Question

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

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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.

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 $A = \left( {\begin{array}{*{20}{c}} 1&2 \\ 3&4 \end{array}} \right)$
${A_{11}} = 4,{A_{12}} = - 3,{A_{21}} = - 2,{A_{22}} = 1$
$adjA = \left( {\begin{array}{*{20}{c}} {{A_{11}}}&{{A_{21}}} \\ {{A_{12}}}&{{A_{22}}} \end{array}} \right)$
$adjA = \left( {\begin{array}{*{20}{c}} 4&{ - 2} \\ { - 3}&1 \end{array}} \right)$