Let $\mathbf A$ be the square matrix:

$\mathbf A = \begin {pmatrix} 1 & 2 & 0 \\ 0 & -1 & 2 \\ -1 & 2 & 0 \end {pmatrix}$

Then the adjugate matrix of $\mathbf A$ is:

$\adj {\mathbf A} = \begin {pmatrix} -4 & 0 & 4 \\ -2 & 0 & -2 \\ -1 & -4 & -1 \end {pmatrix}$

## Proof

For a square matrix $\mathbf A = a_{i j}$ of order $3$, the adjugate matrix of $\mathbf A$ is:

$\adj {\mathbf A} = \begin {pmatrix} A_{11} & A_{21} & A_{31} \\ A_{12} & A_{22} & A_{32} \\ A_{13} & A_{23} & A_{33} \end {pmatrix}$

For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:

 $\ds A_{1 1}$ $=$ $\ds \paren {-1}^{1 + 1} \begin {vmatrix} -1 & 2 \\ 2 & 0 \end {vmatrix}$ $\ds$ $=$ $\ds 1 \times \paren {\paren {-1} \times 0 - 2 \times 2}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -4$

 $\ds A_{1 2}$ $=$ $\ds \paren {-1}^{1 + 2} \begin {vmatrix} 0 & 2 \\ -1 & 0 \end {vmatrix}$ $\ds$ $=$ $\ds -1 \times \paren {2 \times 0 - \paren {-1} \times 2}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -2$

 $\ds A_{1 3}$ $=$ $\ds \paren {-1}^{1 + 3} \begin {vmatrix} 0 & -1 \\ -1 & 2 \end {vmatrix}$ $\ds$ $=$ $\ds 1 \times \paren {0 \times 2 - \paren {-1} \times \paren {-1} }$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -1$

 $\ds A_{2 1}$ $=$ $\ds \paren {-1}^{2 + 1} \begin {vmatrix} 2 & 0 \\ 2 & 0 \end {vmatrix}$ $\ds$ $=$ $\ds -1 \times \paren {2 \times 0 - 2 \times 0}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 0$

 $\ds A_{2 2}$ $=$ $\ds \paren {-1}^{2 + 2} \begin {vmatrix} 1 & 0 \\ -1 & 0 \end {vmatrix}$ $\ds$ $=$ $\ds 1 \times \paren {1 \times 0 - 0 \times \paren {-1} }$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 0$

 $\ds A_{2 3}$ $=$ $\ds \paren {-1}^{2 + 3} \begin {vmatrix} 1 & 2 \\ -1 & 2 \end {vmatrix}$ $\ds$ $=$ $\ds -1 \times \paren {1 \times 2 - \paren {-1} \times 2}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -4$

 $\ds A_{3 1}$ $=$ $\ds \paren {-1}^{3 + 1} \begin {vmatrix} 2 & 0 \\ -1 & 2 \end {vmatrix}$ $\ds$ $=$ $\ds 1 \times \paren {2 \times 2 - \paren {-1} \times 0}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds 4$

 $\ds A_{3 2}$ $=$ $\ds \paren {-1}^{3 + 2} \begin {vmatrix} 1 & 0 \\ 0 & 2 \end {vmatrix}$ $\ds$ $=$ $\ds -1 \times \paren {1 \times 2 - 0 \times 0}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -2$

 $\ds A_{3 3}$ $=$ $\ds \paren {-1}^{3 + 3} \begin {vmatrix} 1 & 2 \\ 0 & -1 \end {vmatrix}$ $\ds$ $=$ $\ds 1 \times \paren {1 \times \paren {-1} - 0 \times 2}$ Definition of Determinant of Order 2 $\ds$ $=$ $\ds -1$

Hence:

$\adj {\mathbf A} = \begin {pmatrix} -4 & 0 & 4 \\ -2 & 0 & -2 \\ -1 & -4 & -1 \end {pmatrix}$

$\Box$

We check this result, using Product of Matrix with Adjugate equals Determinant by Unit Matrix.

First we note that $\map \det {\mathbf A} = -8$, by Expansion Theorem for Determinants, for example expanding using column $1$.

 $\ds \adj {\mathbf A} \mathbf A$ $=$ $\ds \begin {pmatrix} 1 & 2 & 0 \\ 0 & -1 & 2 \\ -1 & 2 & 0 \\ \end {pmatrix} \begin {pmatrix} -4 & 0 & 4 \\ -2 & 0 & -2 \\ -1 & -4 & -1 \\ \end {pmatrix}$ $\ds$ $=$ $\ds \begin {pmatrix} -8 & 0 & 0 \\ 0 & -8 & 0 \\ 0 & 0 & -8 \\ \end {pmatrix}$ $\ds$ $=$ $\ds \map \det {\mathbf A} \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {pmatrix}$

and all is well.

$\blacksquare$