. $${\left( {{A^T}} \right)^2} = {\left( {{A^{ - 1}}} \right)^{ - 1}}$$
D. None of the above
Answer :   $${\left( {{A^T}} \right)^2} = {A^T}$$
Solution :
Let $$A$$ and $$B$$ be two matrices such that $$AB = A$$   and $$BA = B$$
Now, consider $$AB = A$$
Take Transpose on both side
$$\eqalign{ & {\left( {AB} \right)^T} = {A^T} \cr & \Rightarrow {A^T} = {B^T}.{A^T}\,\,\,\,.....\left( 1 \right) \cr & {\text{Now, }}BA = B \cr} $$
Take, Transpose on both side
$$\eqalign{ & {\left( {BA} \right)^T} = {B^T} \cr & \Rightarrow {B^T} = {A^T}.{B^T}\,\,\,\,.....\left( 2 \right) \cr} $$
Now, from equation (1) and (2). we have
$$\eqalign{ & {A^T} = \left( {{A^T}.{B^T}} \right){A^T} \cr & {A^T} = {A^T}\left( {{B^T}{A^T}} \right) \cr & = {A^T}{\left( {AB} \right)^T}\,\,\,\left( {\because {{\left( {AB} \right)}^T} = {B^T} = {B^T}{A^T}} \right) \cr & = {A^T}.{A^T} \cr & {\text{Thus, }}{A^T} = {\left( {{A^T}} \right)^2} \cr} $$

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 $$C$$
Releted Question 2

If $$\omega \left( { \ne 1} \right)$$  is a cube root of unity, then
\[\left| {\begin{array}{*{20}{c}} 1&{1 + i + {\omega ^2}}&{{\omega ^2}}\\ {1 - i}&{ - 1}&{{\omega ^2} - 1}\\ { - i}&{ - i + \omega - 1}&{ - 1} \end{array}} \right|=\]

A. 0
B. 1
C. $$i$$
D. $$\omega $$
Releted Question 3

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

A. no solution
B. unique solution
C. infinitely many solutions
D. finitely many solutions
Releted Question 4

If $$A$$ and $$B$$ are square matrices of equal degree, then which one is correct among the followings?

A. $$A + B = B + A$$
B. $$A + B = A - B$$
C. $$A - B = B - A$$
D. $$AB=BA$$

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Matrices and Determinants


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