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