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Calculateur de matrices

Déterminant, inverse, rang et forme échelonnée réduite d'une matrice jusqu'à 6 × 6, avec chaque opération sur les lignes, ainsi que transposée, produit et somme, en fractions exactes.

Mis à jour Exemples vérifiés : 8

One row per line (or rows separated by ;), entries separated by spaces or commas; fractions like 1/2 work
Essayer
Résultat
[0, 0, 1; −2, 1, 3; 3, −1, −5]
Résultat: [0, 0, 1; −2, 1, 3; 3, −1, −5]
det A
−1
rank A
3
trace A
5

A is invertible (det A = −1); the inverse undoes A, so A × A⁻¹ is the identity matrix.

A⁻¹

001−2133−1−5
A⁻¹ Lignes : 3
Col 1Col 2Col 3
001
−213
3−1−5
Comment le calcul est effectué S

  1. Write [A | I]

    [211100132010100001]\left[\begin{array}{ccc|ccc}2 & 1 & 1 & 1 & 0 & 0 \\ 1 & 3 & 2 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1\end{array}\right]

  2. Column 1

    R1↔R2, R2←R2−2R1, R3←R3−R1  ⟶  [1320100−5−31−200−3−20−11]R_{1} \leftrightarrow R_{2},\ R_{2} \leftarrow R_{2} - 2R_{1},\ R_{3} \leftarrow R_{3} - R_{1} \;\longrightarrow\; \left[\begin{array}{ccc|ccc}1 & 3 & 2 & 0 & 1 & 0 \\ 0 & -5 & -3 & 1 & -2 & 0 \\ 0 & -3 & -2 & 0 & -1 & 1\end{array}\right]

  3. Column 2

    R2←−15 R2, R1←R1−3R2, R3←R3+3R2  ⟶  [101535−1500135−1525000−15−35151]R_{2} \leftarrow -\frac{1}{5}\,R_{2},\ R_{1} \leftarrow R_{1} - 3R_{2},\ R_{3} \leftarrow R_{3} + 3R_{2} \;\longrightarrow\; \left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{5} & \frac{3}{5} & -\frac{1}{5} & 0 \\ 0 & 1 & \frac{3}{5} & -\frac{1}{5} & \frac{2}{5} & 0 \\ 0 & 0 & -\frac{1}{5} & -\frac{3}{5} & \frac{1}{5} & 1\end{array}\right]

  4. Column 3

    R3←−5 R3, R1←R1−15R3, R2←R2−35R3  ⟶  [100001010−2130013−1−5]R_{3} \leftarrow -5\,R_{3},\ R_{1} \leftarrow R_{1} - \frac{1}{5}R_{3},\ R_{2} \leftarrow R_{2} - \frac{3}{5}R_{3} \;\longrightarrow\; \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & -2 & 1 & 3 \\ 0 & 0 & 1 & 3 & -1 & -5\end{array}\right]

  5. The right half is the inverse

    A−1=[001−2133−1−5]A^{-1} = \left[\begin{array}{ccc}0 & 0 & 1 \\ -2 & 1 & 3 \\ 3 & -1 & -5\end{array}\right]

    Check: A × A⁻¹ = I.

À propos de Calculateur de matrices

The calculator works on matrices up to 6 × 6 in exact fractions. The inverse and the reduced row echelon form come from Gauss–Jordan elimination: [A | I] is row-reduced until the left half is the identity, and the right half is then A⁻¹. The determinant is the product of the pivots from forward elimination, negated once for each row swap; the rank is the number of pivots; and a product has entries (AB)ᵢⱼ = Σₖ aᵢₖbₖⱼ.

Linear algebra courses, 3D graphics transforms and systems of equations are the usual uses. The default matrix [2 1 1; 1 3 2; 1 0 0] has determinant −1, so it is invertible, and its inverse [0 0 1; −2 1 3; 3 −1 −5] has whole-number entries.

Type one row per line, with entries separated by spaces or commas; fractions such as 1/2 stay exact. A matrix with determinant 0 is singular: it has no inverse, and its rank is below its size.

Exemples détaillés

Inverse of the default 3 × 3

Calculate
Inverse of A
Matrix A
2 1 1 1 3 2 1 0 0
Résultat
[0, 0, 1; −2, 1, 3; 3, −1, −5]
det A
-1

Source de vérification : Python fractions Gauss–Jordan on [A | I]; det by cofactor expansion = −1

Inverse of [4 7; 2 6]

Calculate
Inverse of A
Matrix A
4 7 2 6
Résultat
[3/5, −7/10; −1/5, 2/5]
det A
10

Source de vérification : (1/(ad − bc))·[d −b; −c a] = (1/10)·[6 −7; −2 4]

Determinant of the default 3 × 3

Calculate
Determinant of A
Matrix A
2 1 1 1 3 2 1 0 0
Résultat
−1
det A
-1
trace A
5

Source de vérification : Cofactor expansion: 2·0 − 1·(0 − 2) + 1·(0 − 3) = −1

Rank of a singular matrix (edge case)

Calculate
Rank of A
Matrix A
1 2 3 2 4 6 1 1 1
Résultat
2
rank A
2
det A
0

Source de vérification : Row 2 = 2 × row 1; Python fractions RREF has 2 pivots

Questions

How do you find the inverse of a matrix?

Write A next to the identity matrix, [A | I], and apply row operations until the left half becomes I; the right half is then A⁻¹. A 2 × 2 matrix [a b; c d] has a shortcut: A⁻¹ = (1/(ad − bc)) × [d −b; −c a]. For [4 7; 2 6], ad − bc = 24 − 14 = 10, so A⁻¹ = [3/5 −7/10; −1/5 2/5].

How do you calculate the determinant of a 3 × 3 matrix?

Expand along a row or column, multiplying each entry by the determinant of its 2 × 2 minor with alternating signs. For [2 1 1; 1 3 2; 1 0 0], the bottom row is quickest because two of its entries are 0: det = 1 × (1 × 2 − 1 × 3) = −1. For larger matrices row reduction reaches the same answer with far fewer operations.

When does a matrix have no inverse?

When its determinant is 0, which happens exactly when one row or column is a combination of the others. In [1 2 3; 2 4 6; 1 1 1], row 2 is twice row 1, so the rank is 2 rather than 3 and the determinant is 0. Such a matrix is called singular, and a system Ax = b built on it has either no solution or infinitely many.

How do you multiply two matrices?

Each entry of AB is a row of A times a column of B, summed: (AB)ᵢⱼ = Σₖ aᵢₖbₖⱼ. For [1 2; 3 4] × [5 6; 7 8] the top-left entry is 1 × 5 + 2 × 7 = 19, and the product is [19 22; 43 50]. A needs as many columns as B has rows, and order matters: here BA = [23 34; 31 46].

What is reduced row echelon form?

A matrix is in reduced row echelon form (RREF) when each non-zero row starts with a 1, that leading 1 is the only non-zero entry in its column, the leading 1s step right going down, and zero rows sit at the bottom. Every matrix has exactly one RREF, and its number of leading 1s is the rank. For an augmented matrix it reads off the solution: [1 0 0 −8; 0 1 0 1; 0 0 1 −2] means x = −8, y = 1, z = −2.

Quelle est la précision de « Calculateur de matrices » ?

La précision dépend de vos données et des hypothèses de la méthode. Le calcul décimal utilise 50 chiffres significatifs, mais les estimations, méthodes numériques et données sources peuvent être moins précises ; l’arrondi affiché ne supprime pas ces limites. Exemples résolus vérifiés à partir de sources indépendantes : 8. Par exemple, « Inverse of the default 3 × 3 » est vérifié à l’aide de Python fractions Gauss–Jordan on [A | I]; det by cofactor expansion = −1.

D’où vient cette méthode ?

Wikipedia — Gaussian elimination (row reduction and the RREF example); Wolfram MathWorld — Matrix Inverse; G. Strang, Introduction to Linear Algebra (5th ed.), chapters 2–3.

À propos de ce calculateur

A−1: [A∣I]→row operations[I∣A−1](AB)ij=∑kaikbkj\begin{gathered} A^{-1}:\ [A \mid I] \xrightarrow{\text{row operations}} [I \mid A^{-1}] \\[10pt] (AB)_{ij} = \sum_k a_{ik} b_{kj} \end{gathered}

Sources

  1. Wikipedia — Gaussian elimination (row reduction and the RREF example)
  2. Wolfram MathWorld — Matrix Inverse
  3. G. Strang, Introduction to Linear Algebra (5th ed.), chapters 2–3

Vérifié avec les références

Ce calculateur comprend 8 exemples résolus dont les réponses proviennent de sources indépendantes. Ils font partie de la suite de tests et peuvent aussi être exécutés ici.

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