Adjugate Matrix/Examples/Arbitrary Matrix 2
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Example of Adjugate Matrix
Let $\mathbf A$ be the square matrix:
- $\mathbf A = \begin {pmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \\ \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
- $\adj {\mathbf A} = \begin {pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end {pmatrix}$
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 $A_{i j}$:
\(\ds A_{1 1}\) | \(=\) | \(\ds \paren {-1}^{1 + 1} \begin {vmatrix} 1 & 0 \\ -1 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {1 \times 0 - \paren {-1} \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
\(\ds A_{1 2}\) | \(=\) | \(\ds \paren {-1}^{1 + 2} \begin {vmatrix} -1 & 0 \\ 0 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {\paren {-1} \times 0 - 0 \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
\(\ds A_{1 3}\) | \(=\) | \(\ds \paren {-1}^{1 + 3} \begin {vmatrix} -1 & 1 \\ 0 & -1 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {\paren {-1} \times \paren {-1} - 0 \times 1}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{2 1}\) | \(=\) | \(\ds \paren {-1}^{2 + 1} \begin {vmatrix} 0 & -1 \\ -1 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {0 \times 0 - \paren {-1} \times \paren {-1} }\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{2 2}\) | \(=\) | \(\ds \paren {-1}^{2 + 2} \begin {vmatrix} 1 & -1 \\ 0 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {1 \times 0 - \paren {-1} \times 0 }\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
\(\ds A_{2 3}\) | \(=\) | \(\ds \paren {-1}^{2 + 3} \begin {vmatrix} 1 & 0 \\ 0 & -1 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {1 \times \paren {-1} - 0 \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{3 1}\) | \(=\) | \(\ds \paren {-1}^{3 + 1} \begin {vmatrix} 0 & -1 \\ 1 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {0 \times 0 - \paren {-1} \times 1}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{3 2}\) | \(=\) | \(\ds \paren {-1}^{3 + 2} \begin {vmatrix} 1 & -1 \\ -1 & 0 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {1 \times 0 - \paren {-1} \times \paren {-1} }\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{3 3}\) | \(=\) | \(\ds \paren {-1}^{3 + 3} \begin {vmatrix} 1 & 0 \\ -1 & 1 \\ \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {1 \times 1 - 0 \times \paren {-1} }\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
Hence:
- $\adj {\mathbf A} = \begin {pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end {pmatrix}$
$\Box$
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 Expansion Theorem for Determinants, expanding using column $3$.
\(\ds \adj {\mathbf A} \mathbf A\) | \(=\) | \(\ds \begin {pmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 0 \\ \end {pmatrix} \begin {pmatrix} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \\ \end {pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \begin {pmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end {pmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \det {\mathbf A} \begin {pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {pmatrix}\) |
and all is well.
$\blacksquare$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.13 \ \text {(b)}$