a) Obtenga los valores de a a a para los que la matriz A A A tenga inversa. La matriz A A A tiene inversa si y solo si su determinante es distinto de cero. Calculamos el determinante de A A A :
A = ( 1 1 − 2 a − 3 a − 1 1 0 2 a ) A = \begin{pmatrix} 1 & 1 & -2 \\ a - 3 & a - 1 & 1 \\ 0 & 2 & a \end{pmatrix} A = 1 a − 3 0 1 a − 1 2 − 2 1 a det ( A ) = 1 ⋅ ∣ a − 1 1 2 a ∣ − 1 ⋅ ∣ a − 3 1 0 a ∣ + ( − 2 ) ⋅ ∣ a − 3 a − 1 0 2 ∣ \det(A) = 1 \cdot \begin{vmatrix} a-1 & 1 \\ 2 & a \end{vmatrix} - 1 \cdot \begin{vmatrix} a-3 & 1 \\ 0 & a \end{vmatrix} + (-2) \cdot \begin{vmatrix} a-3 & a-1 \\ 0 & 2 \end{vmatrix} det ( A ) = 1 ⋅ a − 1 2 1 a − 1 ⋅ a − 3 0 1 a + ( − 2 ) ⋅ a − 3 0 a − 1 2 det ( A ) = 1 ⋅ ( ( a − 1 ) a − 1 ⋅ 2 ) − 1 ⋅ ( ( a − 3 ) a − 1 ⋅ 0 ) − 2 ⋅ ( 2 ( a − 3 ) − ( a − 1 ) ⋅ 0 ) \det(A) = 1 \cdot ((a-1)a - 1 \cdot 2) - 1 \cdot ((a-3)a - 1 \cdot 0) - 2 \cdot (2(a-3) - (a-1) \cdot 0) det ( A ) = 1 ⋅ (( a − 1 ) a − 1 ⋅ 2 ) − 1 ⋅ (( a − 3 ) a − 1 ⋅ 0 ) − 2 ⋅ ( 2 ( a − 3 ) − ( a − 1 ) ⋅ 0 ) \det(A) = (a^2 - a - 2) - (a^2 - 3a) - 2(2a - 6)
det ( A ) = a 2 − a − 2 − a 2 + 3 a − 4 a + 12 \det(A) = a^2 - a - 2 - a^2 + 3a - 4a + 12 det ( A ) = a 2 − a − 2 − a 2 + 3 a − 4 a + 12 \det(A) = (a^2 - a^2) + (-a + 3a - 4a) + (-2 + 12)
det ( A ) = − 2 a + 10 \det(A) = -2a + 10 det ( A ) = − 2 a + 10 Para que la matriz A A A tenga inversa, su determinante debe ser distinto de cero:
− 2 a + 10 ≠ 0 -2a + 10 \neq 0 − 2 a + 10 = 0 − 2 a ≠ − 10 -2a \neq -10 − 2 a = − 10 Por lo tanto, la matriz A A A tiene inversa para todos los valores de a ∈ R a \in \mathbb{R} a ∈ R excepto para a = 5 a=5 a = 5 .
b) Para a = 1 a = 1 a = 1 , resuelva la ecuación X ⋅ A − B = C ⋅ A X \cdot A - B = C \cdot A X ⋅ A − B = C ⋅ A . Sustituimos a = 1 a=1 a = 1 en la matriz A A A :
A = ( 1 1 − 2 1 − 3 1 − 1 1 0 2 1 ) = ( 1 1 − 2 − 2 0 1 0 2 1 ) A = \begin{pmatrix} 1 & 1 & -2 \\ 1 - 3 & 1 - 1 & 1 \\ 0 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 1 & -2 \\ -2 & 0 & 1 \\ 0 & 2 & 1 \end{pmatrix} A = 1 1 − 3 0 1 1 − 1 2 − 2 1 1 = 1 − 2 0 1 0 2 − 2 1 1 Dado que a = 1 ≠ 5 a=1 \neq 5 a = 1 = 5 , la matriz A A A tiene inversa. Reordenamos la ecuación para despejar X X X :
X ⋅ A − B = C ⋅ A X \cdot A - B = C \cdot A X ⋅ A − B = C ⋅ A X ⋅ A = C ⋅ A + B X \cdot A = C \cdot A + B X ⋅ A = C ⋅ A + B Multiplicamos por la derecha por A − 1 A^{-1} A − 1 :
X ⋅ A ⋅ A − 1 = ( C ⋅ A + B ) ⋅ A − 1 X \cdot A \cdot A^{-1} = (C \cdot A + B) \cdot A^{-1} X ⋅ A ⋅ A − 1 = ( C ⋅ A + B ) ⋅ A − 1 X = C ⋅ A ⋅ A − 1 + B ⋅ A − 1 X = C \cdot A \cdot A^{-1} + B \cdot A^{-1} X = C ⋅ A ⋅ A − 1 + B ⋅ A − 1 X = C + B ⋅ A − 1 X = C + B \cdot A^{-1} X = C + B ⋅ A − 1 Calculamos A − 1 A^{-1} A − 1 para a = 1 a=1 a = 1 . El determinante de A A A para a = 1 a=1 a = 1 es:
det ( A ) = − 2 ( 1 ) + 10 = 8 \det(A) = -2(1) + 10 = 8 det ( A ) = − 2 ( 1 ) + 10 = 8 Calculamos la matriz de cofactores de A A A :
Cof ( A ) 11 = ∣ 0 1 2 1 ∣ = − 2 Cof ( A ) 12 = − ∣ − 2 1 0 1 ∣ = 2 Cof ( A ) 13 = ∣ − 2 0 0 2 ∣ = − 4 \text{Cof}(A)_{11} = \begin{vmatrix} 0 & 1 \\ 2 & 1 \end{vmatrix} = -2 \qquad \text{Cof}(A)_{12} = -\begin{vmatrix} -2 & 1 \\ 0 & 1 \end{vmatrix} = 2 \qquad \text{Cof}(A)_{13} = \begin{vmatrix} -2 & 0 \\ 0 & 2 \end{vmatrix} = -4 Cof ( A ) 11 = 0 2 1 1 = − 2 Cof ( A ) 12 = − − 2 0 1 1 = 2 Cof ( A ) 13 = − 2 0 0 2 = − 4 Cof ( A ) 21 = − ∣ 1 − 2 2 1 ∣ = − ( 1 − ( − 4 ) ) = − 5 Cof ( A ) 22 = ∣ 1 − 2 0 1 ∣ = 1 Cof ( A ) 23 = − ∣ 1 1 0 2 ∣ = − ( 2 − 0 ) = − 2 \text{Cof}(A)_{21} = -\begin{vmatrix} 1 & -2 \\ 2 & 1 \end{vmatrix} = -(1 - (-4)) = -5 \qquad \text{Cof}(A)_{22} = \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1 \qquad \text{Cof}(A)_{23} = -\begin{vmatrix} 1 & 1 \\ 0 & 2 \end{vmatrix} = -(2 - 0) = -2 Cof ( A ) 21 = − 1 2 − 2 1 = − ( 1 − ( − 4 )) = − 5 Cof ( A ) 22 = 1 0 − 2 1 = 1 Cof ( A ) 23 = − 1 0 1 2 = − ( 2 − 0 ) = − 2 Cof ( A ) 31 = ∣ 1 − 2 0 1 ∣ = 1 Cof ( A ) 32 = − ∣ 1 − 2 − 2 1 ∣ = − ( 1 − 4 ) = 3 Cof ( A ) 33 = ∣ 1 1 − 2 0 ∣ = 0 − ( − 2 ) = 2 \text{Cof}(A)_{31} = \begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} = 1 \qquad \text{Cof}(A)_{32} = -\begin{vmatrix} 1 & -2 \\ -2 & 1 \end{vmatrix} = -(1 - 4) = 3 \qquad \text{Cof}(A)_{33} = \begin{vmatrix} 1 & 1 \\ -2 & 0 \end{vmatrix} = 0 - (-2) = 2 Cof ( A ) 31 = 1 0 − 2 1 = 1 Cof ( A ) 32 = − 1 − 2 − 2 1 = − ( 1 − 4 ) = 3 Cof ( A ) 33 = 1 − 2 1 0 = 0 − ( − 2 ) = 2 Cof ( A ) = ( − 2 2 − 4 − 5 1 − 2 1 3 2 ) \text{Cof}(A) = \begin{pmatrix} -2 & 2 & -4 \\ -5 & 1 & -2 \\ 1 & 3 & 2 \end{pmatrix} Cof ( A ) = − 2 − 5 1 2 1 3 − 4 − 2 2 La matriz adjunta (transpuesta de la matriz de cofactores) es:
Adj ( A ) = Cof ( A ) t = ( − 2 − 5 1 2 1 3 − 4 − 2 2 ) \text{Adj}(A) = \text{Cof}(A)^t = \begin{pmatrix} -2 & -5 & 1 \\ 2 & 1 & 3 \\ -4 & -2 & 2 \end{pmatrix} Adj ( A ) = Cof ( A ) t = − 2 2 − 4 − 5 1 − 2 1 3 2 La inversa de A A A es:
A − 1 = 1 det ( A ) Adj ( A ) = 1 8 ( − 2 − 5 1 2 1 3 − 4 − 2 2 ) A^{-1} = \frac{1}{\det(A)} \text{Adj}(A) = \frac{1}{8} \begin{pmatrix} -2 & -5 & 1 \\ 2 & 1 & 3 \\ -4 & -2 & 2 \end{pmatrix} A − 1 = det ( A ) 1 Adj ( A ) = 8 1 − 2 2 − 4 − 5 1 − 2 1 3 2 Ahora calculamos B ⋅ A − 1 B \cdot A^{-1} B ⋅ A − 1 :
B ⋅ A − 1 = ( − 1 3 2 ) ⋅ 1 8 ( − 2 − 5 1 2 1 3 − 4 − 2 2 ) B \cdot A^{-1} = (-1 \quad 3 \quad 2) \cdot \frac{1}{8} \begin{pmatrix} -2 & -5 & 1 \\ 2 & 1 & 3 \\ -4 & -2 & 2 \end{pmatrix} B ⋅ A − 1 = ( − 1 3 2 ) ⋅ 8 1 − 2 2 − 4 − 5 1 − 2 1 3 2 B ⋅ A − 1 = 1 8 ( ( − 1 ) ( − 2 ) + 3 ( 2 ) + 2 ( − 4 ) ( − 1 ) ( − 5 ) + 3 ( 1 ) + 2 ( − 2 ) ( − 1 ) ( 1 ) + 3 ( 3 ) + 2 ( 2 ) ) B \cdot A^{-1} = \frac{1}{8} \left( (-1)(-2) + 3(2) + 2(-4) \quad (-1)(-5) + 3(1) + 2(-2) \quad (-1)(1) + 3(3) + 2(2) \right) B ⋅ A − 1 = 8 1 ( ( − 1 ) ( − 2 ) + 3 ( 2 ) + 2 ( − 4 ) ( − 1 ) ( − 5 ) + 3 ( 1 ) + 2 ( − 2 ) ( − 1 ) ( 1 ) + 3 ( 3 ) + 2 ( 2 ) ) B ⋅ A − 1 = 1 8 ( ( 2 + 6 − 8 ) ( 5 + 3 − 4 ) ( − 1 + 9 + 4 ) ) B \cdot A^{-1} = \frac{1}{8} \left( (2+6-8) \quad (5+3-4) \quad (-1+9+4) \right) B ⋅ A − 1 = 8 1 ( ( 2 + 6 − 8 ) ( 5 + 3 − 4 ) ( − 1 + 9 + 4 ) ) B ⋅ A − 1 = 1 8 ( 0 4 12 ) = ( 0 4 8 12 8 ) = ( 0 1 2 3 2 ) B \cdot A^{-1} = \frac{1}{8} \left( 0 \quad 4 \quad 12 \right) = (0 \quad \frac{4}{8} \quad \frac{12}{8}) = (0 \quad \frac{1}{2} \quad \frac{3}{2}) B ⋅ A − 1 = 8 1 ( 0 4 12 ) = ( 0 8 4 8 12 ) = ( 0 2 1 2 3 ) Finalmente, calculamos X = C + B ⋅ A − 1 X = C + B \cdot A^{-1} X = C + B ⋅ A − 1 :
X = ( − 2 1 4 ) + ( 0 1 2 3 2 ) X = (-2 \quad 1 \quad 4) + (0 \quad \frac{1}{2} \quad \frac{3}{2}) X = ( − 2 1 4 ) + ( 0 2 1 2 3 ) X = ( − 2 + 0 1 + 1 2 4 + 3 2 ) X = \left(-2+0 \quad 1+\frac{1}{2} \quad 4+\frac{3}{2}\right) X = ( − 2 + 0 1 + 2 1 4 + 2 3 ) X = ( − 2 2 + 1 2 8 + 3 2 ) X = \left(-2 \quad \frac{2+1}{2} \quad \frac{8+3}{2}\right) X = ( − 2 2 2 + 1 2 8 + 3 ) X = ( − 2 3 2 11 2 ) X = \left(-2 \quad \frac{3}{2} \quad \frac{11}{2}\right) X = ( − 2 2 3 2 11 ) c) Determine razonadamente la dimensión de la matriz D D D que permita realizar la operación B ⋅ A + D ⋅ C t ⋅ B B \cdot A + D \cdot C^t \cdot B B ⋅ A + D ⋅ C t ⋅ B . Analizamos las dimensiones de las matrices dadas:
A es de dimensi o ˊ n 3 × 3 A \text{ es de dimensión } 3 \times 3 A es de dimensi o ˊ n 3 × 3 B es de dimensi o ˊ n 1 × 3 B \text{ es de dimensión } 1 \times 3 B es de dimensi o ˊ n 1 × 3 C es de dimensi o ˊ n 1 × 3 C \text{ es de dimensión } 1 \times 3 C es de dimensi o ˊ n 1 × 3 Para que la operación B ⋅ A + D ⋅ C t ⋅ B B \cdot A + D \cdot C^t \cdot B B ⋅ A + D ⋅ C t ⋅ B sea posible, debemos seguir las reglas de multiplicación y suma de matrices. 1. Dimensión de B ⋅ A B \cdot A B ⋅ A :
dim ( B ) = 1 × 3 \text{dim}(B) = 1 \times 3 dim ( B ) = 1 × 3 dim ( A ) = 3 × 3 \text{dim}(A) = 3 \times 3 dim ( A ) = 3 × 3 El producto B ⋅ A B \cdot A B ⋅ A es posible y su dimensión es 1 × 3 1 \times 3 1 × 3 .
dim ( B ⋅ A ) = 1 × 3 \text{dim}(B \cdot A) = 1 \times 3 dim ( B ⋅ A ) = 1 × 3 2. Dimensión de C t C^t C t :
dim ( C ) = 1 × 3 ⟹ dim ( C t ) = 3 × 1 \text{dim}(C) = 1 \times 3 \implies \text{dim}(C^t) = 3 \times 1 dim ( C ) = 1 × 3 ⟹ dim ( C t ) = 3 × 1 3. Dimensión de C t ⋅ B C^t \cdot B C t ⋅ B :
dim ( C t ) = 3 × 1 \text{dim}(C^t) = 3 \times 1 dim ( C t ) = 3 × 1 dim ( B ) = 1 × 3 \text{dim}(B) = 1 \times 3 dim ( B ) = 1 × 3 El producto C t ⋅ B C^t \cdot B C t ⋅ B es posible y su dimensión es 3 × 3 3 \times 3 3 × 3 .
dim ( C t ⋅ B ) = 3 × 3 \text{dim}(C^t \cdot B) = 3 \times 3 dim ( C t ⋅ B ) = 3 × 3 4. Dimensión de D ⋅ C t ⋅ B D \cdot C^t \cdot B D ⋅ C t ⋅ B : Sea D D D una matriz de dimensión m × n m \times n m × n . Para que el producto D ⋅ ( C t ⋅ B ) D \cdot (C^t \cdot B) D ⋅ ( C t ⋅ B ) sea posible, el número de columnas de D D D debe ser igual al número de filas de ( C t ⋅ B ) (C^t \cdot B) ( C t ⋅ B ) .
Así, la matriz D D D tiene dimensión m × 3 m \times 3 m × 3 . El resultado del producto D ⋅ C t ⋅ B D \cdot C^t \cdot B D ⋅ C t ⋅ B tendrá dimensión m × 3 m \times 3 m × 3 .
dim ( D ⋅ C t ⋅ B ) = m × 3 \text{dim}(D \cdot C^t \cdot B) = m \times 3 dim ( D ⋅ C t ⋅ B ) = m × 3 5. Dimensión de la suma B ⋅ A + D ⋅ C t ⋅ B B \cdot A + D \cdot C^t \cdot B B ⋅ A + D ⋅ C t ⋅ B : Para que la suma de dos matrices sea posible, ambas deben tener la misma dimensión. Hemos determinado que:
dim ( B ⋅ A ) = 1 × 3 \text{dim}(B \cdot A) = 1 \times 3 dim ( B ⋅ A ) = 1 × 3 dim ( D ⋅ C t ⋅ B ) = m × 3 \text{dim}(D \cdot C^t \cdot B) = m \times 3 dim ( D ⋅ C t ⋅ B ) = m × 3 Por lo tanto, para que la suma sea posible, m m m debe ser igual a 1 1 1 .
Concluimos que la dimensión de la matriz D D D debe ser 1 × 3 1 \times 3 1 × 3 .
