This is the trivial case:
Thus the determinant of an order $1$ matrix is that element itself.
\(\ds \begin {vmatrix} a_{1 1} & a_{1 2} \\ a_{2 1} & a_{2 2} \end{vmatrix}\) | \(=\) | \(\ds \map \sgn {1, 2} a_{1 1} a_{2 2} + \map \sgn {2, 1} a_{1 2} a_{2 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_{1 1} a_{2 2} - a_{1 2} a_{2 1}\) |
Let:
$\quad \map \det {\mathbf A} = \begin {vmatrix} a_{1 1} & a_{1 2} & a_{1 3} \\ a_{2 1} & a_{2 2} & a_{2 3} \\ a_{3 1} & a_{3 2} & a_{3 3} \end {vmatrix}$
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:
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.