Determinant/Examples/Order 3

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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

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