What is the adjoint of \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]\]?
A. \[\left[ {\begin{array}{*{20}{c}}4&8&3\\2&1&6\\0&2&1\end{array}} \right]\]
B. \[\left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}} \right]\]
C. \[\left[ {\begin{array}{*{20}{c}}{11}&9&3\\1&2&8\\6&9&1\end{array}} \right]\]
D. \[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&1\\{ - 1}&3&3\\{ - 2}&3&{ - 3}\end{array}} \right]\]

Hint: We will find the cofactors of the given matrix. Then we will find the transpose matrix of the cofactor matrix to find the adjoint matrix.

Formula used:
The adjoint of the matrix \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}&{{a_{13}}}\\{{a_{21}}}&{{a_{22}}}&{{a_{23}}}\\{{a_{31}}}&{{a_{32}}}&{{a_{33}}}\end{array}} \right]\] is \[\left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]where \[{A_{ij}}\] are the cofactors.

Complete step by step solution:
Given matrix is \[A = \left[ {\begin{array}{*{20}{c}}3&{ - 3}&4\\2&{ - 3}&4\\0&{ - 1}&1\end{array}} \right]\].
The cofactors of A are
\[{A_{11}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - 3 \cdot 1 - 4\left( { - 1} \right)\]
\[ = 1\]
\[{A_{12}} = - \left| {\begin{array}{*{20}{c}}2&4\\0&1\end{array}} \right|\]
\[ = - \left( {2 \cdot 1 - 4 \cdot 0} \right)\]
\[ = - 2\]
\[{A_{13}} = \left| {\begin{array}{*{20}{c}}2&3\\0&{ - 1}\end{array}} \right|\]
\[ = 2 \cdot \left( { - 1} \right) - 3 \cdot 0\]
\[ = - 2\]
\[{A_{21}} = - \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 1}&1\end{array}} \right|\]
\[ = - \left[ { - 3 \cdot 1 - 4 \cdot \left( { - 1} \right)} \right]\]
\[ = - 1\]
\[{A_{22}} = \left| {\begin{array}{*{20}{c}}3&4\\0&1\end{array}} \right|\]
\[ = \left[ {3 \cdot 1 - 4 \cdot 0} \right]\]
\[ = 3\]
\[{A_{23}} = - \left| {\begin{array}{*{20}{c}}3&{ - 3}\\0&{ - 1}\end{array}} \right|\]
\[ = - \left[ {3 \cdot \left( { - 1} \right) - \left( { - 3} \right) \cdot 0} \right]\]
\[ = 3\]
\[{A_{31}} = \left| {\begin{array}{*{20}{c}}{ - 3}&4\\{ - 3}&4\end{array}} \right|\]
\[ = \left[ { - 3 \cdot 4 - \left( { - 3} \right) \cdot 4} \right]\]
\[ = 0\]
\[{A_{32}} = - \left| {\begin{array}{*{20}{c}}3&4\\2&4\end{array}} \right|\]
\[ = - \left[ {3 \cdot 4 - 2 \cdot 4} \right]\]
\[ = - 4\]
\[{A_{33}} = \left| {\begin{array}{*{20}{c}}3&{ - 3}\\2&{ - 3}\end{array}} \right|\]
\[ = \left[ {3 \cdot \left( { - 3} \right) - 2 \cdot \left( { - 3} \right)} \right]\]
\[ = - 3\]
The Adjoint of the matrix is
\[Adj\,A = \left[ {\begin{array}{*{20}{c}}{{A_{11}}}&{{A_{21}}}&{{A_{31}}}\\{{A_{12}}}&{{A_{22}}}&{{A_{32}}}\\{{A_{13}}}&{{A_{23}}}&{{A_{33}}}\end{array}} \right]\]
Substitute the value of cofactors
\[ = \left[ {\begin{array}{*{20}{c}}1&{ - 1}&0\\{ - 2}&3&{ - 4}\\{ - 2}&3&{ - 3}\end{array}} \right]\]
Hence option B is the correct option.

Additional information:
We can find the adjoint of a matrix if the matrix is a square matrix. This means the number of rows is equal to the number of columns. The adjoint of a matrix is used to the determine inverse of that matrix. In other words, we can find the inverse of a matrix if the matrix is a square matrix.

Note: Students often do a common mistake to find the adjoint of the matrix. They forgot to transpose the cofactor matrix. To calculate the adjoint we have to find the transpose matrix of the cofactor matrix.