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Algebra of Matrices — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Matrices5 Marks
Question
For the matrices $A$ and $B$, verify that $(AB)^T = B^TA^T$, where $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix},\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
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Answer

Given: $\text{A}=\begin{bmatrix}1&3\\2&4\end{bmatrix}$
$\text{A}^\text{T}=\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\text{B}=\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\text{B}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}$
Now,
$\text{AB}=\begin{bmatrix}1&3\\2&4\end{bmatrix}\begin{bmatrix}1&4\\2&5\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}1+6&4+15\\2+8&8+20\end{bmatrix}$
$\Rightarrow\text{AB}=\begin{bmatrix}7&19\\10&28\end{bmatrix}$
$\Rightarrow(\text{AB})^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(1)$
Also,
$\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1&2\\4&5\end{bmatrix}\begin{bmatrix}1&2\\3&4\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}1+6&2+8\\4+15&8+20\end{bmatrix}$
$\Rightarrow\text{B}^\text{T}\text{A}^\text{T}=\begin{bmatrix}7&10\\19&28\end{bmatrix}\ \dots(2)$
$\therefore\ (\text{AB})^\text{T}=\text{B}^\text{T}\text{A}^\text{T}$ [From eqs. (1) and (2)]

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