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

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}_{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.