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Find adjoint of a matrix $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$

Answer
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Hint: Here we first need to find the adjoint of a matrix by taking the cofactor of every element present in the matrix and then take transpose of the matrix.

“Complete step-by-step answer:”
The adjoint of a matrix $A = {\left( {{a_{ij}}} \right)_{n \times n}}$ is defined as the transpose of the matrix ${\left( {{A_{ij}}} \right)_{n \times n}}$ where $\left( {{A_{ij}}} \right)$ is cofactor of the element $\left( {{a_{ij}}} \right)$
Now let $A = \left( {\begin{array}{*{20}{c}}
  1&2 \\
  3&4
\end{array}} \right)$
So first of all we evaluate the cofactor of the every element,
\[{A_{11}} = 4,{A_{12}} = - 3,{A_{21}} = - 2,{A_{22}} = 1\]
\[adjA = \left( {\begin{array}{*{20}{c}}
  {{A_{11}}}&{{A_{21}}} \\
  {{A_{12}}}&{{A_{22}}}
\end{array}} \right)\]
On putting the value, we have
\[adjA = \left( {\begin{array}{*{20}{c}}
  4&{ - 2} \\
  { - 3}&1
\end{array}} \right)\]
And hence this will be the adjoint of A

NOTE: In this type of question first of all we need to find cofactor of every element and then after putting transpose we can find the adjoint of the given matrix.Students have to take care of signs while calculating the cofactors and while taking transpose of matrix.