# Determinant properties

For every $n \times n$ matrix, we can define its determinant, being a (possibly complex) number. It is defined recursively by taking out one of its rows, say the $i^{\text{th}}$ row= $$$$\text{Det}(A) = \sum_{k=1}^{n} a_{ik} (-1)^{k+i} \text{Det}(A(i,k))$$$$ where the $(n-1)\times(n-1)$ matrix $A(i,k)$ is obtained from $A$ by removing the $i^{\text{th}}$ row and the $k^{\text{th}}$ column. The determinant of a $0\times 0$ matrix is $1$.

Many properties of determinants are easy to derive if $C = AB$ then, $$$$\text{Det}(C) = \text{Det}(AB) = \text{Det}(A)\text{Det}(B).$$$$

If $C = x A$ with $x$ a number and $A$ an $n\times n$ matrix $$$$\text{Det}(C) = \text{Det}(x A) = x^{n}\text{Det}(A).$$$$

Futhermore, the inverse of a matrix $A$ is a matrix $B$ such that $BA = AB = 1$. Let’s choose a notation for B as $B = A^{-1}$. The matrix elements of B are (note: $A(j, i)$, not $A(i, j)$) $$$$b_{ij} = \frac{(-1)^{i+j}\text{Det}(A(j,i))}{\text{Det}(A)}$$$$ from this, it follows that a matrix $A$ has an inverse if $\text{Det}(A)\neq 0$.

Another definition of the determinant makes use of the $\epsilon$ symbol, defined by

• $\epsilon_{i_{1}\cdots i_{n}} = 0$ if in $i_{1}\cdots i_{n}$ two or more of the indicies are equal,
• $\epsilon_{i_{1}\cdots i_{n}} = \pm 1$ if in $i_{1}\cdots i_{n}$ no two indicies are equal, in which case we have
• $\epsilon_{i_{1}\cdots i_{n}} = 1$ if in $i_{1}\cdots i_{n}$ are an even permutation of $123\ldots n$,
• $\epsilon_{i_{1}\cdots i_{n}} = -1$ if in $i_{1}\cdots i_{n}$ are an odd permutation of $123\ldots n$. In particular, $\epsilon_{123\cdots n} = +1$

The determinant of a matrix $A$ is then given by $$$$A_{i_{1}j_{1}}A_{i_{2}j_{2}}\cdots A_{i_{n}j_{n}}\epsilon_{j_{1}\cdots j_{n}} = \text{Det}(A)\epsilon_{i_{1}\cdots i_{n}}$$$$

Published on May 6, 2021

Last revised on May 6, 2021

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