# Determinant of an Integer Matrix

Given a square integer matrix as input, output the determinant of the matrix.

## Rules

• You may assume that all elements in the matrix, the determinant of the matrix, and the total number of elements in the matrix are within the representable range of integers for your language.
• Outputting a decimal/float value with a fractional part of 0 is allowed (e.g. 42.0 instead of 42).
• Builtins are allowed, but you are encouraged to include a solution that does not use builtins.

## Test Cases

[[42]] -> 42
[[2, 3], [1, 4]] -> 5
[[1, 2, 3], [4, 5, 6], [7, 8, 9]] -> 0
[[13, 17, 24], [19, 1, 3], [-5, 4, 0]] -> 1533
[[372, -152, 244], [-97, -191, 185], [-53, -397, -126]] -> 46548380
[[100, -200, 58, 4], [1, -90, -55, -165], [-67, -83, 239, 182], [238, -283, 384, 392]] -> 571026450
[[432, 45, 330, 284, 276], [-492, 497, 133, -289, -28], [-443, -400, 56, 150, -316], [-344, 316, 92, 205, 104], [277, 307, -464, 244, -422]] -> -51446016699154
[[416, 66, 340, 250, -436, -146], [-464, 68, 104, 471, -335, -442], [159, -407, 310, -489, -248, 370], [62, 277, 446, -325, 47, -193], [460, 460, -418, -28, 234, -374], [249, 375, 489, 172, -423, 125]] -> 39153009069988024
[[-246, -142, 378, -156, -373, 444], [186, 186, -23, 50, 349, -413], [216, 1, -418, 38, 47, -192], [109, 345, -356, -296, -47, -498], [-283, 91, 258, 66, -127, 79], [218, 465, -420, -326, -445, 19]] -> -925012040475554
[[-192, 141, -349, 447, -403, -21, 34], [260, -307, -333, -373, -324, 144, -190], [301, 277, 25, 8, -177, 180, 405], [-406, -9, -318, 337, -118, 44, -123], [-207, 33, -189, -229, -196, 58, -491], [-426, 48, -24, 72, -250, 160, 359], [-208, 120, -385, 251, 322, -349, -448]] -> -4248003140052269106
[[80, 159, 362, -30, -24, -493, 410, 249, -11, -109], [-110, -123, -461, -34, -266, 199, -437, 445, 498, 96], [175, -405, 432, -7, 157, 169, 336, -276, 337, -200], [-106, -379, -157, -199, 123, -172, 141, 329, 158, 309], [-316, -239, 327, -29, -482, 294, -86, -326, 490, -295], [64, -201, -155, 238, 131, 182, -487, -462, -312, 196], [-297, -75, -206, 471, -94, -46, -378, 334, 407, -97], [-140, -137, 297, -372, 228, 318, 251, -93, 117, 286], [-95, -300, -419, 41, -140, -205, 29, -481, -372, -49], [-140, -281, -88, -13, -128, -264, 165, 261, -469, -62]] -> 297434936630444226910432057

• – user45941
Nov 10, 2017 at 13:50
• is there a maximum size of matrix that needs to be supported or is it arbitrary? Nov 10, 2017 at 14:59
• @TaylorScott First rule listed: You may assume that all elements in the matrix, the determinant of the matrix, and the total number of elements in the matrix are within the representable range of integers for your language.
– user45941
Nov 10, 2017 at 15:00
• You know you have an interesting challenge when you have 4 Jelly answers consecutively out-golfing each other... Nov 10, 2017 at 18:21

# Jelly, 15 bytes

LŒ!ðŒcIṠ;ị"Pð€S


Try it online!

## How it works

LŒ!ðŒcIṠ;ị"Pð€S   input
L                 length
Œ!               all_permutations
ð        ð€    for each permutation:
Œc                take all unordered pairs
I               calculate the difference between
the two integers of each pair
Ṡ              signum of each difference
(positive -> 1, negative -> -1)
;             append:
ị"             the list of elements generated by taking
each row according to the index specified
by each entry of the permutation
P          product of everything
S   sum


## Why it works -- mathy version

The operator det takes a matrix and returns a scalar. An n-by-n matrix can be thought of as a collection of n vectors of length n, so det is really a function that takes n vectors from ℤn and returns a scalar.

Therefore, I write det(v1, v2, v3, ..., vn) for det [v1 v2 v3 ... vn].

Notice that det is linear in each argument, i.e. det(v1 + λw1, v2, v3, ..., vn) = det(v1, v2, v3, ..., vn) + λ det(w1, v2, v3, ..., vn). Therefore, it is a linear map from (ℤn)n to ℤ.

It suffices to determine the image of the basis under the linear map. The basis of (ℤn)n consists of n-fold tensor products of the basis elements of ℤn, i.e. e.g. e5 ⊗ e3 ⊗ e1 ⊗ e5 ⊗ e1. Tensor products that include identical tensors must be sent to zero, since the determinant of a matrix in which two columns are identical is zero. It remains to check what the tensor products of distinct basis elements are sent to. The indices of the vectors in the tensor product form a bijection, i.e. a permutation, in which even permutations are sent to 1 and odd permutations are sent to -1.

For example, to find the determinant of [[1, 2], [3, 4]]: note that the columns are [1, 3] and [2, 4]. We decompose [1, 3] to give (1 e1 + 3 e2) and (2 e1 + 4 e2). The corresponding element in the tensor product is (1 e1 ⊗ 2 e1 + 1 e1 ⊗ 4 e2 + 3 e2 ⊗ 2 e1 + 3 e2 ⊗ 4 e2), which we simplify to (2 e1 ⊗ e1 + 4 e1 ⊗ e2 + 6 e2 ⊗ e1 + 12 e2 ⊗ e2). Therefore:

det [[1, 2], [3, 4]]
= det(1 e1 + 3 e2, 2 e1 + 4 e2)
= det(2 e1 ⊗ e1 + 4 e1 ⊗ e2 + 6 e2 ⊗ e1 + 12 e2 ⊗ e2)
= det(2 e1 ⊗ e1) + det(4 e1 ⊗ e2) + det(6 e2 ⊗ e1) + det(12 e2 ⊗ e2)
= 2 det(e1 ⊗ e1) + 4 det(e1 ⊗ e2) + 6 det(e2 ⊗ e1) + 12 det( e2 ⊗ e2)
= 2 (0) + 4 (1) + 6 (-1) + 12 (0)
= 4 - 6
= -2

Now it remains to prove that the formula for finding the parity of the permutation is valid. What my code does is essentially find the number of inversions, i.e. the places where an element on the left is bigger than an element on the right (not necessarily consecutively).

For example, in the permutation 3614572, there are 9 inversions (31, 32, 61, 64, 65, 62, 42, 52, 72), so the permutation is odd.

The justification is that each transposition (swapping two elements) either adds one inversion or takes away one inversion, swapping the parity of the number of inversions, and the parity of the permutation is the parity of the number of transpositions needed to achieve the permutation.

Therefore, in conclusion, our formula is given by:

## Why it works -- non-mathy version

where σ is a permutation of 𝕊n the group of all permutations on n letters, and sgn is the sign of the permutation, AKA (-1) raised to the parity of the permutation, and aij is the (ij)th entry in the matrix (i down, j across).

• That "non-mathy version" is pretty damn mathy. Nov 10, 2017 at 19:01
• @MDXF formulas and symbols and numbers hardly constitute maths. Mathematics is the abstraction and the generalization and the logic behind the formal manipulations of the symbols. Nov 10, 2017 at 19:02
• @JAB Jelly implements its own custom code page. (One of these days, TIO will include a link to the code page...) Nov 10, 2017 at 23:16
• @Mego the "sum-of-diagonal-products" only work for 1x1, 2x2, and 3x3 matrices. For bigger matrices, you need to consider all permutations and their parity. Nov 12, 2017 at 7:31
• +1 for actually include the proof in the post instead of saying "because this formula is listed on the page abcxyz it must be true". Nov 13, 2017 at 5:51

# Jelly, 161512 10 bytes

Ḣ×Zß-Ƥ$Ṛḅ-  Uses Laplace expansion. Thanks to @miles for golfing off 3 5 bytes! Try it online! ### How it works Ḣ×Zß-Ƥ$Ṛḅ-  Main link. Argument: M (matrix / 2D array)

Ḣ           Head; pop and yield the first row of M.
J-*×Ḣ€×ÇSµḢḢ$Ṗ?  Try it online! # Explanation œcL’$ṚÑ€         Helper Link; get the next level of subdeterminants (for Laplace Expansion)
œc               Combinations without replacement of length:
L’$Length of input - 1 (this will get all submatrices, except it's reversed) Ṛ Reverse the whole thing Ñ€ Get the determinant of each of these J-*×Ḣ€×ÇSµḢḢ$Ṗ?  Main Link
?  If the next value is truthy
Ṗ   Remove the last element (truthy if the matrix was at least size 2)
J-*×Ḣ€×ÇSµ       Then expand
ḢḢ$Otherwise, get the first element of the first element (m[0][0] in Python) J [1, 2, ..., len(z)] -* (-1 ** z) for each z in the length range × Vectorizing multiply with Ḣ€ The first element of each (this gets the first column); modifies each row (makes it golfier yay) ×Ç Vectorizing multiply with the subdeterminants S Sum  -2 bytes thanks to user202729's solution # R, 32 bytes function(m)Re(prod(eigen(m)$va))


Uses Not a Tree's algorithm of taking the eigenvalues of the matrix and taking the real part of their product.

Try it online!

p%(l:r)=l!!0*f(tail<$>p++r)-(p++[l])%r []%_=1 p%_=0 f=([]%)  Try it online! # C, 176 125 bytes Thanks to @ceilingcat for golfing 42 bytes, and thanks to both @Lynn and @Jonathan Frech for saving a byte each! d(M,n)int*M;{int i=n--,s=*M*!n,c,T[n*n];for(;i--;s+=M[i]*(1-i%2*2)*d(T,n))for(c=n*n;c--;T[c]=M[n-~c+c/n+(c%n>=i)]);return s;}  Calculates the determinant using the Laplace expansion along the first row. Try it online! Unrolled: d(M, n)int*M; { int i=n--, s=*M*!n, c, T[n*n]; for (; i--; s+=M[i]*(1-i%2*2)*d(T,n)) for (c=n*n; c--;) T[c] = M[n-~c+c/n+(c%n>=i)]; return s; }  • (i%2*-2+1)(1-i%2*2) saves one more byte. – Lynn Nov 11, 2017 at 13:34 • n+1+c can be n-~c. Nov 11, 2017 at 22:52 • Suggest i=s instead of return s Nov 25, 2017 at 0:49 # Jelly, 43 bytes Finally I've done writing my non-builtin solution in a golfing language! ḣ⁹’’¤;ṫḊ€Ç×⁸ị@⁹’¤¤ḷ/¤ çÐ€⁸J‘¤µJ-*×NS ÇḢḢ$Ṗ?


Thanks to HyperNeutrino for saving a byte!

Try it online! (spaced code for clarity)

terribly long way to remove n'th elements from a list, will improve later

This answer had been outgolfed by answers of HyperNeutrino, Dennis and Leaky Nun. Jelly is very popular as a golfing language.

Quick explanation:

ÇḢḢ$Ṗ? Main link. ? If Ṗ after remove the last element, the value is not empty (truthy) Ç then execute the last link ḢḢ$      else get the element at index [1, 1].

çÐ€⁸J‘¤µJ-*×NS     Helper link 1, take input as a matrix.
⁸J‘¤            the range [2, 3, ..., n+1]
µ           With the result,
J-*        generate the range [-1, 1, -1, 1, ...] with that length
×N      Multiply by negative
S     Sum

ḣ⁹’’¤;ṫḊ€Ç×⁸ị@⁹’¤¤ḷ/¤    Helper link 2, take left input as a matrix, right input as a number in range [2..n+1]
ḣ
⁹’’¤                    Take head ρ-2 of the matrix
;                   concatenate with
ṫ                  tail ρ (that is, remove item ρ-1)
Ḋ€                Remove first column
Ç               Calculate determinant of remaining matrix
×         ¤    multiply by
ḷ/     the first column,
ị@           row #
⁹’¤        ρ-1 (just removed in determinant calculation routine) of
⁸     ¤       the matrix.


# Octave, 28 bytes

@(x)round(prod(diag(qr(x))))


Try it online!

This uses the QR decomposition of a matrix X into an orthgonal matrix Q and an upper triangular matrix R. The determinant of X is the product of those of Q and R. An orthogonal matrix has unit determinant, and for a triangular matrix the determinant is the product of its diagonal entries. Octave's qr function called with a single output gives R.

The result is rounded to the closest integer. For large input matrices, floating-point inaccuracies may produce an error exceeding 0.5 and thus produce a wrong result.

• That's an interesting way to evade the det builtin. ;) Nov 10, 2017 at 23:49
• @tomsmeding :-) Also, it had already been "used" in Stewie's answer Nov 10, 2017 at 23:53

# MATL, 3 bytes / 5 bytes

### With built-in function

&0|


Try it online!

### Without built-in

Thanks to Misha Lavrov for pointing out a mistake, now corrected

YvpYo


Try it online!

This computes the determinant as the product of the eigenvalues, rounded to the closest integer to avoid floating-point inaccuracies.

Yv       % Implicit input. Push vector containing the eigenvalues
p        % Product
Yo       % Round. Implicit display

• Wouldn't the product of the singular values only tell you the absolute value of the determinant? Nov 10, 2017 at 22:02
• @MishaLavrov You're totally right! Thanks for noticing. I corrected it by using eigenvalues instead of singular values... and that saved 4 bytes \o/ Nov 10, 2017 at 22:12

## CJam (50 45 bytes)

{:A_,{1\$_,,.=1b\)/:CAff*A@zf{\f.*1fb}..-}/;C}


This is an anonymous block (function) which takes a 2D array on the stack and leaves an integer on the stack.

Online test suite

### Dissection

This implements the Faddeev-LeVerrier algorithm, and I think it's the first answer to take that approach.

The objective is to calculate the coefficients $$\c_k\$$ of the characteristic polynomial of the $$\n\times n\$$ matrix $$\A\$$, $$p(\lambda )\equiv \det(\lambda I_{n}-A)=\sum _{k=0}^{n}c_{k}\lambda ^{k}$$ where, evidently, $$\c_n = 1\$$ and $$\c_0 = (−1)^n \det A\$$.

The coefficients are determined recursively from the top down, by dint of the auxiliary matrices $$\M\$$, \begin{aligned}M_{0}&\equiv 0&c_{n}&=1\qquad &(k=0)\\M_{k}&\equiv AM_{k-1}+c_{n-k+1}I\qquad \qquad &c_{n-k}&=-{\frac {1}{k}}\mathrm {tr} (AM_{k})\qquad &k=1,\ldots ,n~.\end{aligned}

The code never works directly with $$\c_{n-k}\$$ and $$\M_k\$$, but always with $$$$-1)^k c_{n-k}\$$ and $$\(-1)^{k+1}AM_k\$$, so the recurrence is $$(-1)^k c_{n-k} = \frac1k \mathrm{tr} ((-1)^{k+1} AM_{k}) \\ (-1)^{k+2} AM_{k+1} = (-1)^k c_{n-k}A - A((-1)^{k+1}A M_k)$$ { e# Define a block :A e# Store the input matrix in A _, e# Take the length of a copy { e# for i = 0 to n-1 e# Stack: (-1)^{i+2}AM_{i+1} i 1_,,.=1b e# Calculate tr((-1)^{i+2}AM_{i+1})$$/:C       e#       Divide by (i+1) and store in C
Aff*        e#       Multiply by A
A@          e#       Push a copy of A, bring (-1)^{i+2}AM_{i+1} to the top
zf{\f.*1fb} e#       Matrix multiplication
..-         e#       Matrix subtraction
}/
;             e#   Pop (-1)^{n+2}AM_{n+1} (which incidentally is 0)
C             e#   Fetch the last stored value of C
}


# Octave, 30 bytes

@(x)-prod(diag([~,l,~]=lu(x)))


Try it online!

or, the boring 4 byte solution (saved 6 bytes thanks to Luis Mendo (forgot the rules regaring builtin functions)):

@det


Coming up! :)

# TI-Basic, 2 bytes

det(Ans


Ah, well.

As a high school student (who's forced to own one of these calculators), this function is hella useful so...

• It's still hella useful in college - linear algebra does not go away Nov 10, 2017 at 14:57
• @TaylorScott In fact, it comes back with a vengeance in differential equations.
– user45941
Nov 10, 2017 at 16:05
• @Mego - you are right on that; though for some reason they let me take all of calc and that before linear :/ Nov 11, 2017 at 1:23
• @TaylorScott Due to an oversight by my university's math department, linalg was not a prerequisite for diffeq when I took it. When my professor realized that, he quickly gave us a 3-day crash course in linalg.
– user45941
Nov 11, 2017 at 19:41

a#((b:c):r)=b*d(a++map tail r)-(a++[c])#r
_#_=0
d[]=1
d l=[]#l


Try it online! (Footer with test cases taken from @totallyhuman's solution.)

d computes the determinant using a Laplace expansion along the first column. Needs three bytes more than the permanent.

# Wolfram Language (Mathematica), 53 52 bytes

1##&@@@(t=Tuples)@#.Signature/@t[Range@Tr[1^#]&/@#]&


Try it online!

Unfortunately, computing the determinant of an n by n matrix this way uses O(nn) memory, which puts large test cases out of reach.

## How it works

The first part, 1##&@@@(t=Tuples)@#, computes all possible products of a term from each row of the given matrix. t[Range@Tr[1^#]&/@#] gives a list of the same length whose elements are things like {3,2,1} or {2,2,3} saying which entry of each row we picked out for the corresponding product.

We apply Signature to the second list, which maps even permutations to 1, odd permutations to -1, and non-permutations to 0. This is precisely the coefficient with which the corresponding product appears in the determinant.

Finally, we take the dot product of the two lists.

If even Signature is too much of a built-in, at 73 bytes we can take

1##&@@@(t=Tuples)@#.(1##&@@Order@@@#~Subsets~{2}&/@t[Range@Tr[1^#]&/@#])&


replacing it by 1##&@@Order@@@#~Subsets~{2}&. This computes Signature of a possibly-permutation by taking the product of Order applied to all pairs of elements of the permutation. Order will give 1 if the pair is in ascending order, -1 if it's in descending order, and 0 if they're equal.

-1 byte thanks to @user202729

• 52 bytes (in case you didn't know this Mathematica golfing tip) Nov 11, 2017 at 14:09
• I did, but somehow forgot about it here. Thanks! Nov 11, 2017 at 16:07

# Python 2, 75 bytes

f=lambda m,p=[]:m[0][0]*f(zip(*p+m[1:])[1:])-f(m[1:],p+m[:1])if m else[]==p


Try it online!

## SageMath, various

Here are a bunch of methods for computing the determinant that I found interesting, all programmed in SageMath. They can all be tried here.

### Builtin, 3 bytes

det


This one isn't too interesting. Sage provides global-level aliases to many common operations that would normally be object methods, so this is shorter than lambda m:m.det().

### Real Part of Product of Eigenvalues, 36 bytes

lambda m:real(prod(m.eigenvalues()))


Unfortunately, eigenvalues is not one of those global-level aliases. That, combined with the fact that Sage doesn't have a neat way to compose functions, means we're stuck with a costly lambda. This function symbolic values which are automatically converted to numeric values when printed, so some floating point inaccuracy may be present in some outputs.

### Product of Diagonal in Jordan Normal Form, 60 bytes

lambda m:prod(m.jordan_form()[x,x]for x in range(m.nrows()))


In Jordan Normal form, an NxN matrix is represented as a block matrix, with N blocks on the diagonal. Each block consists of either a single eigenvalue, or a MxM matrix with a repeated eigenvalue on the diagonal and 1s on the super-diagonal (the diagonal above and to the right of the "main" diagonal). This results in a matrix with all eigenvalues (with multiplicity) on the main diagonal, and some 1s on the super-diagonal corresponding to repeated eigenvalues. This returns the product of the diagonal of the Jordan normal form, which is the product of the eigenvalues (with multiplicty), so this is a more roundabout way of performing the same computation as the previous solution.

Because Sage wants the Jordan normal form to be over the same ring as the original matrix, this only works if all of the eigenvalues are rational. Complex eigenvalues result in an error (unless the original matrix is over the ring CDF (complex double floats) or SR). However, this means that taking the real part is not necessary, compared to the above solution.

### Product of Diagonal in Smith Decomposition

lambda m:prod(m.smith_form()[0].diagonal())


Unlike Jordan normal form, Smith normal form is guaranteed to be over the same field as the original matrix. Rather than computing the eigenvalues and representing them with a block diagonal matrix, Smith decomposition computes the elementary divisors of the matrix (which is a topic a bit too complicated for this post), puts them into a diagonal matrix D, and computes two matrices with unit determinant U and V such that D = U*A*V (where A is the original matrix). Since the determinant of the product of matrices equals the product of the determinants of the matrices (det(A*B*...) = det(A)*det(B)*...), and U and V are defined to have unit determinants, det(D) = det(A). The determinant of a diagonal matrix is simply the product of the elements on the diagonal.

### Laplace Expansion, 109 bytes

lambda m:m.nrows()>1and sum((-1)**j*m[0,j]*L(m[1:,:j].augment(m[1:,j+1:]))for j in range(m.ncols()))or m[0,0]


This performs Laplace expansion along the first row, using a recursive approach. det([[a]]) = a is used for the base case. It should be shorter to use det([[]]) = 1 for the base case, but my attempt at that implementation had a bug that I haven't been able to track down yet.

### Leibniz's Formula, 100 bytes

L2 = lambda m:sum(sgn(p)*prod(m[k,p[k]-1]for k in range(m.ncols()))for p in Permutations(m.ncols()))


This directly implements Leibniz's formula. For a much better explanation of the formula and why it works than I could possibly write, see this excellent answer.

### Real Part of e^(Tr(ln(M))), 48 bytes

lambda m:real(exp(sum(map(ln,m.eigenvalues()))))


This function returns symbolic expressions. To get a numerical approximation, call n(result) before printing.

This is an approach that I haven't seen anyone use yet. I'm going to give a longer, more-detailed explanation for this one.

Let A be a square matrix over the reals. By definition, the determinant of A is equal to the product of the eigenvalues of A. The trace of A is equal to the sum of A's eigenvalues. For real numbers r_1 and r_2, exp(r_1) * exp(r_2) = exp(r_1 + r_2). Since the matrix exponential function is defined to be analogous to the scalar exponential function (especially in the previous identity), and the matrix exponential can be computed by diagonalizing the matrix and applying the scalar exponential function to the eigenvalues on the diagonal, we can say det(exp(A)) = exp(trace(A)) (the product of exp(λ) for each eigenvalue λ of A equals the sum of the eigenvalues of exp(A)). Thus, if we can find a matrix L such that exp(L) = A, we can compute det(A) = exp(trace(L)).

We can find such a matrix L by computing log(A). The matrix logarithm can be computed in the same way as the matrix exponential: form a square diagonal matrix by applying the scalar logarithm function to each eigenvalue of A (this is why we restriced A to the reals). Since we only care about the trace of L, we can skip the construction and just directly sum the exponentials of the eigenvalues. The eigenvalues can be complex, even if the matrix isn't over the complex ring, so we take the real part of the sum.

• The last part is a fascinating idea, but the header and explanation don't match the code, which doesn't take a matrix logarithm. It's just real(prod(m.eigenvalues())) ungolfed. Oct 20, 2018 at 19:18

# Java (OpenJDK 8), 195 192 177 176 bytes

long d(int[][]m){long D=0;for(int i=m.length,l=i-1,t[][]=new int[l][l],j,k;i-->0;D+=m[0][i]*(1-i%2*2)*(l<1?1:d(t)))for(j=l*l;j-->0;)t[j/l][k=j%l]=m[1+j/l][k<i?k:k+1];return D;}


Try it online!

Like many other answers, this also uses the Laplace formula. A slightly less golfed version:

long d(int[][]m){
long D=0;
int i=m.length,l=i-1,t[][]=new int[l][l],j,k
for(;i-->0;)
for(j=l*l;j-->0;)
t[j/l][k=j%l]=m[1+j/l][k<i?k:k+1];
D+=m[0][i]*(1-i%2*2)*(l<1?1:d(t));
return D;
}


# Wolfram Language (Mathematica), 3 bytes

Det


Try it online!

Per the meta consensus, mainly upvote nontrivial solutions that take effort to write.

# Python 2, 95 bytes

-12 bytes thanks to Lynn.

f=lambda m:sum((-1)**i*v*f([j[:i]+j[i+1:]for j in m[1:]])for i,v in enumerate(m[0]))if m else 1


Try it online!

• Here, too, you can use [] as a base case: f=lambda m:sum((-1)**i*v*f([j[:i]+j[i+1:]for j in m[1:]])for i,v in enumerate(m[0]))if m else 1 for 95 bytes!
– Lynn
Nov 11, 2017 at 13:31
• m==[]or sum(...) gives 92 bytes. Oct 22, 2018 at 1:18

# Python 3, 238 bytes, 227 bytes, 224 bytes, 216 bytes

from functools import*
from itertools import*
r=range;n=len;s=sum
f=lambda l:s(reduce(lambda p,m:p*m,[l[a][b]for a,b in zip(r(n(l)),j)])*(-1)**s(s(y<j[x]for y in j[x:])for x in r(n(l)))for j in permutations(r(n(l))))


Try it online!

My solution uses the definition of a determinant for calculations. Unfortunately, the complexity of this algorithm is n! and I can not show the passage of the last test, but in theory this is possible.

# Java 8, 266261259 258 bytes

long d(int[][]m){long r=0;int i=0,j,k,l=m.length,t[][]=new int[l-1][l-1],q=m[0][0];if(l<3)return l<2?q:q*m[1][1]-m[0][1]*m[1][0];for(;i<l;r+=m[0][i]*(1-i++%2*2)*d(t))for(j=0;++j<l;)for(k=l;k-->0;){q=m[j][k];if(k<i)t[j-1][k]=q;if(k>i)t[j-1][k-1]=q;}return r;}


Look mom, no build-ins.. because Java has none.. >.>

-7 bytes thanks to @ceilingcat.

Explanation:

Try it here. (Only the last test case is too big to fit in a long of size 263-1.)

long d(int[][]m){             // Method with integer-matrix parameter and long return-type
long r=0;                   //  Return-long, starting at 0
int i=0,j,k,                //  Index-integers
l=m.length,             //  Dimensions of the square matrix
t[][]=new int[l-1][l-1],//  Temp-matrix, one size smaller than m
q=m[0][0];              //  The first value in the matrix (to reduce bytes)
if(l<3)                     //  If the dimensions are 1 or 2:
return l<2?               //   If the dimensions are 1:
q                       //    Simply return the only item in it
:                        //   Else (the dimensions are 2):
q*m[1][1]-m[0][1]*m[1][0];
//    Calculate the determinant of the 2x2 matrix
//  If the dimensions are 3 or larger:
for(;i<l;                   //  Loop (1) from 0 to l (exclusive)
r+=                     //    After every iteration: add the following to the result:
m[0][i]              //     The item in the first row and i'th column,
*(1-i++%2*2)         //     multiplied by 1 if i is even; -1 if odd,
*d(t))               //     multiplied by a recursive call with the temp-matrix
for(j=0;                  //   Reset index j to 0
++j<l;)               //   Inner loop (2) from 0 to l (exclusive)
for(k=l;k-->0;){        //    Inner loop (3) from l-1 to 0 (inclusive)
q=m[j][k];            //     Set the integer at location j,k to reduce bytes
if(k<i)               //     If k is smaller than i:
t[j-1][k]=q;        //      Set this integer at location j-1,k
if(k>i)               //     Else-if k is larger than i:
t[j-1][k-1]=q;      //      Set this integer at location j-1,k-1
//     Else: k and i are equals: do nothing (implicit)
}                       //    End of inner loop (3)
//   End of inner loop (2) (implicit / single-line body)
//  End of loop (1) (implicit / single-line body)
return r;                   //  Return the result-long
}                             // End of method


# JavaScript (ES6), 91

Recursive Laplace

q=(a,s=1)=>+a||a.reduce((v,[r],i)=>v-(s=-s)*r*q(a.map(r=>r.slice(1)).filter((r,j)=>j-i)),0)


Less golfed

q = (a,s=1) => // s used as a local variable
a[1] // check if a is a single element array
// if array, recursive call expanding along 1st column
? a.reduce((v,[r],i) => v-(s=-s)*r*q(a.map(r=>r.slice(1)).filter((r,j)=>j-i)),0)
: +a // single element, convert to number


Test

q=(a,s=1)=>+a||a.reduce((v,[r],i)=>v-(s=-s)*r*q(a.map(r=>r.slice(1)).filter((r,j)=>j-i)),0)

TestCases=[[42]] -> 42
[[2, 3], [1, 4]] -> 5
[[1, 2, 3], [4, 5, 6], [7, 8, 9]] -> 0
[[13, 17, 24], [19, 1, 3], [-5, 4, 0]] -> 1533
[[372, -152, 244], [-97, -191, 185], [-53, -397, -126]] -> 46548380
[[100, -200, 58, 4], [1, -90, -55, -165], [-67, -83, 239, 182], [238, -283, 384, 392]] -> 571026450
[[432, 45, 330, 284, 276], [-492, 497, 133, -289, -28], [-443, -400, 56, 150, -316], [-344, 316, 92, 205, 104], [277, 307, -464, 244, -422]] -> -51446016699154
.split('\n')

TestCases.forEach(r=>{
[a,k] = r.split (' -> ')
a = eval(a)
d = q(a)
console.log('Test '+(k==d ? 'OK':'KO')+
'\nMatrix '+a.join('|')+
'\nResult '+d+
'\nCheck  '+k)
})

• 83 bytes with the same behavior Nov 14, 2018 at 10:04
• Or 85 bytes to support the empty matrix (whose determinant should be 1). Nov 14, 2018 at 10:05
• (I used the same optimizations in this answer, which is derived from yours.) Nov 14, 2018 at 10:07

# Pari/GP, 6 bytes

matdet


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# Pari/GP, 52 bytes

f(a)=if(#a,sum(n=1,#a,-(-1)^n*a[1,n]*f(a[^1,^n])),1)


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# Python 2 + numpy, 29 bytes

from numpy.linalg import*
det


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# Julia, 3 bytes

det


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# J, 5 bytes

-/ .*


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