Determinant

The area of the parallelogram is the absolute value of the determinant of the matrix formed by the vectors representing the parallelogram's sides.

The Leibniz formula for the determinant of a 2 × 2 matrix is

${\displaystyle {\begin{vmatrix}a&b\\c&d\end{vmatrix}}=ad-bc.}$

If the matrix entries are real numbers, the matrix A can be used to represent two linear maps: one that maps the standard basis vectors to the rows of A, and one that maps them to the columns of A. In either case, the images of the basis vectors form a parallelogram that represents the image of the unit square under the mapping. The parallelogram defined by the rows of the above matrix is the one with vertices at (0, 0), (a, b), (a + c, b + d), and (c, d), as shown in the accompanying diagram.

The absolute value of ad − bc is the area of the parallelogram, and thus represents the scale factor by which areas are transformed by A. (The parallelogram formed by the columns of A is in general a different parallelogram, but since the determinant is symmetric with respect to rows and columns, the area will be the same.)

The absolute value of the determinant together with the sign becomes the oriented area of the parallelogram. The oriented area is the same as the usual area, except that it is negative when the angle from the first to the second vector defining the parallelogram turns in a clockwise direction (which is opposite to the direction one would get for the identity matrix).

To show that ad − bc is the signed area, one may consider a matrix containing two vectors u ≡ (a, b) and v ≡ (c, d) representing the parallelogram's sides. The signed area can be expressed as |u| |v| sin θ for the angle θ between the vectors, which is simply base times height, the length of one vector times the perpendicular component of the other. Due to the sine this already is the signed area, yet it may be expressed more conveniently using the cosine of the complementary angle to a perpendicular vector, e.g. u^⊥ = (−b, a), so that |u^⊥| |v| cos θ′, which can be determined by the pattern of the scalar product to be equal to ad − bc:

${\displaystyle {\text{Signed area}}=|{\boldsymbol {u}}|\,|{\boldsymbol {v}}|\,\sin \,\theta =\left|{\boldsymbol {u}}^{\perp }\right|\,\left|{\boldsymbol {v}}\right|\,\cos \,\theta '={\begin{pmatrix}-b\\a\end{pmatrix}}\cdot {\begin{pmatrix}c\\d\end{pmatrix}}=ad-bc.}$

Thus the determinant gives the scaling factor and the orientation induced by the mapping represented by A. When the determinant is equal to one, the linear mapping defined by the matrix is equi-areal and orientation-preserving.

The object known as the bivector is related to these ideas. In 2D, it can be interpreted as an oriented plane segment formed by imagining two vectors each with origin (0, 0), and coordinates (a, b) and (c, d). The bivector magnitude (denoted by (a, b) ∧ (c, d)) is the signed area, which is also the determinant ad − bc.[3]

The Laplace formula for the determinant of a 3 × 3 matrix is

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

this can be expanded out to give the Leibniz formula.

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

${\displaystyle {\begin{aligned}{\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}}&=a(ei-fh)-b(di-fg)+c(dh-eg)\\&=aei+bfg+cdh-ceg-bdi-afh.\end{aligned}}}$

The rule of Sarrus is a mnemonic for the 3 × 3 matrix 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:

${\displaystyle {\begin{aligned}~~~~~~{\begin{vmatrix}a&b&c\\e&f&g\\h&i&j\end{vmatrix}}=\end{aligned}}}$ ${\displaystyle \qquad =\color {red}{afj+bgh+cei}\color {blue}{-hfc-iga-jeb}}$

This scheme for calculating the determinant of a 3 × 3 matrix does not carry over into higher dimensions.

The determinant of a matrix of arbitrary size can be defined by the Leibniz formula or the Laplace formula.

The Leibniz formula for the determinant of an n × n matrix A is

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

Here the sum is computed over all permutations σ of the set {1, 2, ..., n}. A permutation is a function that reorders this set of integers. The value in the ith position after the reordering σ is denoted by σ_i. For example, for n = 3, the original sequence 1, 2, 3 might be reordered to σ = [2, 3, 1], with σ₁ = 2, σ₂ = 3, and σ₃ = 1. The set of all such permutations (also known as the symmetric group on n elements) is denoted by S_n. For each permutation σ, sgn(σ) denotes the signature of σ, a value that is +1 whenever the reordering given by σ can be achieved by successively interchanging two entries an even number of times, and −1 whenever it can be achieved by an odd number of such interchanges.

In any of the ${\displaystyle n!}$ summands, the term

${\displaystyle \prod _{i=1}^{n}a_{i,\sigma _{i}}}$

is notation for the product of the entries at positions (i, σ_i), where i ranges from 1 to n:

${\displaystyle a_{1,\sigma _{1}}\cdot a_{2,\sigma _{2}}\cdots a_{n,\sigma _{n}}.}$

For example, the determinant of a 3 × 3 matrix A (n = 3) is

${\displaystyle {\begin{aligned}&\sum _{\sigma \in S_{n}}\operatorname {sgn}(\sigma )\prod _{i=1}^{n}a_{i,\sigma _{i}}\\={}&\operatorname {sgn}([1,2,3])\prod _{i=1}^{n}a_{i,[1,2,3]_{i}}+\operatorname {sgn}([1,3,2])\prod _{i=1}^{n}a_{i,[1,3,2]_{i}}+\operatorname {sgn}([2,1,3])\prod _{i=1}^{n}a_{i,[2,1,3]_{i}}+{}\\&\operatorname {sgn}([2,3,1])\prod _{i=1}^{n}a_{i,[2,3,1]_{i}}+\operatorname {sgn}([3,1,2])\prod _{i=1}^{n}a_{i,[3,1,2]_{i}}+\operatorname {sgn}([3,2,1])\prod _{i=1}^{n}a_{i,[3,2,1]_{i}}\\={}&\prod _{i=1}^{n}a_{i,[1,2,3]_{i}}-\prod _{i=1}^{n}a_{i,[1,3,2]_{i}}-\prod _{i=1}^{n}a_{i,[2,1,3]_{i}}+\prod _{i=1}^{n}a_{i,[2,3,1]_{i}}+\prod _{i=1}^{n}a_{i,[3,1,2]_{i}}-\prod _{i=1}^{n}a_{i,[3,2,1]_{i}}\\[2pt]={}&a_{1,1}a_{2,2}a_{3,3}-a_{1,1}a_{2,3}a_{3,2}-a_{1,2}a_{2,1}a_{3,3}+a_{1,2}a_{2,3}a_{3,1}+a_{1,3}a_{2,1}a_{3,2}-a_{1,3}a_{2,2}a_{3,1}.\end{aligned}}}$

It is sometimes useful to extend the Leibniz formula to a summation in which not only permutations, but all sequences of n indices in the range 1, ..., n occur, ensuring that the contribution of a sequence will be zero unless it denotes a permutation. Thus the totally antisymmetric Levi-Civita symbol ${\displaystyle \varepsilon _{i_{1},\cdots ,i_{n}}}$ extends the signature of a permutation, by setting ${\displaystyle \varepsilon _{\sigma (1),\cdots ,\sigma (n)}=\operatorname {sgn} (\sigma )}$ for any permutation σ of n, and ${\displaystyle \varepsilon _{i_{1},\cdots ,i_{n}}=0}$ when no permutation σ exists such that ${\displaystyle \sigma (j)=i_{j}}$ for ${\displaystyle j=1,\ldots ,n}$ (or equivalently, whenever some pair of indices are equal). The determinant for an n × n matrix can then be expressed using an n-fold summation as

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

or using two epsilon symbols as

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

where now each i_r and each j_r should be summed over 1, ..., n.

However, through the use of tensor notation and the suppression of the summation symbol (Einstein's summation convention) we can obtain a much more compact expression of the determinant of the second order system of ${\displaystyle n=3}$ dimensions, ${\displaystyle a_{n}^{m}}$;

${\displaystyle \det(a_{n}^{m})e_{rst}=e_{ijk}a_{r}^{i}a_{s}^{j}a_{t}^{k}}$

where ${\displaystyle e_{rst}}$ and ${\displaystyle e_{ijk}}$ represent 'e-systems' that take on the values 0, +1 and −1 given the number of permutations of ${\displaystyle ijk}$ and ${\displaystyle rst}$. More specifically, ${\displaystyle e_{ijk}}$ is equal to 0 when there is a repeated index in ${\displaystyle ijk}$; +1 when an even number of permutations of ${\displaystyle ijk}$ is present; −1 when an odd number of permutations of ${\displaystyle ijk}$ is present. The number of indices present in the e-systems is equal to ${\displaystyle n}$ and thus can be generalized in this manner.[4]

The following identity holds for a Schur complement of a square matrix:

The Schur complement arises as the result of performing a block Gaussian elimination by multiplying the matrix M from the right with a block lower triangular matrix

${\displaystyle L={\begin{bmatrix}I_{p}&0\\-D^{-1}C&I_{q}\end{bmatrix}}.}$

Here I_p denotes the p×p identity matrix. After multiplication with the matrix L, the Schur complement appears in the upper p×p block. The product matrix is

${\displaystyle {\begin{aligned}ML&={\begin{bmatrix}A&B\\C&D\end{bmatrix}}{\begin{bmatrix}I_{p}&0\\-D^{-1}C&I_{q}\end{bmatrix}}={\begin{bmatrix}A-BD^{-1}C&B\\0&D\end{bmatrix}}\\[5pt]&={\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}}.\end{aligned}}}$

That is, we have effected a Gaussian decomposition

${\displaystyle {\begin{bmatrix}A&B\\C&D\end{bmatrix}}={\begin{bmatrix}I_{p}&BD^{-1}\\0&I_{q}\end{bmatrix}}{\begin{bmatrix}A-BD^{-1}C&0\\0&D\end{bmatrix}}{\begin{bmatrix}I_{p}&0\\D^{-1}C&I_{q}\end{bmatrix}}.}$

The first and last matrices on the RHS have determinant unity, so we have

${\displaystyle \det {\begin{bmatrix}A&B\\C&D\end{bmatrix}}=\det(D)\cdot \det \left(A-BD^{-1}C\right).}$

This is Schur's determinant identity.

The determinant of a matrix product of square matrices equals the product of their determinants:

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

Thus the determinant is a multiplicative map. This property is a consequence of the characterization given above of the determinant as the unique n-linear alternating function of the columns with value 1 on the identity matrix, since the function M_n(K) → K that maps M ↦ det(AM) can easily be seen to be n-linear and alternating in the columns of M, and takes the value det(A) at the identity. The formula can be generalized to (square) products of rectangular matrices, giving the Cauchy–Binet formula, which also provides an independent proof of the multiplicative property.

The determinant det(A) of a matrix A is non-zero if and only if A is invertible or, yet another equivalent statement, if its rank equals the size of the matrix. If so, 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 determinant one still have this property. Thus, the set of such matrices (of fixed size n) form a group known as the special linear group. 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.

Laplace expansion expresses the determinant of a matrix in terms of its minors. The minor M_i,j is defined to be the determinant of the (n−1) × (n−1)-matrix that results from A by removing the ith row and the jth column. The expression (−1)^i+j M_i,j is known as a cofactor. For every i, one has the equality

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

which is called the Laplace expansion along the ith row. Similarly, the Laplace expansion along the jth column is the equality

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

For example, the Laplace expansion of the 3 × 3 matrix

${\displaystyle A={\begin{bmatrix}-2&2&-3\\-1&1&3\\2&0&-1\end{bmatrix}},}$

along the second column (j = 2 and the sum runs over i) is given by,

${\displaystyle {\begin{aligned}\det(A)&=(-1)^{1+2}\cdot 2\cdot {\begin{vmatrix}-1&3\\2&-1\end{vmatrix}}+(-1)^{2+2}\cdot 1\cdot {\begin{vmatrix}-2&-3\\[4pt]2&-1\end{vmatrix}}+(-1)^{3+2}\cdot 0\cdot {\begin{vmatrix}-2&-3\\-1&3\end{vmatrix}}\\[4pt]&=(-2)\cdot ((-1)\cdot (-1)-2\cdot 3)+1\cdot ((-2)\cdot (-1)-2\cdot (-3))\\[4pt]&=(-2)\cdot (-5)+8=18.\end{aligned}}}$

Laplace expansion can be used iteratively for computing determinants, but this is efficient for small matrices and sparse matrices only, since for general matrices this requires to compute an exponential number of determinants, even if care is taken to compute each minor only once. The adjugate matrix adj(A) is the transpose of the matrix of the cofactors, that is,

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

For every matrix, one has[8]

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

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

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

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.

Let A be an arbitrary n×n matrix of complex numbers with 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 the determinant of A is the product of all 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.

Conversely, determinants can be used to find the eigenvalues of the matrix A: they are the solutions of the characteristic equation

${\displaystyle \det(A-\lambda I)=0~,}$

where I is the identity matrix of the same dimension as A and λ is a (scalar) number which solves the equation (there are no more than n solutions, where n is the dimension of A).

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

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

being positive, for all k between 1 and n.

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[10]

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

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 KL-divergence between two multivariate normal distributions.

Also,

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

These inequalities can be proved by bringing the matrix A to the diagonal form. 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.

For a matrix equation

${\displaystyle Ax=b}$, given that A has a nonzero determinant,

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 A_i is the matrix formed by replacing the ith column of A by the column vector 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

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

It has recently been shown that Cramer's rule can be implemented in O(n³) time,[11] which is comparable to more common methods of solving systems of linear equations, such as LU, QR, or singular value decomposition.

Suppose A, B, C, and D are matrices of dimension n × n, n × m, m × n, and m × m, respectively. Then

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

This can be seen from the Leibniz formula for determinants, or from a decomposition like (for the former case)

${\displaystyle {\begin{pmatrix}A&0\\C&D\end{pmatrix}}={\begin{pmatrix}A&0\\C&I_{m}\end{pmatrix}}{\begin{pmatrix}I_{n}&0\\0&D\end{pmatrix}}.}$

When A is invertible, one has

${\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(A)\times \det \left(D-CA^{-1}B\right).}$

as can be seen by employing the decomposition

${\displaystyle {\begin{pmatrix}A&B\\C&D\end{pmatrix}}={\begin{pmatrix}A&0\\C&I_{m}\end{pmatrix}}{\begin{pmatrix}I_{n}&A^{-1}B\\0&D-CA^{-1}B\end{pmatrix}}.}$

When D is invertible, a similar identity with ${\displaystyle \det(D)}$ factored out can be derived analogously,[12] that is,

${\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\det(D)\times \det \left(A-BD^{-1}C\right).}$

When the blocks are square matrices of the same order further formulas hold. For example, if C and D commute (i.e., CD = DC), then the following formula comparable to the determinant of a 2 × 2 matrix holds:[13]

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

Generally, if all pairs of n × n matrices of the np × np block matrix commute, then the determinant of the block matrix is equal to the determinant of the matrix obtained by computing the determinant of the block matrix considering its entries as the entries of a p × p matrix.[14] As the previous formula shows, for p = 2, this criterion is sufficient, but not necessary.

When A = D and B = C, the blocks are square matrices of the same order and the following formula holds (even if A and B do not commute)

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

When D is a 1×1 matrix, B is a column vector, and C is a row vector then

${\displaystyle \det {\begin{pmatrix}A&B\\C&D\end{pmatrix}}=\left(D-CA^{-1}B\right)\det(A)\,.}$

Let ${\displaystyle s}$ be a scalar complex number. If a block matrix is square, its characteristic polynomial can be factored with

${\displaystyle {\begin{aligned}\det {\begin{pmatrix}A-sI&B\\C&D-sI\end{pmatrix}}&=\det(A-sI)\det \left(D-{\frac {C\operatorname {adj} (A-sI)B}{\det(A-sI)}}-sI\right)\quad {\text{if}}\quad s\notin \operatorname {eig} (A)\\&=\det(A-sI)^{1-n-m}\det \left(\det(A-sI)D-C\operatorname {adj} (A-sI)B-\det(A-sI)sI\right)\end{aligned}}}$

It can be seen, e.g. using the Leibniz formula, that the determinant of real (or analogously for complex) square matrices is a polynomial function from R^n×n to R, and so it is everywhere differentiable. Its derivative can be expressed using Jacobi's formula:[15]

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

where adj(A) denotes the adjugate of A. In particular, if 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 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 the tangent space of certain matrix Lie groups.

If the matrix A is written as ${\displaystyle A={\begin{bmatrix}\mathbf {a} &\mathbf {b} &\mathbf {c} \end{bmatrix}}}$ where 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:

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

The above identities concerning the determinant of products and inverses of matrices imply that similar matrices have the same determinant: two matrices A and B are similar, if there exists an invertible matrix X such that A = X⁻¹BX. Indeed, repeatedly applying the above identities yields

${\displaystyle \det(A)=\det(X)^{-1}\det(B)\det(X)=\det(B)\det(X)^{-1}\det(X)=\det(B).}$

The determinant is therefore also called a similarity invariant. The determinant of a linear transformation

${\displaystyle T:V\rightarrow V}$

for some finite-dimensional vector space V is defined to be the determinant of the matrix describing it, with respect to an arbitrary choice of basis in V. By the similarity invariance, this determinant is independent of the choice of the basis for V and therefore only depends on the endomorphism T.

The determinant of a linear transformation T : V → V of an n-dimensional vector space V can be formulated in a coordinate-free manner by considering the nth exterior power ΛⁿV of V. T induces a linear map

${\displaystyle {\begin{aligned}\Lambda ^{n}T:\Lambda ^{n}V&\rightarrow \Lambda ^{n}V\\v_{1}\wedge v_{2}\wedge \dots \wedge v_{n}&\mapsto Tv_{1}\wedge Tv_{2}\wedge \dots \wedge Tv_{n}.\end{aligned}}}$

As ΛⁿV is one-dimensional, the map ΛⁿT is given by multiplying with some scalar. This scalar coincides with the determinant of T, that is to say

${\displaystyle \left(\Lambda ^{n}T\right)\left(v_{1}\wedge \dots \wedge v_{n}\right)=\det(T)\cdot v_{1}\wedge \dots \wedge v_{n}.}$

This definition agrees with the more concrete coordinate-dependent definition. In particular, for a square ${\displaystyle n\times n}$ matrix A whose columns are ${\displaystyle \mathbf {a} _{1},\ldots ,\mathbf {a} _{n}}$, its determinant satisfies ${\displaystyle \mathbf {a} _{1}\wedge \dots \wedge \mathbf {a} _{n}=\det(A)\cdot \mathbf {e} _{1}\wedge \dots \wedge \mathbf {e} _{n}}$, where ${\displaystyle \{\mathbf {e} _{1},\dots ,\mathbf {e} _{n}\}}$ is the standard basis of ${\displaystyle \mathbb {R} ^{n}}$. This follows from the characterization of the determinant given above. For example, switching two columns changes the sign of the determinant; likewise, permuting the vectors in the exterior product v₁ ∧ v₂ ∧ v₃ ∧ ... ∧ v_n to v₂ ∧ v₁ ∧ v₃ ∧ ... ∧ v_n, say, also changes its sign.

For this reason, the highest non-zero exterior power Λⁿ(V) is sometimes also called the determinant of V and similarly for more involved objects such as vector bundles or chain complexes of vector spaces. Minors of a matrix can also be cast in this setting, by considering lower alternating forms Λ^kV with k < n.

The determinant can also be characterized as the unique function

${\displaystyle D:M_{n}(K)\to K}$

from the set of all n × n matrices with entries in a field K to that field satisfying the following three properties: first, D is an n-linear function: considering all but one column of A fixed, the determinant is linear in the remaining column, that is

${\displaystyle D(v_{1},\dots ,v_{i-1},av_{i}+bw,v_{i+1},\dots ,v_{n})=aD(v_{1},\dots ,v_{i-1},v_{i},v_{i+1},\dots ,v_{n})+bD(v_{1},\dots ,v_{i-1},w,v_{i+1},\dots ,v_{n})}$

for any column vectors v₁, ..., v_n, and w and any scalars (elements of K) a and b. Second, D is an alternating function: for any matrix A with two identical columns, D(A) = 0. Finally, D(I_n) = 1, where I_n is the identity matrix.

This fact also implies that every other n-linear alternating function F: M_n(K) → K satisfies

${\displaystyle F(M)=F(I)D(M).}$

This definition can also be extended where K is a commutative ring R, in which case a matrix is invertible if and only if its determinant is an invertible element in R. For example, a matrix A with entries in Z, the integers, is invertible (in the sense that there exists an inverse matrix with integer entries) if the determinant is +1 or −1. Such a matrix is called unimodular.

The determinant defines a mapping

${\displaystyle \operatorname {GL} _{n}(R)\rightarrow R^{\times },}$

between the group of invertible n × n matrices with entries in R and the multiplicative group of units in R. Since it respects the multiplication in both groups, this map is a group homomorphism. Secondly, given a ring homomorphism f: R → S, there is a map GL_n(f): GL_n(R) → GL_n(S) given by replacing all entries in R by their images under f. The determinant respects these maps, i.e., given a matrix A = (a_i,j) with entries in R, the identity

${\displaystyle f(\det((a_{i,j})))=\det((f(a_{i,j})))}$