Which of the following is the reversal law of transposes?

The correct option is (AB)’=B’A’.

The reversal law of transposes is a fundamental property in matrix algebra. It states that the transpose of the product of two matrices is equal to the product of their transposes taken in the reverse order.

Let $A$ and $B$ be two matrices such that the product $AB$ is defined. Let the dimensions of matrix $A$ be $m\times n$ and the dimensions of matrix $B$ be $n\times p$. The product matrix $C=AB$ will have dimensions of $m\times p$, and its transpose $C^T=(AB{)}^{\prime }$ will have dimensions $p\times m$.

To see why the order must reverse:
- The transpose of $A$, denoted as $A\prime$, has dimensions $n\times m$.
- The transpose of $B$, denoted as $B\prime$, has dimensions $p\times n$.
If we try to multiply $A\prime B\prime$, a matrix of size $(n\times m)$ is multiplied by a matrix of size $(p\times n)$. This multiplication is not defined unless $m=p$.
However, if we multiply them in the reverse order as $B\prime A\prime$, we multiply a matrix of size $(p\times n)$ by a matrix of size $(n\times m)$. The inner dimensions match, and the resulting product matrix has dimensions $p\times m$, which matches the dimensions of $(AB{)}^{\prime }$.

By comparing the individual elements, the entry in the $i$-th row and $j$-th column of $(AB{)}^{\prime }$ is the entry in the $j$-th row and $i$-th column of $AB$, which is given by:
$\sum _{k=1}^n{A}_{jk}{B}_{ki}$
This is identical to the corresponding dot product of the $i$-th row of $B\prime$ (which is the $i$-th column of $B$) and the $j$-th column of $A\prime$ (which is the $j$-th row of $A$):
$\sum _{k=1}^n(B\prime {)}_{ik}(A\prime {)}_{kj}$
Therefore, the equality holds, validating the relation $(AB{)}^{\prime }=B\prime A\prime$.