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Calculadora de matrices

Determinante, inversa, rango y forma escalonada reducida de una matriz de hasta 6 × 6, con cada operación de fila, además de transpuesta, producto y suma en fracciones exactas.

Actualizado Ejemplos verificados: 8

One row per line (or rows separated by ;), entries separated by spaces or commas; fractions like 1/2 work
Probar
Resultado
[0, 0, 1; −2, 1, 3; 3, −1, −5]
Resultado: [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 Filas: 3
Col 1Col 2Col 3
001
−213
3−1−5
Cómo se calcula 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.

Acerca de Calculadora 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.

Ejemplos resueltos

Inverse of the default 3 × 3

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

Fuente de comprobación: 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
Resultado
[3/5, −7/10; −1/5, 2/5]
det A
10

Fuente de comprobación: (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
Resultado
−1
det A
-1
trace A
5

Fuente de comprobación: 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
Resultado
2
rank A
2
det A
0

Fuente de comprobación: Row 2 = 2 × row 1; Python fractions RREF has 2 pivots

Preguntas

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.

¿Qué precisión tiene «Calculadora de matrices»?

La precisión depende de tus datos y de los supuestos del método. El cálculo decimal usa 50 cifras significativas, pero las estimaciones, los métodos numéricos y los datos de origen pueden ser menos precisos; el redondeo mostrado no elimina esos límites. Ejemplos resueltos comprobados con fuentes independientes: 8. Por ejemplo, «Inverse of the default 3 × 3» se comprueba con Python fractions Gauss–Jordan on [A | I]; det by cofactor expansion = −1.

¿De dónde procede el método?

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

Acerca de esta calculadora

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}

Fuentes

  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

Verificado con las referencias

Esta calculadora incluye 8 ejemplos resueltos con respuestas de fuentes independientes. Forman parte del conjunto de pruebas y también puedes ejecutarlos aquí.

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