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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.
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Resultados del cálculo

1. Definición y fórmulas

The matrix sine of a square matrix $A$ is a matrix function defined by the usual Taylor series:

$$ \sin(A)=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}A^{2k+1}=A-\frac{A^3}{3!}+\frac{A^5}{5!}-\cdots $$

Alternative (complex-exponential) expression:

$$ \sin(A)=\frac{e^{iA}-e^{-iA}}{2i}. $$

If $A$ is diagonalizable $A = VDV^{-1}$, then $\sin(A)=V\sin(D)V^{-1}$ where $\sin(D)$ applies $\sin(\cdot)$ to diagonal entries. The series always converges for every square matrix $A$.

2. Observaciones útiles para el cálculo

  • Diagonal / diagonalizable matrices: compute $\sin$ on eigenvalues and transform back: $\sin(A)=V\sin(D)V^{-1}$.
  • Special nilpotent matrices: if $N^m=0$ then series truncates and computation is finite.
  • Powers that square to $-I$: if $A^2=-\alpha^2 I$ (rotation-like), then $\sin(A)=\frac{\sin(\alpha)}{\alpha}A$ (special case with $\alpha\neq0$).
  • Numerical methods: use Schur decomposition or scaling-and-squaring with Padé approximants for stability in code.

3. Ejemplos resueltos

Ejemplo 1: matriz diagonal

Compute $\sin(A)$ for

$$ A=\begin{pmatrix}\pi/2 & 0\\[6pt]0 & \pi\end{pmatrix} $$

Caso diagonal: aplica el seno a las entradas diagonales

$$ \sin(A)=\begin{pmatrix}\sin(\pi/2) & 0\\[6pt]0 & \sin(\pi)\end{pmatrix}=\begin{pmatrix}1 & 0\\[6pt]0 & 0\end{pmatrix}. $$

Input hint for calculator: enter matrix rows as π/2 0 and 0 π.

Ejemplo 2: matriz generadora de rotación

Let

$$ A=\begin{pmatrix}0 & -1\\[6pt]1 & 0\end{pmatrix}. $$

Observe $A^2=-I$, so powers cycle: $A^{2k+1}=(-1)^k A$. Thus the series becomes

$$ \sin(A)=A\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}=A\sin(1). $$

So

$$ \sin(A)=\sin(1)\begin{pmatrix}0 & -1\\[6pt]1 & 0\end{pmatrix}. $$

Input hint: enter 0 -1 and 1 0 for matrix A.

4. Errores comunes

  • Applying sine elementwise: $\sin(A)\neq(\sin(a_{ij}))$ in general — use powers or diagonalization.
  • Assuming $\sin(A+B)=\sin(A)+\sin(B)$: matrix sine is not linear in that way (unless $A$ and $B$ commute and extra conditions hold).
  • Truncating series carelessly: for numerical work, naive truncation can cause large errors; prefer stable algorithms.
  • Using diagonalization without checking: not all matrices are diagonalizable; use Schur/Jordan-based methods if needed.
  • Mishandling scaling factors: for $A^2=-\alpha^2 I$ use $\sin(A)=\frac{\sin(\alpha)}{\alpha}A$ (avoid division by zero when $\alpha=0$ — treat separately).

5. Problemas de práctica (respuestas ocultas)

Try these exercises. Click Mostrar respuesta to reveal the result.

Ejercicio 1

$$ A=\begin{pmatrix}0 & 1\\[6pt]-1 & 0\end{pmatrix} $$
Mostrar respuesta
$$ \sin(A)=\sin(1)\begin{pmatrix}0 & 1\\[6pt]-1 & 0\end{pmatrix}. $$

Because $A^2=-I$ and $\sin(A)=A\sin(1)$ (see Example 2).

Ejercicio 2

$$ A=\begin{pmatrix}1 & 0\\[6pt]0 & 2\end{pmatrix} $$
Mostrar respuesta
$$ \sin(A)=\begin{pmatrix}\sin(1) & 0\\[6pt]0 & \sin(2)\end{pmatrix}. $$

Diagonal case: apply sine to each eigenvalue.

Ejercicio 3

$$ A=\begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix} $$
Mostrar respuesta

Here $A^2=0$, so the sine series truncates:

$$ \sin(A)=A=\begin{pmatrix}0 & 1\\[6pt]0 & 0\end{pmatrix}. $$

Nilpotent matrices often yield finite polynomial results.

Ejercicio 4

$$ A=\begin{pmatrix}\pi/6 & 0 & 0\\[6pt]0 & \pi/3 & 0\\[6pt]0 & 0 & \pi/2\end{pmatrix} $$
Mostrar respuesta
$$ \sin(A)=\begin{pmatrix}\sin(\pi/6) & 0 & 0\\[6pt]0 & \sin(\pi/3) & 0\\[6pt]0 & 0 & \sin(\pi/2)\end{pmatrix} = \begin{pmatrix}1/2 & 0 & 0\\[6pt]0 & \sqrt{3}/2 & 0\\[6pt]0 & 0 & 1\end{pmatrix}. $$

Diagonal example with common angle values.

Si estás explorando el seno matricial, también te conviene revisar el rango de una matriz, la SVD y la RREF.

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