Determinant

The determinant has several key properties that can be proved by direct evaluation of the definition for ${\displaystyle 2\times 2}$-matrices, and that continue to hold for determinants of larger matrices. They are as follows:[1] first, the determinant of the identity matrix ${\displaystyle {\begin{pmatrix}1&0\\0&1\end{pmatrix}}}$ is 1. Second, the determinant is zero if two rows are the same:

${\displaystyle {\begin{vmatrix}a&b\\a&b\end{vmatrix}}=ab-ba=0.}$

This holds similarly if the two columns are the same. Moreover,

${\displaystyle {\begin{vmatrix}a&b+b'\\c&d+d'\end{vmatrix}}=a(d+d')-(b+b')c={\begin{vmatrix}a&b\\c&d\end{vmatrix}}+{\begin{vmatrix}a&b'\\c&d'\end{vmatrix}}.}$

Finally, if any column is multiplied by some number ${\displaystyle r}$ (i.e., all entries in that column are multiplied by that number), the determinant is also multiplied by that number:

${\displaystyle {\begin{vmatrix}r\cdot a&b\\r\cdot c&d\end{vmatrix}}=rad-brc=r(ad-bc)=r\cdot {\begin{vmatrix}a&b\\c&d\end{vmatrix}}.}$

The Leibniz formula for the determinant of a 3 × 3 matrix is the following:

${\displaystyle {\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}=aei+bfg+cdh-ceg-bdi-afh.\ }$

In this expression, each term has one factor from each row, all in different columns, arranged in increasing row order. For example, bdi has b from the first row second column, d from the second row first column, and i from the third row third column. The signs are determined by how many transpositions of factors are necessary to arrange the factors in increasing order of their columns (given that the terms are arranged left-to-right in increasing row order): positive for an even number of transpositions and negative for an odd number. For the example of bdi, the single transposition of bd to db gives dbi, whose three factors are from the first, second and third columns respectively; this is an odd number of transpositions, so the term appears with negative sign.

Rule of Sarrus

The rule of Sarrus is a mnemonic for the expanded form of this determinant: the sum of the products of three diagonal north-west to south-east lines of matrix elements, minus the sum of the products of three diagonal south-west to north-east lines of elements, when the copies of the first two columns of the matrix are written beside it as in the illustration. This scheme for calculating the determinant of a 3 × 3 matrix does not carry over into higher dimensions.

Generalizing the above to higher dimensions, the determinant of an ${\displaystyle n\times n}$ matrix is an expression involving permutations and their signatures. A permutation of the set ${\displaystyle \{1,2,\dots ,n\}}$ is a bijective function ${\displaystyle \sigma }$ from this set to itself, with values ${\displaystyle \sigma (1),\sigma (2),\ldots ,\sigma (n)}$ exhausting the entire set. The set of all such permutations, called the symmetric group, is commonly denoted ${\displaystyle S_{n}}$. The signature ${\displaystyle \operatorname {sgn}(\sigma )}$ of a permutation ${\displaystyle \sigma }$ is ${\displaystyle +1,}$ if the permutation can be obtained with an even number of transpositions (exchanges of two entries); otherwise, it is ${\displaystyle -1.}$

Given a matrix

${\displaystyle A={\begin{bmatrix}a_{1,1}\ldots a_{1,n}\\\vdots \qquad \vdots \\a_{n,1}\ldots a_{n,n}\end{bmatrix}},}$

the Leibniz formula for its determinant is, using sigma notation for the sum,

${\displaystyle \det(A)={\begin{vmatrix}a_{1,1}\ldots a_{1,n}\\\vdots \qquad \vdots \\a_{n,1}\ldots a_{n,n}\end{vmatrix}}=\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )a_{1,\sigma (1)}\cdots a_{n,\sigma (n)}.}$

Using pi notation for the product, this can be shortened into

${\displaystyle \det(A)=\sum _{\sigma \in S_{n}}\left(\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{i,\sigma (i)}\right)}$.

The Levi-Civita symbol ${\displaystyle \varepsilon _{i_{1},\ldots ,i_{n}}}$ is defined on the n-tuples of integers in ${\displaystyle \{1,\ldots ,n\}}$ as 0 if two of the integers are equal, and otherwise as the signature of the permutation defined by the n-tuple of integers. With the Levi-Civita symbol, the Leibniz formula becomes

${\displaystyle \det(A)=\sum _{i_{1},i_{2},\ldots ,i_{n}}\varepsilon _{i_{1}\cdots i_{n}}a_{1,i_{1}}\!\cdots a_{n,i_{n}},}$

where the sum is taken over all n-tuples of integers in ${\displaystyle \{1,\ldots ,n\}.}$ [4][5]

The determinant can be characterized by the following three key properties. To state these, it is convenient to regard an ${\displaystyle n\times n}$-matrix A as being composed of its ${\displaystyle n}$ columns, so denoted as

${\displaystyle A={\big (}a_{1},\dots ,a_{n}{\big )},}$

where the column vector ${\displaystyle a_{i}}$ (for each i) is composed of the entries of the matrix in the i-th column.

1. ${\displaystyle \det \left(I\right)=1}$, where ${\displaystyle I}$ is an identity matrix.
2. The determinant is multilinear: if the jth column of a matrix ${\displaystyle A}$ is written as a linear combination ${\displaystyle a_{j}=r\cdot v+w}$ of two column vectors v and w and a number r, then the determinant of A is expressible as a similar linear combination:

   ${\displaystyle {\begin{aligned}|A|&={\big |}a_{1},\dots ,a_{j-1},r\cdot v+w,a_{j+1},\dots ,a_{n}|\\&=r\cdot |a_{1},\dots ,v,\dots a_{n}|+|a_{1},\dots ,w,\dots ,a_{n}|\end{aligned}}}$

3. The determinant is alternating: whenever two columns of a matrix are identical, its determinant is 0:

   ${\displaystyle |a_{1},\dots ,v,\dots ,v,\dots ,a_{n}|=0.}$

If the determinant is defined using the Leibniz formula as above, these three properties can be proved by direct inspection of that formula. Some authors also approach the determinant directly using these three properties: it can be shown that there is exactly one function that assigns to any ${\displaystyle n\times n}$-matrix A a number that satisfies these three properties.[6] This also shows that this more abstract approach to the determinant yields the same definition as the one using the Leibniz formula.

To see this it suffices to expand the determinant by multi-linearity in the columns into a (huge) linear combination of determinants of matrices in which each column is a standard basis vector. These determinants are either 0 (by property 9) or else ±1 (by properties 1 and 12 below), so the linear combination gives the expression above in terms of the Levi-Civita symbol. While less technical in appearance, this characterization cannot entirely replace the Leibniz formula in defining the determinant, since without it the existence of an appropriate function is not clear.

These rules have several further consequences:

• The determinant is a homogeneous function, i.e.,
  $${\displaystyle \det(cA)=c^{n}\det(A)}$$

  
  (for an ${\displaystyle n\times n}$ matrix ${\displaystyle A}$).
• Interchanging any pair of columns of a matrix multiplies its determinant by −1. This follows from the determinant being multilinear and alternating (properties 2 and 3 above):
  $${\displaystyle |a_{1},\dots ,a_{j},\dots a_{i},\dots ,a_{n}|=-|a_{1},\dots ,a_{i},\dots ,a_{j},\dots ,a_{n}|.}$$

  
  This formula can be applied iteratively when several columns are swapped. For example
  $${\displaystyle |a_{3},a_{1},a_{2},a_{4}\dots ,a_{n}|=-|a_{1},a_{3},a_{2},a_{4},\dots ,a_{n}|=|a_{1},a_{2},a_{3},a_{4},\dots ,a_{n}|.}$$

  
  Yet more generally, any permutation of the columns multiplies the determinant by the sign of the permutation.
• If some column can be expressed as a linear combination of the other columns (i.e. the columns of the matrix form a linearly dependent set), the determinant is 0. As a special case, this includes: if some column is such that all its entries are zero, then the determinant of that matrix is 0.
• Adding a scalar multiple of one column to another column does not change the value of the determinant. This is a consequence of multilinearity and being alternative: by multilinearity the determinant changes by a multiple of the determinant of a matrix with two equal columns, which determinant is 0, since the determinant is alternating.
• If ${\displaystyle A}$ is a triangular matrix, i.e. ${\displaystyle a_{ij}=0}$, whenever ${\displaystyle i>j}$ or, alternatively, whenever ${\displaystyle i<j}$, then its determinant equals the product of the diagonal entries:
  $${\displaystyle \det(A)=a_{11}a_{22}\cdots a_{nn}=\prod _{i=1}^{n}a_{ii}.}$$

  
  Indeed, such a matrix can be reduced, by appropriately adding multiples of the columns with fewer nonzero entries to those with more entries, to a diagonal matrix (without changing the determinant). For such a matrix, using the linearity in each column reduces to the identity matrix, in which case the stated formula holds by the very first characterizing property of determinants. Alternatively, this formula can also be deduced from the Leibniz formula, since the only permutation ${\displaystyle \sigma }$ which gives a non-zero contribution is the identity permutation.

These characterizing properties and their consequences listed above are both theoretically significant, but can also be used to compute determinants for concrete matrices. In fact, Gaussian elimination can be applied to bring any matrix into upper triangular form, and the steps in this algorithm affect the determinant in a controlled way. The following concrete example illustrates the computation of the determinant of the matrix ${\displaystyle A}$ using that method:

${\displaystyle A={\begin{bmatrix}-2&-1&2\\2&1&4\\-3&3&-1\end{bmatrix}}.}$

Computation of the determinant of matrix ${\displaystyle A}$

Matrix	${\displaystyle B={\begin{bmatrix}-3&-1&2\\3&1&4\\0&3&-1\end{bmatrix}}}$	${\displaystyle C={\begin{bmatrix}-3&5&2\\3&13&4\\0&0&-1\end{bmatrix}}}$	${\displaystyle D={\begin{bmatrix}5&-3&2\\13&3&4\\0&0&-1\end{bmatrix}}}$	${\displaystyle E={\begin{bmatrix}18&-3&2\\0&3&4\\0&0&-1\end{bmatrix}}}$
Obtained by	add the second column to the first	add 3 times the third column to the second	swap the first two columns	add ${\displaystyle -{\frac {13}{3}}}$ times the second column to the first
Determinant	${\displaystyle |A|=|B|}$	${\displaystyle |B|=|C|}$	${\displaystyle |D|=-|C|}$	${\displaystyle |E|=|D|}$

Combining these equalities gives ${\displaystyle |A|=-|E|=-(18\cdot 3\cdot (-1))=54.}$

The determinant of the transpose of ${\displaystyle A}$ equals the determinant of A:

${\displaystyle \det \left(A^{\textsf {T}}\right)=\det(A)}$.

This can be proven by inspecting the Leibniz formula.[7] This implies that in all the properties mentioned above, the word "column" can be replaced by "row" throughout. For example, viewing an n × n matrix as being composed of n rows, the determinant is an n-linear function.

The determinant is a multiplicative map, i.e., for square matrices ${\displaystyle A}$ and ${\displaystyle B}$ of equal size, the determinant of a matrix product equals the product of their determinants:

${\displaystyle \det(AB)=\det(A)\det(B)}$

This key fact can be proven by observing that, for a fixed matrix ${\displaystyle B}$, both sides of the equation are alternating and multilinear as a function depending on the columns of ${\displaystyle A}$. Moreover, they both take the value ${\displaystyle \det B}$ when ${\displaystyle A}$ is the identity matrix. The above-mentioned unique characterization of alternating multilinear maps therefore shows this claim.[8]

A matrix ${\displaystyle A}$ with entries in a field is invertible precisely if its determinant is nonzero. This follows from the multiplicativity of the determinant and the formula for the inverse involving the adjugate matrix mentioned below. In this event, the determinant of the inverse matrix is given by

${\displaystyle \det \left(A^{-1}\right)={\frac {1}{\det(A)}}=[\det(A)]^{-1}}$.

In particular, products and inverses of matrices with non-zero determinant (respectively, determinant one) still have this property. Thus, the set of such matrices (of fixed size ${\displaystyle n}$ over a field ${\displaystyle K}$) forms a group known as the general linear group ${\displaystyle \operatorname {GL} _{n}(K)}$ (respectively, a subgroup called the special linear group ${\displaystyle \operatorname {SL} _{n}(K)\subset \operatorname {GL} _{n}(K)}$. More generally, the word "special" indicates the subgroup of another matrix group of matrices of determinant one. Examples include the special orthogonal group (which if n is 2 or 3 consists of all rotation matrices), and the special unitary group.

Because the determinant respects multiplication and inverses, it is in fact a group homomorphism from ${\displaystyle \operatorname {GL} _{n}(K)}$ into the multiplicative group ${\displaystyle K^{\times }}$ of nonzero elements of ${\displaystyle K}$. This homomorphism is surjective and its kernel is ${\displaystyle \operatorname {SL} _{n}(K)}$ (the matrices with determinant one). Hence, by the first isomorphism theorem, this shows that ${\displaystyle \operatorname {SL} _{n}(K)}$ is a normal subgroup of ${\displaystyle \operatorname {GL} _{n}(K)}$, and that the quotient group ${\displaystyle \operatorname {GL} _{n}(K)/\operatorname {SL} _{n}(K)}$ is isomorphic to ${\displaystyle K^{\times }}$.

The Cauchy–Binet formula is a generalization of that product formula for rectangular matrices. This formula can also be recast as a multiplicative formula for compound matrices whose entries are the determinants of all quadratic submatrices of a given matrix.[9][10]

Laplace expansion expresses the determinant of a matrix ${\displaystyle A}$ recursively in terms of determinants of smaller matrices, known as its minors. The minor ${\displaystyle M_{i,j}}$ is defined to be the determinant of the ${\displaystyle (n-1)\times (n-1)}$-matrix that results from ${\displaystyle A}$ by removing the ${\displaystyle i}$-th row and the ${\displaystyle j}$-th column. The expression ${\displaystyle (-1)^{i+j}M_{i,j}}$ is known as a cofactor. For every ${\displaystyle i}$, one has the equality

${\displaystyle \det(A)=\sum _{j=1}^{n}(-1)^{i+j}a_{i,j}M_{i,j},}$

which is called the Laplace expansion along the ith row. For example, the Laplace expansion along the first row (${\displaystyle i=1}$) gives the following formula:

${\displaystyle {\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}=a{\begin{vmatrix}e&f\\h&i\end{vmatrix}}-b{\begin{vmatrix}d&f\\g&i\end{vmatrix}}+c{\begin{vmatrix}d&e\\g&h\end{vmatrix}}}$

Unwinding the determinants of these ${\displaystyle 2\times 2}$-matrices gives back the Leibniz formula mentioned above. Similarly, the Laplace expansion along the ${\displaystyle j}$-th column is the equality

${\displaystyle \det(A)=\sum _{i=1}^{n}(-1)^{i+j}a_{i,j}M_{i,j}.}$

Laplace expansion can be used iteratively for computing determinants, but this approach is inefficient for large matrices. However, it is useful for computing the determinants of highly symmetric matrix such as the Vandermonde matrix

$${\displaystyle {\begin{vmatrix}1&1&1&\cdots &1\\x_{1}&x_{2}&x_{3}&\cdots &x_{n}\\x_{1}^{2}&x_{2}^{2}&x_{3}^{2}&\cdots &x_{n}^{2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\x_{1}^{n-1}&x_{2}^{n-1}&x_{3}^{n-1}&\cdots &x_{n}^{n-1}\end{vmatrix}}=\prod _{1\leq i<j\leq n}\left(x_{j}-x_{i}\right).}$$

The n-term Laplace expansion along a row or column can be generalized to write an n x n determinant as a sum of ${\displaystyle {\tbinom {n}{k}}}$ terms, each the product of the determinant of a k x k submatrix and the determinant of the complementary (n−k) x (n−k) submatrix.

The adjugate matrix ${\displaystyle \operatorname {adj} (A)}$ is the transpose of the matrix of the cofactors, that is,

${\displaystyle (\operatorname {adj} (A))_{i,j}=(-1)^{i+j}M_{ji}.}$

For every matrix, one has[11]

${\displaystyle (\det A)I=A\operatorname {adj} A=(\operatorname {adj} A)\,A.}$

Thus the adjugate matrix can be used for expressing the inverse of a nonsingular matrix:

${\displaystyle A^{-1}={\frac {1}{\det A}}\operatorname {adj} A.}$

The formula for the determinant of a ${\displaystyle 2\times 2}$-matrix above continues to hold, under appropriate further assumptions, for a block matrix, i.e., a matrix composed of four submatrices ${\displaystyle A,B,C,D}$ of dimension ${\displaystyle m\times m}$, ${\displaystyle m\times n}$, ${\displaystyle n\times m}$ and ${\displaystyle n\times n}$, respectively. The easiest such formula, which can be proven using either the Leibniz formula or a factorization involving the Schur complement, is

${\displaystyle \det {\begin{pmatrix}A&0\\C&D\end{pmatrix}}=\det(A)\det(D)=\det {\begin{pmatrix}A&B\\0&D\end{pmatrix}}.}$

If ${\displaystyle A}$ is invertible, then it follows with results from the section on multiplicativity that

${\displaystyle {\begin{aligned}\det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}&=\det(A)\det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}\underbrace {\det {\begin{pmatrix}A^{-1}&-A^{-1}B\\0&I_{n}\end{pmatrix}}} _{=\,\det(A^{-1})\,=\,(\det A)^{-1}}\\&=\det(A)\det {\begin{pmatrix}I_{m}&0\\CA^{-1}&D-CA^{-1}B\end{pmatrix}}\\&=\det(A)\det(D-CA^{-1}B),\end{aligned}}}$

which simplifies to ${\displaystyle \det(A)(D-CA^{-1}B)}$ when ${\displaystyle D}$ is a ${\displaystyle 1\times 1}$-matrix.

A similar result holds when ${\displaystyle D}$ is invertible, namely

${\displaystyle {\begin{aligned}\det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}&=\det(D)\det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}\underbrace {\det {\begin{pmatrix}I_{m}&0\\-D^{-1}C&D^{-1}\end{pmatrix}}} _{=\,\det(D^{-1})\,=\,(\det D)^{-1}}\\&=\det(D)\det {\begin{pmatrix}A-BD^{-1}C&BD^{-1}\\0&I_{n}\end{pmatrix}}\\&=\det(D)\det(A-BD^{-1}C).\end{aligned}}}$

Both results can be combined to derive Sylvester's determinant theorem, which is also stated below.

If the blocks are square matrices of the same size further formulas hold. For example, if ${\displaystyle C}$ and ${\displaystyle D}$ commute (i.e., ${\displaystyle CD=DC}$), then[12]

${\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(AD-BC).}$

This formula has been generalized to matrices composed of more than ${\displaystyle 2\times 2}$ blocks, again under appropriate commutativity conditions among the individual blocks.[13]

For ${\displaystyle A=D}$ and ${\displaystyle B=C}$, the following formula holds (even if ${\displaystyle A}$ and ${\displaystyle B}$ do not commute)

${\displaystyle \det {\begin{pmatrix}A&B\\B&A\end{pmatrix}}=\det(A-B)\det(A+B).}$

Sylvester's determinant theorem states that for A, an m × n matrix, and B, an n × m matrix (so that A and B have dimensions allowing them to be multiplied in either order forming a square matrix):

${\displaystyle \det \left(I_{\mathit {m}}+AB\right)=\det \left(I_{\mathit {n}}+BA\right),}$

where I_m and I_n are the m × m and n × n identity matrices, respectively.

From this general result several consequences follow.

The determinant of the sum ${\displaystyle A+B}$ of two square matrices of the same size is not in general expressible in terms of the determinants of A and of B. However, for positive semidefinite matrices ${\displaystyle A}$, ${\displaystyle B}$ and ${\displaystyle C}$ of equal size,

$${\displaystyle \det(A+B+C)+\det(C)\geq \det(A+C)+\det(B+C){\text{,}}}$$

with the corollary[15][16]

$${\displaystyle \det(A+B)\geq \det(A)+\det(B){\text{.}}}$$

Conversely, if ${\displaystyle A}$ and ${\displaystyle B}$ are Hermitian, positive-definite, and size ${\displaystyle n\times n}$, then the determinant has concave ${\displaystyle n}$^th root;[17] this implies

$${\displaystyle {\sqrt[{n}]{\det {\!(A+B)}}}\geq {\sqrt[{n}]{\det {\!(A)}}}+{\sqrt[{n}]{\det {\!(B)}}}}$$

by homogeneity.

For the special case of ${\displaystyle 2\times 2}$ matrices with complex entries, the determinant of the sum can be written in terms of determinants and traces in the following identity:

${\displaystyle \det(A+B)=\det(A)+\det(B)+{\text{tr}}(A){\text{tr}}(B)-{\text{tr}}(AB).}$

This has an application to ${\displaystyle 2\times 2}$ matrix algebras. For example, consider the complex numbers as a matrix algebra. The complex numbers have a representation as matrices of the form

$${\displaystyle aI+b\mathbf {i}$$ :=a{\begin{pmatrix}1&0\\0&1\end{pmatrix}}+b{\begin{pmatrix}0&-1\\1&0\end{pmatrix}}}

with ${\displaystyle a}$ and ${\displaystyle b}$ real. Since ${\displaystyle {\text{tr}}(\mathbf {i} )=0}$, taking ${\displaystyle A=aI}$ and ${\displaystyle B=b\mathbf {i} }$ in the above identity gives

${\displaystyle \det(aI+b\mathbf {i} )=a^{2}\det(I)+b^{2}\det(\mathbf {i} )=a^{2}+b^{2}.}$

This result followed just from ${\displaystyle {\text{tr}}(\mathbf {i} )=0}$ and ${\displaystyle \det(I)=\det(\mathbf {i} )=1}$.

The determinant is closely related to two other central concepts in linear algebra, the eigenvalues and the characteristic polynomial of a matrix. Let ${\displaystyle A}$ be an ${\displaystyle n\times n}$-matrix with complex entries. Then, by the Fundamental Theorem of Algebra, ${\displaystyle A}$ must have exactly n eigenvalues ${\displaystyle \lambda _{1},\lambda _{2},\ldots ,\lambda _{n}}$. (Here it is understood that an eigenvalue with algebraic multiplicity μ occurs μ times in this list.) Then, it turns out the determinant of A is equal to the product of these eigenvalues,

${\displaystyle \det(A)=\prod _{i=1}^{n}\lambda _{i}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}.}$

The product of all non-zero eigenvalues is referred to as pseudo-determinant.

From this, one immediately sees that the determinant of a matrix ${\displaystyle A}$ is zero iff ${\displaystyle 0}$ is an eigenvalue of ${\displaystyle A}$. In other words, ${\displaystyle A}$ is invertible iff ${\displaystyle 0}$ is not an eigenvalue of ${\displaystyle A}$.

The characteristic polynomial is defined as[18]

${\displaystyle \chi _{A}(t)=\det(t\cdot I-A).}$

Here, ${\displaystyle t}$ is the indeterminate of the polynomial and ${\displaystyle I}$ is the identity matrix of the same size as ${\displaystyle A}$. By means of this polynomial, determinants can be used to find the eigenvalues of the matrix ${\displaystyle A}$: they are precisely the roots of this polynomial, i.e., those complex numbers ${\displaystyle \lambda }$ such that

${\displaystyle \chi _{A}(\lambda )=0.}$

A Hermitian matrix is positive definite if all its eigenvalues are positive. Sylvester's criterion asserts that this is equivalent to the determinants of the submatrices

${\displaystyle A_{k}:={\begin{bmatrix}a_{1,1}&a_{1,2}&\cdots &a_{1,k}\\a_{2,1}&a_{2,2}&\cdots &a_{2,k}\\\vdots &\vdots &\ddots &\vdots \\a_{k,1}&a_{k,2}&\cdots &a_{k,k}\end{bmatrix}}}$

being positive, for all ${\displaystyle k}$ between ${\displaystyle 1}$ and ${\displaystyle n}$.[19]

The trace tr(A) is by definition the sum of the diagonal entries of A and also equals the sum of the eigenvalues. Thus, for complex matrices A,

${\displaystyle \det(\exp(A))=\exp(\operatorname {tr} (A))}$

or, for real matrices A,

${\displaystyle \operatorname {tr} (A)=\log(\det(\exp(A))).}$

Here exp(A) denotes the matrix exponential of A, because every eigenvalue λ of A corresponds to the eigenvalue exp(λ) of exp(A). In particular, given any logarithm of A, that is, any matrix L satisfying

${\displaystyle \exp(L)=A}$

the determinant of A is given by

${\displaystyle \det(A)=\exp(\operatorname {tr} (L)).}$

For example, for n = 2, n = 3, and n = 4, respectively,

${\displaystyle {\begin{aligned}\det(A)&={\frac {1}{2}}\left(\left(\operatorname {tr} (A)\right)^{2}-\operatorname {tr} \left(A^{2}\right)\right),\\\det(A)&={\frac {1}{6}}\left(\left(\operatorname {tr} (A)\right)^{3}-3\operatorname {tr} (A)~\operatorname {tr} \left(A^{2}\right)+2\operatorname {tr} \left(A^{3}\right)\right),\\\det(A)&={\frac {1}{24}}\left(\left(\operatorname {tr} (A)\right)^{4}-6\operatorname {tr} \left(A^{2}\right)\left(\operatorname {tr} (A)\right)^{2}+3\left(\operatorname {tr} \left(A^{2}\right)\right)^{2}+8\operatorname {tr} \left(A^{3}\right)~\operatorname {tr} (A)-6\operatorname {tr} \left(A^{4}\right)\right).\end{aligned}}}$

cf. Cayley-Hamilton theorem. Such expressions are deducible from combinatorial arguments, Newton's identities, or the Faddeev–LeVerrier algorithm. That is, for generic n, detA = (−1)ⁿc₀ the signed constant term of the characteristic polynomial, determined recursively from

${\displaystyle c_{n}=1;~~~c_{n-m}=-{\frac {1}{m}}\sum _{k=1}^{m}c_{n-m+k}\operatorname {tr} \left(A^{k}\right)~~(1\leq m\leq n)~.}$

In the general case, this may also be obtained from[20]

${\displaystyle \det(A)=\sum _{\begin{array}{c}k_{1},k_{2},\ldots ,k_{n}\geq 0\\k_{1}+2k_{2}+\cdots +nk_{n}=n\end{array}}\prod _{l=1}^{n}{\frac {(-1)^{k_{l}+1}}{l^{k_{l}}k_{l}!}}\operatorname {tr} \left(A^{l}\right)^{k_{l}},}$

where the sum is taken over the set of all integers k_l ≥ 0 satisfying the equation

${\displaystyle \sum _{l=1}^{n}lk_{l}=n.}$

The formula can be expressed in terms of the complete exponential Bell polynomial of n arguments s_l = −(l – 1)! tr(A^l) as

${\displaystyle \det(A)={\frac {(-1)^{n}}{n!}}B_{n}(s_{1},s_{2},\ldots ,s_{n}).}$

This formula can also be used to find the determinant of a matrix A^I_J with multidimensional indices I = (i₁, i₂, ..., i_r) and J = (j₁, j₂, ..., j_r). The product and trace of such matrices are defined in a natural way as

${\displaystyle (AB)_{J}^{I}=\sum _{K}A_{K}^{I}B_{J}^{K},\operatorname {tr} (A)=\sum _{I}A_{I}^{I}.}$

An important arbitrary dimension n identity can be obtained from the Mercator series expansion of the logarithm when the expansion converges. If every eigenvalue of A is less than 1 in absolute value,

${\displaystyle \det(I+A)=\sum _{k=0}^{\infty }{\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}}{j}}\operatorname {tr} \left(A^{j}\right)\right)^{k}\,,}$

where I is the identity matrix. More generally, if

${\displaystyle \sum _{k=0}^{\infty }{\frac {1}{k!}}\left(-\sum _{j=1}^{\infty }{\frac {(-1)^{j}s^{j}}{j}}\operatorname {tr} \left(A^{j}\right)\right)^{k}\,,}$

is expanded as a formal power series in s then all coefficients of s^m for m > n are zero and the remaining polynomial is det(I + sA).

For a positive definite matrix A, the trace operator gives the following tight lower and upper bounds on the log determinant

${\displaystyle \operatorname {tr} \left(I-A^{-1}\right)\leq \log \det(A)\leq \operatorname {tr} (A-I)}$

with equality if and only if A = I. This relationship can be derived via the formula for the Kullback-Leibler divergence between two multivariate normal distributions.

Also,

${\displaystyle {\frac {n}{\operatorname {tr} \left(A^{-1}\right)}}\leq \det(A)^{\frac {1}{n}}\leq {\frac {1}{n}}\operatorname {tr} (A)\leq {\sqrt {{\frac {1}{n}}\operatorname {tr} \left(A^{2}\right)}}.}$

These inequalities can be proved by expressing the traces and the determinant in terms of the eigenvalues. As such, they represent the well-known fact that the harmonic mean is less than the geometric mean, which is less than the arithmetic mean, which is, in turn, less than the root mean square.

The Leibniz formula shows that the determinant of real (or analogously for complex) square matrices is a polynomial function from ${\displaystyle \mathbf {R} ^{n\times n}}$ to ${\displaystyle \mathbf {R} }$. In particular, it is everywhere differentiable. Its derivative can be expressed using Jacobi's formula:[21]

${\displaystyle {\frac {d\det(A)}{d\alpha }}=\operatorname {tr} \left(\operatorname {adj} (A){\frac {dA}{d\alpha }}\right).}$

where ${\displaystyle \operatorname {adj} (A)}$ denotes the adjugate of ${\displaystyle A}$. In particular, if ${\displaystyle A}$ is invertible, we have

${\displaystyle {\frac {d\det(A)}{d\alpha }}=\det(A)\operatorname {tr} \left(A^{-1}{\frac {dA}{d\alpha }}\right).}$

Expressed in terms of the entries of ${\displaystyle A}$, these are

${\displaystyle {\frac {\partial \det(A)}{\partial A_{ij}}}=\operatorname {adj} (A)_{ji}=\det(A)\left(A^{-1}\right)_{ji}.}$

Yet another equivalent formulation is

${\displaystyle \det(A+\epsilon X)-\det(A)=\operatorname {tr} (\operatorname {adj} (A)X)\epsilon +O\left(\epsilon ^{2}\right)=\det(A)\operatorname {tr} \left(A^{-1}X\right)\epsilon +O\left(\epsilon ^{2}\right)}$,

using big O notation. The special case where ${\displaystyle A=I}$, the identity matrix, yields

${\displaystyle \det(I+\epsilon X)=1+\operatorname {tr} (X)\epsilon +O\left(\epsilon ^{2}\right).}$

This identity is used in describing Lie algebras associated to certain matrix Lie groups. For example, the special linear group ${\displaystyle \operatorname {SL} _{n}}$ is defined by the equation ${\displaystyle \det A=1}$. The above formula shows that its Lie algebra is the special linear Lie algebra ${\displaystyle {\mathfrak {sl}}_{n}}$ consisting of those matrices having trace zero.

Writing a ${\displaystyle 3\times 3}$-matrix as ${\displaystyle A={\begin{bmatrix}a&b&c\end{bmatrix}}}$ where ${\displaystyle a,b,c}$ are column vectors of length 3, then the gradient over one of the three vectors may be written as the cross product of the other two:

${\displaystyle {\begin{aligned}\nabla _{\mathbf {a} }\det(A)&=\mathbf {b} \times \mathbf {c} \\\nabla _{\mathbf {b} }\det(A)&=\mathbf {c} \times \mathbf {a} \\\nabla _{\mathbf {c} }\det(A)&=\mathbf {a} \times \mathbf {b} .\end{aligned}}}$

Determinants can be used to describe the solutions of a linear system of equations, written in matrix form as ${\displaystyle Ax=b}$. This equation has a unique solution ${\displaystyle x}$ if and only if ${\displaystyle \det(A)}$ is nonzero. In this case, the solution is given by Cramer's rule:

${\displaystyle x_{i}={\frac {\det(A_{i})}{\det(A)}}\qquad i=1,2,3,\ldots ,n}$

where ${\displaystyle A_{i}}$ is the matrix formed by replacing the ${\displaystyle i}$-th column of ${\displaystyle A}$ by the column vector ${\displaystyle b}$. This follows immediately by column expansion of the determinant, i.e.

${\displaystyle \det(A_{i})=\det {\begin{bmatrix}a_{1}&\ldots &b&\ldots &a_{n}\end{bmatrix}}=\sum _{j=1}^{n}x_{j}\det {\begin{bmatrix}a_{1}&\ldots &a_{i-1}&a_{j}&a_{i+1}&\ldots &a_{n}\end{bmatrix}}=x_{i}\det(A)}$

where the vectors ${\displaystyle a_{j}}$ are the columns of A. The rule is also implied by the identity

${\displaystyle A\,\operatorname {adj} (A)=\operatorname {adj} (A)\,A=\det(A)\,I_{n}.}$

Cramer's rule can be implemented in ${\displaystyle \operatorname {O} (n^{3})}$ time, which is comparable to more common methods of solving systems of linear equations, such as LU, QR, or singular value decomposition.[37]

Determinants can be used to characterize linearly dependent vectors: ${\displaystyle \det A}$ is zero if and only if the column vectors (or, equivalently, the row vectors) of the matrix ${\displaystyle A}$ are linearly dependent.[38] For example, given two linearly independent vectors ${\displaystyle v_{1},v_{2}\in \mathbf {R} ^{3}}$, a third vector ${\displaystyle v_{3}}$ lies in the plane spanned by the former two vectors exactly if the determinant of the ${\displaystyle 3\times 3}$-matrix consisting of the three vectors is zero. The same idea is also used in the theory of differential equations: given functions ${\displaystyle f_{1}(x),\dots ,f_{n}(x)}$ (supposed to be ${\displaystyle n-1}$ times differentiable), the Wronskian is defined to be

${\displaystyle W(f_{1},\ldots ,f_{n})(x)={\begin{vmatrix}f_{1}(x)&f_{2}(x)&\cdots &f_{n}(x)\\f_{1}'(x)&f_{2}'(x)&\cdots &f_{n}'(x)\\\vdots &\vdots &\ddots &\vdots \\f_{1}^{(n-1)}(x)&f_{2}^{(n-1)}(x)&\cdots &f_{n}^{(n-1)}(x)\end{vmatrix}}.}$