If\[A=\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right)\]. Verify that \[A\left( adjA \right)=\left( adjA \right)A=\left| A \right|I\] where I is the order 2 Identity matrix.

Hint: We calculate the determinant of A. If \[A=\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)\] then its determinant is given by
\[\left| A \right|=ad-bc\] and then we proceed to find the value of adj(A) and then substitute in the given term from the left hand side to verify the result.

Complete step-by-step answer:
Given \[A=\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right)\]
First of all, we calculate determinant of A,
If \[A=\left( \begin{matrix}
   a & b \\
   c & d \\
\end{matrix} \right)\] then its determinant is given by
\[\left| A \right|=ad-bc\]
Hence using the formula in \[A=\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right)\] we get,
\[\left| A \right|=4-6=-2\]
Now we see the process to calculate adj(A) of a matrix A of order 2, there is a direct substitution to calculate the Adjoint of the Matrix A
If we are given matrix\[A=\left( \begin{matrix}
   {{a}_{11}} & {{a}_{12}} \\
   {{a}_{21}} & {{a}_{22}} \\
\end{matrix} \right)\] then \[(adjA)=\left( \begin{matrix}
   {{a}_{22}} & -{{a}_{12}} \\
   -{{a}_{21}} & {{a}_{11}} \\
\end{matrix} \right)\] , that is adj(A) of a order 2 matrix is obtained by interchanging the elements at the main( or principal) diagonal and taking negative of the elements at non principal diagonal.
Applying the above procedure for the above given matrix \[A=\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right)\] we get, \[(adjA)=\left( \begin{matrix}
   4 & -2 \\
   -3 & 1 \\
\end{matrix} \right)\]
Now we will verify the given expression\[A\left( adjA \right)=\left( adjA \right)A=\left| A \right|I\]
To do this we compute \[\left( adjA \right)A\]and \[A\left( adjA \right)\]separately and then check if they are equal
Computing the left-hand side of the above equation we have,
\[\begin{align}
  & A\left( adjA \right)=\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
   4 & -2 \\
   -3 & 1 \\
\end{matrix} \right) \\
 & \\
 & \Rightarrow A\left( adjA \right)=\left( \begin{matrix}
   -2 & 0 \\
   0 & -2 \\
\end{matrix} \right) \\
 & \\
 & \Rightarrow A\left( adjA \right)=(-2)\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)............(i) \\
\end{align}\]

Now similarly we calculate (adjA)A

\[\begin{align}
  & \Rightarrow \left( adjA \right)A=\left( \begin{matrix}
   4 & -2 \\
   -3 & 1 \\
\end{matrix} \right)\left( \begin{matrix}
   1 & 2 \\
   3 & 4 \\
\end{matrix} \right) \\
 & \\
 & \Rightarrow \left( adjA \right)A=\left( \begin{matrix}
   -2 & 0 \\
   0 & -2 \\
\end{matrix} \right) \\
 & \\
 & \Rightarrow \left( adjA \right)A=(-2)\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)..........(ii) \\
\end{align}\]
From (i) and (ii) we get,
\[A(adjA)=(adjA)A=(-2)\left( \begin{matrix}
   1 & 0 \\
   0 & 1 \\
\end{matrix} \right)=\left| A \right|I\]
Hence, we have obtained the result.

Note: The possibility of the mistake is that A(adjA) and (adjA)A can be taken same without calculating separately which is wrong as Matrices do not commute with respect to multiplication. There can be chances when A(adjA) and (adjA)A are different.