Adjugate Matrix/Examples/Arbitrary Matrix 4
Jump to navigation
Jump to search
Example of Adjugate Matrix
Let $\mathbf A$ be the square matrix:
- $\mathbf A = \begin {pmatrix} -1 & 2 & 0 \\ 0 & 1 & 3 \\ 2 & -3 & 3 \end {pmatrix}$
Then the adjugate matrix of $\mathbf A$ is:
- $\adj {\mathbf A} = \begin {pmatrix} 12 & 6 & -2 \\ -6 & -3 & 1 \\ 6 & 3 & -1 \end {pmatrix}$
Proof
For a square matrix $\mathbf A = a_{i j}$ of order $3$, the adjugate matrix of $\mathbf A$ is:
- $\adj {\mathbf A} = \begin {pmatrix} A_{1 1} & A_{2 1} & A_{3 1} \\ A_{1 2} & A_{2 2} & A_{3 2} \\ A_{1 3} & A_{2 3} & A_{3 3} \end {pmatrix}$
For each $a_{i j}$ in $\mathbf A$, we calculate the cofactors $A_{i j}$:
\(\ds A_{1 1}\) | \(=\) | \(\ds \paren {-1}^{1 + 1} \begin {vmatrix} 1 & 3 \\ -3 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {1 \times 3 - \paren {-3} \times 3}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 12\) |
\(\ds A_{1 2}\) | \(=\) | \(\ds \paren {-1}^{1 + 2} \begin {vmatrix} 0 & 3 \\ 2 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {0 \times 3 - 3 \times 2}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 6\) |
\(\ds A_{1 3}\) | \(=\) | \(\ds \paren {-1}^{1 + 3} \begin {vmatrix} 0 & 1 \\ 2 & -3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {0 \times \paren {-3} - 2 \times 1}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds -2\) |
\(\ds A_{2 1}\) | \(=\) | \(\ds \paren {-1}^{2 + 1} \begin {vmatrix} 2 & 0 \\ -3 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {2 \times 3 - \paren {-3} \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds -6\) |
\(\ds A_{2 2}\) | \(=\) | \(\ds \paren {-1}^{2 + 2} \begin {vmatrix} -1 & 0 \\ 2 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {\paren {-1} \times 3 - 2 \times 1}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds -3\) |
\(\ds A_{2 3}\) | \(=\) | \(\ds \paren {-1}^{2 + 3} \begin {vmatrix} -1 & 2 \\ 2 & -3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {\paren {-1} \times \paren {-3} - 2 \times 2}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 1\) |
\(\ds A_{3 1}\) | \(=\) | \(\ds \paren {-1}^{3 + 1} \begin {vmatrix} 2 & 0 \\ 1 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {2 \times 3 - 1 \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 6\) |
\(\ds A_{3 2}\) | \(=\) | \(\ds \paren {-1}^{3 + 2} \begin {vmatrix} -1 & 0 \\ 0 & 3 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -1 \times \paren {\paren {-1} \times 3 - 0 \times 0}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds 3\) |
\(\ds A_{3 3}\) | \(=\) | \(\ds \paren {-1}^{3 + 3} \begin {vmatrix} -1 & 2 \\ 0 & 1 \end {vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times \paren {\paren {-1} \times 1 - 0 \times 2}\) | Determinant of Order 2 | |||||||||||
\(\ds \) | \(=\) | \(\ds -1\) |
Hence:
- $\adj {\mathbf A} = \begin {pmatrix} 12 & 6 & -2 \\ -6 & -3 & 1 \\ 6 & 3 & -1 \end {pmatrix}$
$\blacksquare$
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: Exercises: $1.14 \ \text {(a)}$