a) Determine los valores de a a a para que la matriz A A A sea no invertible. Una matriz es no invertible (o singular) si su determinante es igual a cero. Calculamos el determinante de la matriz A A A :
det ( A ) = ∣ a 2 0 8 a 0 0 0 a ∣ \det(A) = \begin{vmatrix} a & 2 & 0 \\ 8 & a & 0 \\ 0 & 0 & a \end{vmatrix} det ( A ) = a 8 0 2 a 0 0 0 a det ( A ) = a ⋅ ∣ a 0 0 a ∣ − 2 ⋅ ∣ 8 0 0 a ∣ + 0 ⋅ ∣ 8 a 0 0 ∣ \det(A) = a \cdot \begin{vmatrix} a & 0 \\ 0 & a \end{vmatrix} - 2 \cdot \begin{vmatrix} 8 & 0 \\ 0 & a \end{vmatrix} + 0 \cdot \begin{vmatrix} 8 & a \\ 0 & 0 \end{vmatrix} det ( A ) = a ⋅ a 0 0 a − 2 ⋅ 8 0 0 a + 0 ⋅ 8 0 a 0 det ( A ) = a ( a ⋅ a − 0 ⋅ 0 ) − 2 ( 8 ⋅ a − 0 ⋅ 0 ) + 0 \det(A) = a(a \cdot a - 0 \cdot 0) - 2(8 \cdot a - 0 \cdot 0) + 0 det ( A ) = a ( a ⋅ a − 0 ⋅ 0 ) − 2 ( 8 ⋅ a − 0 ⋅ 0 ) + 0 \det(A) = a(a^2) - 2(8a) = a^3 - 16a
Ahora, igualamos el determinante a cero para encontrar los valores de a a a :
a 3 − 16 a = 0 a^3 - 16a = 0 a 3 − 16 a = 0 a ( a 2 − 16 ) = 0 a(a^2 - 16) = 0 a ( a 2 − 16 ) = 0 a ( a − 4 ) ( a + 4 ) = 0 a(a - 4)(a + 4) = 0 a ( a − 4 ) ( a + 4 ) = 0 Los valores de a a a para los cuales la matriz A A A no es invertible son a = 0 a = 0 a = 0 , a = 4 a = 4 a = 4 y a = − 4 a = -4 a = − 4 .
b) Para a = 5 a = 5 a = 5 , calcule la inversa de la matriz A A A . Para a = 5 a = 5 a = 5 , la matriz A A A es:
A = ( 5 2 0 8 5 0 0 0 5 ) A = \begin{pmatrix} 5 & 2 & 0 \\ 8 & 5 & 0 \\ 0 & 0 & 5 \end{pmatrix} A = 5 8 0 2 5 0 0 0 5 El determinante para a = 5 a=5 a = 5 es:
det ( A ) = 5 3 − 16 ( 5 ) = 125 − 80 = 45 \det(A) = 5^3 - 16(5) = 125 - 80 = 45 det ( A ) = 5 3 − 16 ( 5 ) = 125 − 80 = 45 Calculamos la matriz de cofactores C C C :
C 11 = ∣ 5 0 0 5 ∣ = 25 C 12 = − ∣ 8 0 0 5 ∣ = − 40 C 13 = ∣ 8 5 0 0 ∣ = 0 C_{11} = \begin{vmatrix} 5 & 0 \\ 0 & 5 \end{vmatrix} = 25 \quad C_{12} = -\begin{vmatrix} 8 & 0 \\ 0 & 5 \end{vmatrix} = -40 \quad C_{13} = \begin{vmatrix} 8 & 5 \\ 0 & 0 \end{vmatrix} = 0 C 11 = 5 0 0 5 = 25 C 12 = − 8 0 0 5 = − 40 C 13 = 8 0 5 0 = 0 C 21 = − ∣ 2 0 0 5 ∣ = − 10 C 22 = ∣ 5 0 0 5 ∣ = 25 C 23 = − ∣ 5 2 0 0 ∣ = 0 C_{21} = -\begin{vmatrix} 2 & 0 \\ 0 & 5 \end{vmatrix} = -10 \quad C_{22} = \begin{vmatrix} 5 & 0 \\ 0 & 5 \end{vmatrix} = 25 \quad C_{23} = -\begin{vmatrix} 5 & 2 \\ 0 & 0 \end{vmatrix} = 0 C 21 = − 2 0 0 5 = − 10 C 22 = 5 0 0 5 = 25 C 23 = − 5 0 2 0 = 0 C 31 = ∣ 2 0 5 0 ∣ = 0 C 32 = − ∣ 5 0 8 0 ∣ = 0 C 33 = ∣ 5 2 8 5 ∣ = 25 − 16 = 9 C_{31} = \begin{vmatrix} 2 & 0 \\ 5 & 0 \end{vmatrix} = 0 \quad C_{32} = -\begin{vmatrix} 5 & 0 \\ 8 & 0 \end{vmatrix} = 0 \quad C_{33} = \begin{vmatrix} 5 & 2 \\ 8 & 5 \end{vmatrix} = 25 - 16 = 9 C 31 = 2 5 0 0 = 0 C 32 = − 5 8 0 0 = 0 C 33 = 5 8 2 5 = 25 − 16 = 9 La matriz de cofactores es:
C = ( 25 − 40 0 − 10 25 0 0 0 9 ) C = \begin{pmatrix} 25 & -40 & 0 \\ -10 & 25 & 0 \\ 0 & 0 & 9 \end{pmatrix} C = 25 − 10 0 − 40 25 0 0 0 9 La matriz adjunta es la traspuesta de la matriz de cofactores:
adj ( A ) = C T = ( 25 − 10 0 − 40 25 0 0 0 9 ) \text{adj}(A) = C^T = \begin{pmatrix} 25 & -10 & 0 \\ -40 & 25 & 0 \\ 0 & 0 & 9 \end{pmatrix} adj ( A ) = C T = 25 − 40 0 − 10 25 0 0 0 9 Finalmente, la matriz inversa A − 1 A^{-1} A − 1 se calcula como 1 det ( A ) ⋅ adj ( A ) \frac{1}{\det(A)} \cdot \text{adj}(A) d e t ( A ) 1 ⋅ adj ( A ) :
A − 1 = 1 45 ( 25 − 10 0 − 40 25 0 0 0 9 ) A^{-1} = \frac{1}{45} \begin{pmatrix} 25 & -10 & 0 \\ -40 & 25 & 0 \\ 0 & 0 & 9 \end{pmatrix} A − 1 = 45 1 25 − 40 0 − 10 25 0 0 0 9 A − 1 = ( 25 / 45 − 10 / 45 0 / 45 − 40 / 45 25 / 45 0 / 45 0 / 45 0 / 45 9 / 45 ) = ( 5 / 9 − 2 / 9 0 − 8 / 9 5 / 9 0 0 0 1 / 5 ) A^{-1} = \begin{pmatrix} 25/45 & -10/45 & 0/45 \\ -40/45 & 25/45 & 0/45 \\ 0/45 & 0/45 & 9/45 \end{pmatrix} = \begin{pmatrix} 5/9 & -2/9 & 0 \\ -8/9 & 5/9 & 0 \\ 0 & 0 & 1/5 \end{pmatrix} A − 1 = 25/45 − 40/45 0/45 − 10/45 25/45 0/45 0/45 0/45 9/45 = 5/9 − 8/9 0 − 2/9 5/9 0 0 0 1/5 c) Para a = 5 a = 5 a = 5 , resuelva la ecuación matricial A ⋅ X = B A \cdot X = B A ⋅ X = B . Para resolver la ecuación A ⋅ X = B A \cdot X = B A ⋅ X = B , podemos multiplicar por la izquierda por la matriz inversa A − 1 A^{-1} A − 1 (calculada en el apartado anterior, ya que a = 5 a=5 a = 5 ):
A − 1 ⋅ A ⋅ X = A − 1 ⋅ B A^{-1} \cdot A \cdot X = A^{-1} \cdot B A − 1 ⋅ A ⋅ X = A − 1 ⋅ B I ⋅ X = A − 1 ⋅ B I \cdot X = A^{-1} \cdot B I ⋅ X = A − 1 ⋅ B X = A − 1 ⋅ B X = A^{-1} \cdot B X = A − 1 ⋅ B Sustituimos A − 1 A^{-1} A − 1 y B B B :
X = ( 5 / 9 − 2 / 9 0 − 8 / 9 5 / 9 0 0 0 1 / 5 ) ( 1 − 2 10 ) X = \begin{pmatrix} 5/9 & -2/9 & 0 \\ -8/9 & 5/9 & 0 \\ 0 & 0 & 1/5 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 10 \end{pmatrix} X = 5/9 − 8/9 0 − 2/9 5/9 0 0 0 1/5 1 − 2 10 X = ( ( 5 / 9 ) ( 1 ) + ( − 2 / 9 ) ( − 2 ) + ( 0 ) ( 10 ) ( − 8 / 9 ) ( 1 ) + ( 5 / 9 ) ( − 2 ) + ( 0 ) ( 10 ) ( 0 ) ( 1 ) + ( 0 ) ( − 2 ) + ( 1 / 5 ) ( 10 ) ) X = \begin{pmatrix} (5/9)(1) + (-2/9)(-2) + (0)(10) \\ (-8/9)(1) + (5/9)(-2) + (0)(10) \\ (0)(1) + (0)(-2) + (1/5)(10) \end{pmatrix} X = ( 5/9 ) ( 1 ) + ( − 2/9 ) ( − 2 ) + ( 0 ) ( 10 ) ( − 8/9 ) ( 1 ) + ( 5/9 ) ( − 2 ) + ( 0 ) ( 10 ) ( 0 ) ( 1 ) + ( 0 ) ( − 2 ) + ( 1/5 ) ( 10 ) X = ( 5 / 9 + 4 / 9 + 0 − 8 / 9 − 10 / 9 + 0 0 + 0 + 10 / 5 ) X = \begin{pmatrix} 5/9 + 4/9 + 0 \\ -8/9 - 10/9 + 0 \\ 0 + 0 + 10/5 \end{pmatrix} X = 5/9 + 4/9 + 0 − 8/9 − 10/9 + 0 0 + 0 + 10/5 X = ( 9 / 9 − 18 / 9 2 ) = ( 1 − 2 2 ) X = \begin{pmatrix} 9/9 \\ -18/9 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix} X = 9/9 − 18/9 2 = 1 − 2 2 
