The identity element in the group $M = \left\{ {\left[ {\begin{array}{*{20}{c}}
  x&x \\
  x&x
\end{array}} \right],x \in \mathbb{R};x \ne 0} \right\}$ with respect to matrix multiplication is
A. $\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right]$
B. $\dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right]$
C. $\left[ {\begin{array}{*{20}{c}}
  1&0 \\
  0&1
\end{array}} \right]$
D. $\left[ {\begin{array}{*{20}{c}}
  0&1 \\
  1&0
\end{array}} \right]$

Hint: We will be using the concept of matrix multiplication and scalar multiplication. The process of multiplying two matrices, commonly referred to as matrix multiplication, generates a single matrix. The result of multiplying a real number by a matrix is known as a scalar multiplication. Each entry of the matrix is multiplied by the specified scalar in scalar multiplication.
Formula Used: If we have two matrices $A = {\left[ {{a_{ij}}} \right]_{m \times n}}$ and $B = {\left[ {{b_{ij}}} \right]_{n \times p}}$ then their matrix multiplication will be equal to $C = {\left[ {{c_{ij}}} \right]_{m \times p}}$ , where ${c_{ij}} = {a_{i1}}{b_{1j}} + {a_{i2}}{b_{2j}} + \ldots + {a_{in}}{b_{nj}}$ .

Complete step-by-step solution:
We have group $M = \left[ {\begin{array}{*{20}{c}}
  x&x \\
  x&x
\end{array}} \right]$ , $\forall x \in \mathbb{R}$ , $x \ne 0$ .
Suppose a matrix $P = \left[ {\begin{array}{*{20}{c}}
  a&a \\
  a&a
\end{array}} \right]$ be an identity element of the group which can also be written as
$P = a\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right]$
We can obtain $P$ by substituting $x = a$ , where $a \ne 0$ .
If we take the product of the matrices $M$ and $P$ i.e., $MP$ which implies,
$MP = \left[ {\begin{array}{*{20}{c}}
  x&x \\
  x&x
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&a \\
  a&a
\end{array}} \right] = M$ , as $P$ is an identity element of the group.
Similarly, $PM = M$
$ \Rightarrow MP = PM = M$
$\left[ {\begin{array}{*{20}{c}}
  x&x \\
  x&x
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
  a&a \\
  a&a
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
  x&x \\
  x&x
\end{array}} \right]$
Solving this system of equations, we get
$2ax = x$
$ \Rightarrow a = \dfrac{1}{2}$
Hence, the identity element of the group becomes $P = \dfrac{1}{2}\left[ {\begin{array}{*{20}{c}}
  1&1 \\
  1&1
\end{array}} \right]$ .
Hence, the correct option is B.

Note: To multiply two matrices, we must ensure that the number of rows in the second matrix equals the number of columns in the first matrix. As a result, a specific number of rows from the first matrix and a specific number of columns from the second matrix will be present in the final matrix. While during scalar multiplication, the order of the matrix does not matter.