Inverse of a 3 by 3 Matrix

To find the inverse of a matrix -which must be a squarer matrix:
1. Form the matrix of cofactors
2. Find the adjoint matrix by transposing the matrix of cofactors
3. Divide each element of the matrix by the determinant
Example: Consider the matrix  
\[A=\left( \begin{array}{ccc} 2 & 3 & 0 \\ 0 & 1 & 5 \\ 4 & -2 & 1 \end{array} \right)\]

The cofactor  
\[\]
 of a square matrix  
\[A\]
  is the determinant of the submatrix obtained by deleting the ith row and jth column, multiplied by  
\[(-1)^{i+j}\]
.
\[A_{11}=(-1)^2 \left| \begin{array}{cc} 1 & 5 \\ -2 & 1 \end{array} \right| = 1 \times 1 - 5 \times (-2)=11\]

\[A_{12}=(-1)^3 \left| \begin{array}{cc} 0 & 5 \\ 4 & 1 \end{array} \right| = -(0 \times 1 - 5 \times 4)=19\]

\[A_{13}=(-1)^4 \left| \begin{array}{cc} 0 & 1 \\ 4 & -2 \end{array} \right| = 0 \times -2 - 1 \times 4=-4\]

\[A_{21}=(-1)^3 \left| \begin{array}{cc} 3 & 0 \\ -2 & 1 \end{array} \right| = -(3 \times 1 - 0 \times -2)=-3\]

\[A_{22}=(-1)^4 \left| \begin{array}{cc} 2 & 0 \\ 4 & 1 \end{array} \right| = 2 \times 1 - 0 \times 4=2\]

\[A_{23}=(-1)^5 \left| \begin{array}{cc} 2 & 3 \\ 4 & -2 \end{array} \right| = -( 2 \times -2 - 3 \times 4)=16\]

\[A_{31}=(-1)^4 \left| \begin{array}{cc} 3 & 0 \\ 1 & 5 \end{array} \right| = 3 \times 5 - 0 \times 1=15\]

\[A_{32}=(-1)^5 \left| \begin{array}{cc} 2 & 0 \\ 0 & 5 \end{array} \right| = -(2 \times 5 - 0 \times 0)=-10\]

\[A_{33}=(-1)^6 \left| \begin{array}{cc} 2 & 3 \\ 0 & 1 \end{array} \right| = 2 \times 1 - 3 \times 0=2\]

The matrix of cofactors is  
\[\left( \begin{array}{ccc} 11 & 19 & -4 \\ -3 & 2 & 16 \\ 15 & -10 & 2 \end{array} \right)\]

The adjoint matrix is the transpose of the matrix of cofactors, and is equal to  
\[\left( \begin{array}{ccc} 11 & -3 & 15 \\ 19 & 2 & -10 \\ 4 & 16 & 2 \end{array} \right)\]

The determinant of  
\[A\]
  is 82.
The inverse is then  
\[A^{-1}= \frac{1}{82} \left( \begin{array}{ccc} 11 & -3 & 15 \\ 19 & 2 & -10 \\ 4 & 16 & 2 \end{array} \right)\]

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