T1: Matrices y determinantes · Operaciones con matrices y ecuaciones matriciales — MATEMATICAS APLICADAS A LAS CIENCIAS SOCIALES II PEvAU Andaluc%25252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525C3%25252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525252525ADa 2022

Operaciones con matrices y ecuaciones matriciales

Problema

2022 · Ordinaria · Reserva

BLOQUE A

Se consideran las matrices

A = \begin{pmatrix} 1 & 0 & 1 \\ -1 & -1 & 1 \\ 2 & -1 & 0 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix}, C = \begin{pmatrix} 3 \\ -7 \\ -2 \end{pmatrix}

a) Razone si se pueden efectuar las siguientes operaciones y realice las que sean posibles:

C \cdot A, A + B, C^t \cdot B^t

.b) Resuelva la ecuación matricial

A \cdot X = B \cdot X + C

MatricesEcuación matricialInversa de una matriz

Resolución del ejercicio

Se tienen las matrices:

A = \begin{pmatrix} 1 & 0 & 1 \\ -1 & -1 & 1 \\ 2 & -1 & 0 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix}, C = \begin{pmatrix} 3 \\ -7 \\ -2 \end{pmatrix}

Las dimensiones de estas matrices son:

\text{dim}(A) = 3 \times 3 \quad \text{dim}(B) = 3 \times 3 \quad \text{dim}(C) = 3 \times 1

a) Razone si se pueden efectuar las siguientes operaciones y realice las que sean posibles:

C \cdot A, A + B, C^t \cdot B^t

Para la operación $C \cdot A$ :

\text{dim}(C) = 3 \times 1 \quad \text{dim}(A) = 3 \times 3

El número de columnas de $C$ (1) no coincide con el número de filas de $A$ (3). Por lo tanto, la operación $C \cdot A$ no se puede efectuar.Para la operación $A + B$ :

\text{dim}(A) = 3 \times 3 \quad \text{dim}(B) = 3 \times 3

Ambas matrices tienen las mismas dimensiones. Por lo tanto, la operación $A + B$ se puede efectuar. El resultado es:

A + B = \begin{pmatrix} 1 & 0 & 1 \\ -1 & -1 & 1 \\ 2 & -1 & 0 \end{pmatrix} + \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix} = \begin{pmatrix} 1+1 & 0-1 & 1+1 \\ -1-1 & -1-1 & 1-1 \\ 2+1 & -1-1 & 0+1 \end{pmatrix} = \begin{pmatrix} 2 & -1 & 2 \\ -2 & -2 & 0 \\ 3 & -2 & 1 \end{pmatrix}

Para la operación $C^t \cdot B^t$ :Primero, calculamos las matrices transpuestas:

C^t = \begin{pmatrix} 3 & -7 & -2 \end{pmatrix} \quad \text{dim}(C^t) = 1 \times 3

B^t = \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix}^t = \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix} \quad \text{dim}(B^t) = 3 \times 3

El número de columnas de $C^t$ (3) coincide con el número de filas de $B^t$ (3). Por lo tanto, la operación $C^t \cdot B^t$ se puede efectuar. El resultado es:

C^t \cdot B^t = \begin{pmatrix} 3 & -7 & -2 \end{pmatrix} \cdot \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix}

= \begin{pmatrix} 3(1)+(-7)(-1)+(-2)(1) & 3(-1)+(-7)(-1)+(-2)(-1) & 3(1)+(-7)(-1)+(-2)(1) \end{pmatrix}

= \begin{pmatrix} 3+7-2 & -3+7+2 & 3+7-2 \end{pmatrix} = \begin{pmatrix} 8 & 6 & 8 \end{pmatrix}

b) Resuelva la ecuación matricial

A \cdot X = B \cdot X + C

Reordenamos la ecuación para aislar $X$ :

A \cdot X - B \cdot X = C

(A - B) \cdot X = C

Sea $D = A - B$ . Calculamos $D$ :

D = \begin{pmatrix} 1 & 0 & 1 \\ -1 & -1 & 1 \\ 2 & -1 & 0 \end{pmatrix} - \begin{pmatrix} 1 & -1 & 1 \\ -1 & -1 & -1 \\ 1 & -1 & 1 \end{pmatrix} = \begin{pmatrix} 1-1 & 0-(-1) & 1-1 \\ -1-(-1) & -1-(-1) & 1-(-1) \\ 2-1 & -1-(-1) & 0-1 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \\ 1 & 0 & -1 \end{pmatrix}

Ahora, la ecuación es $D \cdot X = C$ . Para resolverla, necesitamos la inversa de $D$ , $D^{-1}$ . Primero, calculamos el determinante de $D$ para verificar si es invertible:

\det(D) = 0 \cdot \det\begin{pmatrix} 0 & 2 \\ 0 & -1 \end{pmatrix} - 1 \cdot \det\begin{pmatrix} 0 & 2 \\ 1 & -1 \end{pmatrix} + 0 \cdot \det\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}

\det(D) = 0 - 1(0 - 2) + 0 = -1(-2) = 2

Dado que $\det(D) = 2 \neq 0$ , la matriz $D$ es invertible.Calculamos la matriz de cofactores de $D$ :

C_{11} = 0 \quad C_{12} = -(-2) = 2 \quad C_{13} = 0

C_{21} = -(-1) = 1 \quad C_{22} = 0 \quad C_{23} = -(-1) = 1

C_{31} = 2 \quad C_{32} = -0 = 0 \quad C_{33} = 0

\text{cof}(D) = \begin{pmatrix} 0 & 2 & 0 \\ 1 & 0 & 1 \\ 2 & 0 & 0 \end{pmatrix}

La matriz adjunta es la transpuesta de la matriz de cofactores:

\text{adj}(D) = (\text{cof}(D))^t = \begin{pmatrix} 0 & 1 & 2 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}

La matriz inversa $D^{-1}$ es:

D^{-1} = \frac{1}{\det(D)} \text{adj}(D) = \frac{1}{2} \begin{pmatrix} 0 & 1 & 2 \\ 2 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1/2 & 1 \\ 1 & 0 & 0 \\ 0 & 1/2 & 0 \end{pmatrix}

Finalmente, resolvemos para $X = D^{-1} \cdot C$ :

X = \begin{pmatrix} 0 & 1/2 & 1 \\ 1 & 0 & 0 \\ 0 & 1/2 & 0 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -7 \\ -2 \end{pmatrix}

X = \begin{pmatrix} (0)(3) + (1/2)(-7) + (1)(-2) \\ (1)(3) + (0)(-7) + (0)(-2) \\ (0)(3) + (1/2)(-7) + (0)(-2) \end{pmatrix}

X = \begin{pmatrix} 0 - 7/2 - 2 \\ 3 + 0 + 0 \\ 0 - 7/2 + 0 \end{pmatrix} = \begin{pmatrix} -11/2 \\ 3 \\ -7/2 \end{pmatrix}