If A and B are square matrices then (AB)’ =

The correct option is B’A’.

To understand why this is correct, we can analyze the definition of the transpose of a matrix and the rule for matrix multiplication.

The transpose of a matrix is an operation that flips a matrix over its diagonal, switching its row and column indices. If a matrix has a representation of elements, the transpose operation interchanges the rows and columns.

Let $A$ and $B$ be two square matrices of order $n\times n$. Let the element in the $i$-th row and $j$-th column of matrix $A$ be denoted as $a_{ij}$, and for matrix $B$ be denoted as $b_{ij}$.

By the definition of matrix multiplication, the element in the $i$-th row and $j$-th column of the product matrix $C=AB$ is given by:

$$c_{ij}=\sum _{k=1}^n{a}_{ik}{b}_{kj}$$

Now, let us find the transpose of this product matrix, denoted by ${\left(AB\right)}^{\prime }$. The element in the $i$-th row and $j$-th column of ${\left(AB\right)}^{\prime }$ corresponds to the element in the $j$-th row and $i$-th column of the original product matrix $AB$:

$${\left[{\left(AB\right)}^{\prime }\right]}_{ij}=c_{ji}=\sum _{k=1}^n{a}_{jk}{b}_{ki}$$

Next, let us consider the transposes of individual matrices, $A^{\prime }$ and $B^{\prime }$. Their elements are defined as:

$${\left(A^{\prime }\right)}_{ij}=a_{ji}$$

and

$${\left(B^{\prime }\right)}_{ij}=b_{ji}$$

Now, let us calculate the product of these transposes in reverse order, which is $B^{\prime }{A}^{\prime }$. The element in the $i$-th row and $j$-th column of this product is:

$${\left[B^{\prime }{A}^{\prime }\right]}_{ij}=\sum _{k=1}^n{\left(B^{\prime }\right)}_{ik}{\left(A^{\prime }\right)}_{kj}$$

Substituting the element values of the transposed matrices into this expression gives:

$${\left[B^{\prime }{A}^{\prime }\right]}_{ij}=\sum _{k=1}^n{b}_{ki}{a}_{jk}$$

Since the multiplication of scalar elements (numbers) is commutative, we can rearrange the order of terms inside the summation:

$${\left[B^{\prime }{A}^{\prime }\right]}_{ij}=\sum _{k=1}^n{a}_{jk}{b}_{ki}$$

Comparing the expressions for ${\left[{\left(AB\right)}^{\prime }\right]}_{ij}$ and ${\left[B^{\prime }{A}^{\prime }\right]}_{ij}$, we observe that they are identical for all $i$ and $j$:

$${\left[{\left(AB\right)}^{\prime }\right]}_{ij}={\left[B^{\prime }{A}^{\prime }\right]}_{ij}$$

Therefore, we establish the fundamental algebraic property of matrix transposes, known as the Reversal Law:

$${\left(AB\right)}^{\prime }=B^{\prime }{A}^{\prime }$$