Determinant/Examples/Order 3
Example of Determinant
Let $\mathbf A = \sqbrk a_3$ be a square matrix of order $3$.
That is, let:
$\quad \mathbf A = \begin {bmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end {bmatrix}$
The determinant of $\mathbf A$ is given by:
$\quad \map \det {\mathbf A} = \begin {vmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end {vmatrix}$
Then:
\(\ds \map \det {\mathbf A}\) | \(=\) | \(\ds a_{1 1} \begin {vmatrix} a_{2 2} & a_{2 3} \\ a_{3 2} & a_{3 3} \end {vmatrix} - a_{1 2} \begin {vmatrix} a_{2 1} & a_{2 3} \\ a_{3 1} & a_{3 3} \end {vmatrix} + a_{1 3} \begin {vmatrix} a_{2 1} & a_{2 2} \\ a_{3 1} & a_{3 2} \end{vmatrix}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map \sgn {1, 2, 3} a_{1 1} a_{2 2} a_{3 3}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \map \sgn {1, 3, 2} a_{1 1} a_{2 3} a_{3 2}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \map \sgn {2, 1, 3} a_{1 2} a_{2 1} a_{3 3}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \map \sgn {2, 3, 1} a_{1 2} a_{2 3} a_{3 1}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \map \sgn {3, 1, 2} a_{1 3} a_{2 1} a_{3 2}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds \map \sgn {3, 2, 1} a_{1 3} a_{2 2} a_{3 1}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds a_{1 1} a_{2 2} a_{3 3}\) | ||||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds a_{1 1} a_{2 3} a_{3 2}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds a_{1 2} a_{2 1} a_{3 3}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds a_{1 2} a_{2 3} a_{3 1}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds + \, \) | \(\ds a_{1 3} a_{2 1} a_{3 2}\) | |||||||||||
\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds a_{1 3} a_{2 2} a_{3 1}\) |
and thence in a single expression as:
- $\ds \map \det {\mathbf A} = \frac 1 6 \sum_{i \mathop = 1}^3 \sum_{j \mathop = 1}^3 \sum_{k \mathop = 1}^3 \sum_{r \mathop = 1}^3 \sum_{s \mathop = 1}^3 \sum_{t \mathop = 1}^3 \map \sgn {i, j, k} \map \sgn {r, s, t} a_{i r} a_{j s} a_{k t}$
where $\map \sgn {i, j, k}$ is the sign of the permutation $\tuple {i, j, k}$ of the set $\set {1, 2, 3}$.
The values of the various instances of $\map \sgn {\lambda_1, \lambda_2, \lambda_3}$ are obtained by applications of Parity of K-Cycle.
Using Einstein Summation Convention
The determinant of a square matrix of order $3$ $\mathbf A$ can be expressed using the Einstein summation convention as:
- $\map \det {\mathbf A} = \dfrac 1 6 \map \sgn {i, j, k} \map \sgn {r, s, t} a_{i r} a_{j s} a_{k t}$
Note that there are $6$ indices which appear twice, and so $6$ summations are assumed.
Sources
- 1998: Richard Kaye and Robert Wilson: Linear Algebra ... (previous) ... (next): Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.6$ Determinant and trace
- 1980: A.J.M. Spencer: Continuum Mechanics ... (previous) ... (next): $2.1$: Matrices: $(2.9)$
- 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): determinant