Verify that if A, B then

Question

Question:

Verify that if $ m{A} = \left[ \begin{matrix} 1 & 2 \ 0 & 1 \ \end{matrix} ight]$, $ m{B} = \left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} ight]$ then

(i) $( m{A}^{t})^{t} = m{A}$

(i) $( m{B}^{t})^{t} = m{B}$

Accepted Answer

Solution:

(i) To prove: $( m{A}^{t})^{t} = m{A}$

Given

$ m{A} = \left[ \begin{matrix} 1 & 2 \ 0 & 1 \ \end{matrix} ight]$

$ m{A}^{t} = \left[ \begin{matrix} 1 & 0 \ 2 & 1 \ \end{matrix} ight]$

Taking transpose of $ m{A}^{t}$, we will get

$( m{A}^{t})^{t} = \left[ \begin{matrix} 1 & 2 \ 0 & 1 \ \end{matrix} ight]= m{A}$

Hence proved: $( m{A}^{t})^{t} = m{A}$

(ii) To prove: $( m{B}^{t})^{t} = m{B}$

Given

$ m{B} = \left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} ight]$

$ m{B}^{t} = \left[ \begin{matrix} 1 & 2 \ 1 & 0 \ \end{matrix} ight]$

Taking transpose of $ m{B}^{t}$, we will get

$( m{B}^{t})^{t} = \left[ \begin{matrix} 1 & 1 \ 2 & 0 \ \end{matrix} ight] = m{B}$

Hence proved: $( m{B}^{t})^{t} = m{B}$