Calculadora de matrices

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A (3×3)
Filas: 3
Columnas: 3
B (3×3)
Filas: 3
Columnas: 3
Supports: fractions (1/2), decimals (0.5), constants (pi, e). Empty cells are treated as 0.

Resultados del cálculo

1. Definición y fórmula: ¿qué es el rango de una matriz?

The rank of a matrix is the number of linearly independent rows or columns. Equivalently, the rank is the number of pivot positions in the matrix's reduced row echelon form (RREF).

Rank applies to both square and rectangular matrices. Row rank equals column rank.

Método práctico (resumen):

  1. Compute the RREF of matrix $A$ (Gauss–Jordan elimination).
  2. Count the number of nonzero (pivot) rows in RREF.
  3. That count is $\operatorname{rank}(A)$.

2. Cómo calcular el rango (método rápido)

  • Convert $A$ to RREF using elementary row operations.
  • Count pivot rows (rows with leading 1's).
  • Alternatively, count maximal number of linearly independent columns (column rank).

Rango = número de pivotes = número de filas no nulas en la RREF.

3. Ejemplos resueltos

Ejemplo 1

Calcula el rango de

$$A=\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix}$$

Reducción por filas (resumen):

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 1 & 1 & 1 \end{bmatrix} \xrightarrow{R_2-2R_1,\;R_3-R_1} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 0 & 0 \\ 0 & -1 & -2 \end{bmatrix} \xrightarrow{R_3 \leftrightarrow R_2} \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 0 \end{bmatrix}$$

There are 2 pivot rows, so

$$\operatorname{rank}(A)=2$$

Ejemplo 2

Halla el rango de

$$B=\begin{bmatrix} 3 & 0 & 6 \\ 1 & -1 & 2 \\ 2 & -1 & 4 \end{bmatrix}$$

Reducción por filas (resumen):

$$\begin{bmatrix} 3 & 0 & 6 \\ 1 & -1 & 2 \\ 2 & -1 & 4 \end{bmatrix} \xrightarrow{R_1\leftrightarrow R_2 \text{ (or scale/elim.)}} \begin{bmatrix} 1 & 0 & 2 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$

Two pivot rows ⇒

$$\operatorname{rank}(B)=2$$

4. Common mistakes (pitfalls)

  • Counting nonzero entries: Rank counts independent rows/columns, not nonzero entries.
  • Stopping too early in elimination: REF might hide dependencies; RREF clarifies pivots.
  • Mixing determinant logic: For square matrices, $\det\neq 0$ ⇒ full rank, but determinant is not defined for non-square matrices.
  • Assuming visually independent: Rows that look different may still be linearly dependent—use elimination to be sure.

5. Practice problems (click to show answers)

Try these. Answers are hidden — click Mostrar respuesta to reveal each rank.

Ejercicio 1

$$M_1=\begin{bmatrix} 1 & 3 & 5 \\ 2 & 6 & 10 \\ 1 & 1 & 2 \end{bmatrix}$$
Mostrar respuesta
$$\operatorname{rank}(M_1)=2$$

Rows 1 and 2 are dependent (row2 = 2·row1), so at most 2 independent rows; RREF confirms 2 pivots.

Ejercicio 2

$$M_2=\begin{bmatrix} 0 & 1 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 0 \end{bmatrix}$$
Mostrar respuesta
$$\operatorname{rank}(M_2)=1$$

Column 2 is a multiple of column 1? Here row 2 = 2·row1, so only one independent row remains.

Ejercicio 3

$$M_3=\begin{bmatrix} 4 & 2 \\ 2 & 1 \\ 6 & 3 \end{bmatrix}$$
Mostrar respuesta
$$\operatorname{rank}(M_3)=1$$

Each row is proportional (row3 = 1.5·row1, row2 = 0.5·row1), so only one independent row.

Ejercicio 4

$$M_4=\begin{bmatrix} 1 & -1 & 2 \\ 3 & -3 & 6 \\ 5 & -5 & 10 \end{bmatrix}$$
Mostrar respuesta
$$\operatorname{rank}(M_4)=1$$

All rows are scalar multiples of the first row (row2 = 3·row1, row3 = 5·row1), so rank = 1.


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