Question

If \[{{A}^{T}}\,\]and\[\,{{B}^{T}}\]are transpose matrices for the square matrices A and B respectively then\[{{(AB)}^{T}}\]is equal to
1. \[{{A}^{T}}{{B}^{T}}\]
2. \[A{{B}^{T}}\]
3. \[B{{A}^{T}}\]
4. \[{{B}^{T}}{{A}^{T}}\]

Answer 1

Hint: So, in this question we will use the concept of a multiplication property of a matrix. And we know that this property can only be satisfied only when the\[A\]and\[B\]matrices are square matrices. In a square matrix the number of rows and columns will be the same.

Complete step by step answer:
In question we have to find the \[{{(AB)}^{T}}\]
Before finding that first we need to understand the concept of transpose of a matrix
In transpose of a matrix rows and columns are interchanges of square matrix A and B.
Consider an example for understanding the multiplication property of a matrix \[{{A}_{2\times 2}}\] and \[{{B}_{2\times 2}}\]
  \[{{A}_{2\times 2}}=\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]---(1)\]
\[{{B}_{2\times 2}}=\left[ \begin{matrix}
   {{a}^{'}}_{11} & {{a}^{'}}_{12} \\
   {{a}^{'}}_{21} & {{a}^{'}}_{22} \\
\end{matrix} \right]---(2)\]
We have to take the multiplication of two matrix \[{{A}_{2\times 2}}\] and \[{{B}_{2\times 2}}\]
\[AB=\left[ \begin{matrix}
   (({{a}_{11}}\times {{a}^{'}}_{11})+({{a}_{12}}\times {{a}^{'}}_{21})) & (({{a}_{11}}\times {{a}^{'}}_{12})+({{a}_{12}}\times {{a}^{'}}_{22})) \\
   (({{a}_{21}}\times {{a}^{'}}_{11})+({{a}_{22}}\times {{a}^{'}}_{21})) & (({{a}_{21}}\times {{a}^{'}}_{12})+({{a}_{22}}\times {{a}^{'}}_{22})) \\
\end{matrix} \right]\]
Now, we have to take the transpose of a matrix \[{{(AB)}^{T}}\]
\[{{(AB)}^{T}}={{\left[ \begin{matrix}
   (({{a}_{11}}\times {{a}^{'}}_{11})+({{a}_{12}}\times {{a}^{'}}_{21})) & (({{a}_{11}}\times {{a}^{'}}_{12})+({{a}_{12}}\times {{a}^{'}}_{22})) \\
   (({{a}_{21}}\times {{a}^{'}}_{11})+({{a}_{22}}\times {{a}^{'}}_{21})) & (({{a}_{21}}\times {{a}^{'}}_{12})+({{a}_{22}}\times {{a}^{'}}_{22})) \\
\end{matrix} \right]}^{T}}\]
That means we have to interchange the rows and column of a matrix
\[{{(AB)}^{T}}=\left[ \begin{matrix}
   (({{a}_{11}}\times {{a}^{'}}_{11})+({{a}_{12}}\times {{a}^{'}}_{21})) & (({{a}_{21}}\times {{a}^{'}}_{11})+({{a}_{22}}\times {{a}^{'}}_{21})) \\
   (({{a}_{11}}\times {{a}^{'}}_{12})+({{a}_{12}}\times {{a}^{'}}_{22})) & (({{a}_{21}}\times {{a}^{'}}_{12})+({{a}_{22}}\times {{a}^{'}}_{22})) \\
\end{matrix} \right]---(3)\]

We have to take the transpose of an individual matrix \[{{A}^{T}}\,\] and \[\,{{B}^{T}}\].
\[\therefore {{A}^{T}}={{\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]}^{T}}\]
By interchange the rows and columns
\[{{A}^{T}}=\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{21}} \\
   {{a}_{12}} & {{a}_{22}} \\
\end{matrix} \right]--(4)\]
Similarly for B matrix
\[{{B}^{T}}={{\left[ \begin{matrix}
   {{a}^{'}}_{11} & {{a}^{'}}_{12} \\
   {{a}^{'}}_{21} & {{a}^{'}}_{22} \\
\end{matrix} \right]}^{T}}=\left[ \begin{matrix}
   {{a}^{'}}_{11} & {{a}^{'}}_{21} \\
   {{a}^{'}}_{12} & {{a}^{'}}_{22} \\
\end{matrix} \right]--(5)\]
By multiplying equation \[(4)\] and equation\[(5)\] we get:
\[{{B}^{T}}{{A}^{T}}={{\left[ \begin{matrix}
   {{a}^{'}}_{11} & {{a}^{'}}_{12} \\
   {{a}^{'}}_{21} & {{a}^{'}}_{22} \\
\end{matrix} \right]}^{T}}{{\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right]}^{T}}\]
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   {{a}^{'}}_{11} & {{a}^{'}}_{21} \\
   {{a}^{'}}_{12} & {{a}^{'}}_{22} \\
\end{matrix} \right]\left[ \begin{matrix}
   {{a}_{11}} & {{a}_{21}} \\
   {{a}_{12}} & {{a}_{22}} \\
\end{matrix} \right]--(5)\]
By simplifying this we get:
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   (({{a}_{11}}\times {{a}^{'}}_{11})+({{a}_{12}}\times {{a}^{'}}_{21})) & (({{a}_{21}}\times {{a}^{'}}_{11})+({{a}_{22}}\times {{a}^{'}}_{21})) \\
   (({{a}_{11}}\times {{a}^{'}}_{12})+({{a}_{12}}\times {{a}^{'}}_{22})) & (({{a}_{21}}\times {{a}^{'}}_{12})+({{a}_{22}}\times {{a}^{'}}_{22})) \\
\end{matrix} \right]---(6)\]
By comparing the equation\[(3)\]and equation\[(6)\]
\[\therefore \,\text{LHS=RHS}\]
\[{{(AB)}^{T}}={{B}^{T}}{{A}^{T}}\]
Consider an example for understanding this concept more clearly
\[A=\left[ \begin{matrix}
   1 & 0 \\
   3 & 2 \\
\end{matrix} \right]\] \[B=\left[ \begin{matrix}
   2 & 4 \\
   5 & 1 \\
\end{matrix} \right]\]
By multiplying these two matrix and then take the transpose we get:
\[AB=\left[ \begin{matrix}
   1 & 0 \\
   3 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
   2 & 4 \\
   5 & 1 \\
\end{matrix} \right]\]
By simplifying we get:
\[AB=\left[ \begin{matrix}
   ((1\times 2)+(0\times 5)) & ((1\times 4)+(0\times 1)) \\
   ((3\times 2)+(2\times 5)) & ((3\times 4)+(2\times 1)) \\
\end{matrix} \right]\]
By solving this we get:
\[AB=\left[ \begin{matrix}
   2+0 & 4+0 \\
   6+10 & 12+2 \\
\end{matrix} \right]\]
\[\therefore AB=\left[ \begin{matrix}
   2 & 4 \\
   16 & 14 \\
\end{matrix} \right]\]
Now, take the transpose of above matrix interchange the rows and columns
\[\therefore {{(AB)}^{T}}=\left[ \begin{matrix}
   2 & 16 \\
   4 & 14 \\
\end{matrix} \right]--(7)\]
Take the transpose of individual matrix by interchange the rows and columns of individual matrix
\[{{A}^{T}}=\left[ \begin{matrix}
   1 & 3 \\
   0 & 2 \\
\end{matrix} \right]\] \[{{B}^{T}}=\left[ \begin{matrix}
   2 & 5 \\
   4 & 1 \\
\end{matrix} \right]\]
By multiplying these above two matrix
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   2 & 5 \\
   4 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
   1 & 3 \\
   0 & 2 \\
\end{matrix} \right]\]
By further solving this we get:
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   ((1\times 2)+(0\times 5)) & ((3\times 2)+(2\times 5)) \\
   ((1\times 4)+(0\times 1)) & ((3\times 4)+(2\times 1)) \\
\end{matrix} \right]\]
By further solving and simplifying we get:
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   2+0 & 6+10 \\
   4+0 & 12+2 \\
\end{matrix} \right]\]
\[{{B}^{T}}{{A}^{T}}=\left[ \begin{matrix}
   2 & 16 \\
   4 & 14 \\
\end{matrix} \right]---(8)\]
By comparing the equation \[(7)\]and equation \[(8)\]
\[\therefore \,\text{LHS=RHS}\]
\[{{(AB)}^{T}}={{B}^{T}}{{A}^{T}}\]

So, the correct answer is “Option 4”.

Note: In this particular question we have to use the multiplication property of a property above example which is taken in the solution only to understand the multiplication property. Be careful while multiplying two matrices because if you multiply \[BA\] instead of \[AB\] then you will not satisfy its property. Multiplication property of a matrix can only be proved when there is a square matrix of A and B.