Question

For the matrices $A$and $B$, verify that ${\left( {AB} \right)^T} = {B^T}{A^T}$, where $A = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right]$, \[B = \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  2&5
\end{array}} \right]\]

Answer 1

Hint: Here, we are required to verify that ${\left( {AB} \right)^T} = {B^T} \cdot {A^T}$. We are given the two matrices $A$ and $B$. Now, in order to verify this, we will first of all multiply both the matrices to find $AB$. Now, interchanging its rows and columns will give us ${\left( {AB} \right)^T}$. Similarly, we will find the individual transpose of both the matrices and multiply them together to verify that LHS $ = $ RHS.

Complete step-by-step answer:
Given matrices are:
$A = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right]$ and \[B = \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  2&5
\end{array}} \right]\]
Now, in order to verify ${\left( {AB} \right)^T} = {B^T}{A^T}$
First of all, we will multiply the matrices $A$ and $B$
$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&4 \\
  2&5
\end{array}} \right]$
$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  {\left( {1 \times 1} \right) + \left( {3 \times 2} \right)}&{\left( {1 \times 4} \right) + \left( {3 \times 5} \right)} \\
  {\left( {2 \times 1} \right) + \left( {4 \times 2} \right)}&{\left( {2 \times 4} \right) + \left( {4 \times 5} \right)}
\end{array}} \right]$
Solving further,
$ \Rightarrow AB = \left[ {\begin{array}{*{20}{c}}
  7&{19} \\
  {10}&{28}
\end{array}} \right]$
Now, when we do the transpose of any given matrix, then, its rows becomes the columns and the columns become the rows. Hence, the rows and columns interchange with each other.
Hence, ${\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  7&{10} \\
  {19}&{28}
\end{array}} \right]$
Therefore, LHS $ = {\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  7&{10} \\
  {19}&{28}
\end{array}} \right]$
Now, since,
$A = \left[ {\begin{array}{*{20}{c}}
  1&3 \\
  2&4
\end{array}} \right]$
Hence, ${A^T} = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right]$………………………………….$\left( 1 \right)$
Also, \[B = \left[ {\begin{array}{*{20}{c}}
  1&4 \\
  2&5
\end{array}} \right]\]
Hence, \[{B^T} = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  4&5
\end{array}} \right]\]……………………………$\left( 2 \right)$
Now, RHS $ = {B^T}{A^T}$
Hence, substituting the values from $\left( 1 \right)$ and $\left( 2 \right)$, we get,
RHS$ = {B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  1&2 \\
  4&5
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right]$
$ \Rightarrow {B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  {\left( {1 \times 1} \right) + \left( {2 \times 3} \right)}&{\left( {1 \times 2} \right) + \left( {2 \times 4} \right)} \\
  {\left( {4 \times 1} \right) + \left( {5 \times 3} \right)}&{\left( {4 \times 2} \right) + \left( {5 \times 4} \right)}
\end{array}} \right]$
Solving further, we get,
$ \Rightarrow {B^T}{A^T} = \left[ {\begin{array}{*{20}{c}}
  7&{10} \\
  {19}&{28}
\end{array}} \right]$
Also, we have proved that ${\left( {AB} \right)^T} = \left[ {\begin{array}{*{20}{c}}
  7&{10} \\
  {19}&{28}
\end{array}} \right]$
Clearly, LHS $ = $ RHS
Hence, ${\left( {AB} \right)^T} = {B^T}{A^T}$
Thus, verified

Note: A matrix is a rectangular array or table of numbers, symbols or expressions, arranged in rows and columns. Now, in order to multiply two matrices it is really important that the number of columns in the first matrix must be equal to the number of rows in the second matrix. But since, in this question, we had both the matrices of order $2 \times 2$, i.e. square matrices, there was no need to check for this multiplication rule before multiplying them together. Also, when we interchange the rows with columns, we find the transpose of that particular matrix.