Let $\mathbf A$ be the square matrix:

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

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

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

## Proof

For a square matrix $\mathbf A = a_{i j}$ of order $4$, the adjugate matrix of $\mathbf A$ is:

$\adj {\mathbf A} = \begin {pmatrix} A_{11} & A_{21} & A_{31} & A_{41} \\ A_{12} & A_{22} & A_{32} & A_{42} \\ A_{13} & A_{23} & A_{33} & A_{43} \\ A_{14} & A_{24} & A_{34} & A_{44} \end {pmatrix}$

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

 $\ds A_{1 1}$ $=$ $\ds \paren {-1}^{1 + 1} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

 $\ds A_{1 2}$ $=$ $\ds \paren {-1}^{1 + 2} \begin {vmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{1 3}$ $=$ $\ds \paren {-1}^{1 + 3} \begin {vmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{1 4}$ $=$ $\ds \paren {-1}^{1 + 4} \begin {vmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{2 1}$ $=$ $\ds \paren {-1}^{2 + 1} \begin {vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds -1$ Determinant of Upper Triangular Matrix

 $\ds A_{2 2}$ $=$ $\ds \paren {-1}^{2 + 2} \begin {vmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

 $\ds A_{2 3}$ $=$ $\ds \paren {-1}^{2 + 3} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{2 4}$ $=$ $\ds \paren {-1}^{2 + 4} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{3 1}$ $=$ $\ds \paren {-1}^{3 + 1} \begin {vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

 $\ds A_{3 2}$ $=$ $\ds \paren {-1}^{3 + 2} \begin {vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds -1$ Determinant of Upper Triangular Matrix

 $\ds A_{3 3}$ $=$ $\ds \paren {-1}^{3 + 3} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

 $\ds A_{3 4}$ $=$ $\ds \paren {-1}^{3 + 4} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 0$ Determinant of Upper Triangular Matrix

 $\ds A_{4 1}$ $=$ $\ds \paren {-1}^{4 + 1} \begin {vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds -1$ Determinant of Upper Triangular Matrix

 $\ds A_{4 2}$ $=$ $\ds \paren {-1}^{4 + 2} \begin {vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 1 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

 $\ds A_{4 3}$ $=$ $\ds \paren {-1}^{4 + 3} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds -1$ Determinant of Upper Triangular Matrix

 $\ds A_{4 4}$ $=$ $\ds \paren {-1}^{4 + 4} \begin {vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end {vmatrix}$ $\ds$ $=$ $\ds 1$ Determinant of Upper Triangular Matrix

Hence:

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

$\Box$

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

First we note that $\map \det {\mathbf A} = 1$, by Determinant of Upper Triangular Matrix.

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

and all is well.

$\blacksquare$