If \[A = \left[ {\begin{array}{*{20}{l}}3&2\\1&4\end{array}} \right]\] then \[A({\mathop{\rm adj}\nolimits} A) = \]
A. \[\left[ {\begin{array}{*{20}{c}}{10}&0\\0&{10}\end{array}} \right]\]
В. \[\left[ {\begin{array}{*{20}{c}}0&{10}\\{10}&0\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{10}&1\\1&{10}\end{array}} \right]\]
D. None of these

Hint: We must first find the adjoint of a matrix by taking the cofactor of each element in the matrix and then taking the matrix's transpose. The adjoint of a matrix \[A = {(a_{ij})_{n \times n}}\] is defined as the transpose of the matrix \[{(A_{ij})_{n \times n}}\] where \[(A_{ij})\] is the element's cofactor \[(a_{ij})\] and then we have to multiply the determinant A which is given in the question with the adjoint matrix of A we found to get the matrix of \[A({\mathop{\rm adj}\nolimits} A)\]

Formula Used:
Cofactor can be found using
\[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\]

Complete Step-by-Step Solution:The transpose of a square matrix's cofactor matrix is its adjugate or adjunct.
We have been already known that the cofactor matrix is made up of all the cofactors of the provided matrix, which are determined using the formula \[{C_{ij}} = {( - 1)^{i + j}}{M_{ij}}\] where minor is \[{M_{ij}}\]
We have to calculate all the cofactors of the given matrix \[A = \left[ {\begin{array}{*{20}{l}}3&2\\1&4\end{array}} \right]\] we have
\[{C_{11}} = {( - 1)^{1:1}}|4| = 4\]
\[{C_{12}} = {( - 1)^{1:2}}|1| = - 1\]
\[{C_{21}} = {( - 1)^{2 + 1}}|2| = - 2\]
\[{C_{22}} = {( - 1)^2} \cdot 2|3| = 3\]
Thus, the cofactor matrix is
\[\left[ {\begin{array}{*{20}{l}}4&{ - 1}\\{ - 2}&3\end{array}} \right]\]
The cofactor’s transpose matrix is,
Here the rows becomes columns and the columns becomes rows, we have
\[\left[ {\begin{array}{*{20}{l}}4&{ - 2}\\{ - 1}&3\end{array}} \right]\]
Therefore, the adjoint of matrix A is
\[\left[ {\begin{array}{*{20}{l}}4&{ - 2}\\{ - 1}&3\end{array}} \right]\]
Now, we have to multiply the matrix A and adjoint of matrix ‘A’, we have
\[ = \left[ {\begin{array}{*{20}{l}}3&2\\1&4\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{l}}4&{ - 2}\\{ - 1}&3\end{array}} \right]\]
Multiply the rows of the first matrix to the columns of the second matrix
\[ = \left[ {\begin{array}{*{20}{c}}{3 \cdot \:4 + 2\left( { - 1} \right)}&{3\left( { - 2} \right) + 2 \cdot \:3}\\{1 \cdot \:4 + 4\left( { - 1} \right)}&{1 \cdot \left( { - 2} \right) + 4 \cdot \:3}\end{array}} \right]\]
Now, we have to solve each element by simplifying it, we get
\[ = \left[ {\begin{array}{*{20}{l}}{10}&0\\0&{10}\end{array}} \right]\]
Therefore, if \[A = \left[ {\begin{array}{*{20}{l}}3&2\\1&4\end{array}} \right]\] then \[A({\mathop{\rm adj}\nolimits} A) = \left[ {\begin{array}{*{20}{l}}{10}&0\\0&{10}\end{array}} \right]\]
Hence, the option A is correct

Note: In this type of problems, we must first determine the cofactor of each element before using transpose to determine the adjoint of the given matrix. When computing cofactors and transposing the matrix, students must keep signs in mind. When evaluating the adjoint matrix, the value must be the transpose of the obtained value.