Ejercicio de Matrices

Ejercicios_Resueltos matrices matriz inversa producto_matrices

Sean las matrices
$A = \left( \begin{array}{ccc} -3 & 2 & 2 \\ 1 & -1 & 0 \\ 0 & 1 & 0 \end{array} \right)$ $\qquad$
$B = \left( \begin{array}{ccc} 2 & 1 & 0 \\ -1 & 1 & -1 \\ 2 & 0 & 1 \end{array} \right)$
– Halla $A^{-1}$ y $B^{-1}$
– Calcula la inversa de $A \cdot B$
– Comprueba que $(A \cdot B)^{-1} = B^{-1} \cdot A^{-1}$

SOLUCIÓN

Matriz inversa

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

$\det(A) = 2$

$\det(A) = 2 \neq 0 \implies \exists\, A^{-1}$

Método de Gauss-Jordan

Matriz ampliada A :

$\left[\begin{array}{ccc|ccc}-3 & 2 & 2 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$

$F_1 = -\dfrac{1}{3}\cdot F_1$

$\left[\begin{array}{ccc|ccc}1 & -\frac{2}{3} & -\frac{2}{3} & -\frac{1}{3} & 0 & 0 \\ 1 & -1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$

$F_2 = F_2 - F_1$

$\left[\begin{array}{ccc|ccc}1 & -\frac{2}{3} & -\frac{2}{3} & -\frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{3} & \frac{2}{3} & \frac{1}{3} & 1 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$

$F_2 = -3\cdot F_2$

$\left[\begin{array}{ccc|ccc}1 & -\frac{2}{3} & -\frac{2}{3} & -\frac{1}{3} & 0 & 0 \\ 0 & 1 & -2 & -1 & -3 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$

$F_1 = F_1 + \frac{2}{3}\cdot F_2$

$\left[\begin{array}{ccc|ccc}1 & 0 & -2 & -1 & -2 & 0 \\ 0 & 1 & -2 & -1 & -3 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1\end{array}\right]$

$F_3 = F_3 - F_2$

$\left[\begin{array}{ccc|ccc}1 & 0 & -2 & -1 & -2 & 0 \\ 0 & 1 & -2 & -1 & -3 & 0 \\ 0 & 0 & 2 & 1 & 3 & 1\end{array}\right]$

$F_3 = \dfrac{1}{2}\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & -2 & -1 & -2 & 0 \\ 0 & 1 & -2 & -1 & -3 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{array}\right]$

$F_1 = F_1 + 2\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & -2 & -1 & -3 & 0 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{array}\right]$

$F_2 = F_2 + 2\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & \frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{array}\right]$

$A^{-1} = \begin{pmatrix}0 & 1 & 1 \\ 0 & 0 & 1 \\ \frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{pmatrix}$

Usando determinantes: A⁻¹ = (1/|A|) · Adj(A)

$A^{-1}=\dfrac{1}{|A|}\cdot\left(\text{Adj}\,A\right)^t$

Paso 1 · Determinante de A:

$|A| = 2$

Paso 2 · Cofactores:

$C_{11}=(-1)^{1+1}\cdot\left|\begin{array}{cc}-1 & 0 \\ 1 & 0\end{array}\right|=(+1)\cdot0=0$

$C_{12}=(-1)^{1+2}\cdot\left|\begin{array}{cc}1 & 0 \\ 0 & 0\end{array}\right|=(-1)\cdot0=0$

$C_{13}=(-1)^{1+3}\cdot\left|\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right|=(+1)\cdot1=1$

$C_{21}=(-1)^{2+1}\cdot\left|\begin{array}{cc}2 & 2 \\ 1 & 0\end{array}\right|=(-1)\cdot-2=2$

$C_{22}=(-1)^{2+2}\cdot\left|\begin{array}{cc}-3 & 2 \\ 0 & 0\end{array}\right|=(+1)\cdot0=0$

$C_{23}=(-1)^{2+3}\cdot\left|\begin{array}{cc}-3 & 2 \\ 0 & 1\end{array}\right|=(-1)\cdot-3=3$

$C_{31}=(-1)^{3+1}\cdot\left|\begin{array}{cc}2 & 2 \\ -1 & 0\end{array}\right|=(+1)\cdot2=2$

$C_{32}=(-1)^{3+2}\cdot\left|\begin{array}{cc}-3 & 2 \\ 1 & 0\end{array}\right|=(-1)\cdot-2=2$

$C_{33}=(-1)^{3+3}\cdot\left|\begin{array}{cc}-3 & 2 \\ 1 & -1\end{array}\right|=(+1)\cdot1=1$

Paso 3 · Matriz de cofactores:

$C = \begin{pmatrix}0 & 0 & 1 \\ 2 & 0 & 3 \\ 2 & 2 & 1\end{pmatrix}$

Paso 4 · Adjunta (traspuesta de C):

$\text{Adj}\,A = C^t = \begin{pmatrix}0 & 2 & 2 \\ 0 & 0 & 2 \\ 1 & 3 & 1\end{pmatrix}$

Paso 5 · Inversa:

$A^{-1}=\dfrac{1}{2}\begin{pmatrix}0 & 2 & 2 \\ 0 & 0 & 2 \\ 1 & 3 & 1\end{pmatrix}=\begin{pmatrix}0 & 1 & 1 \\ 0 & 0 & 1 \\ \frac{1}{2} & \frac{3}{2} & \frac{1}{2}\end{pmatrix}$

Matriz inversa

$A = \begin{pmatrix}2 & 1 & 0 \\ -1 & 1 & -1 \\ 2 & 0 & 1\end{pmatrix}$

$\det(A) = 1$

$\det(A) = 1 \neq 0 \implies \exists\, A^{-1}$

Método de Gauss-Jordan

Matriz ampliada A :

$\left[\begin{array}{ccc|ccc}2 & 1 & 0 & 1 & 0 & 0 \\ -1 & 1 & -1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 0 & 0 & 1\end{array}\right]$

$F_1 = \dfrac{1}{2}\cdot F_1$

$\left[\begin{array}{ccc|ccc}1 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ -1 & 1 & -1 & 0 & 1 & 0 \\ 2 & 0 & 1 & 0 & 0 & 1\end{array}\right]$

$F_2 = F_2 + F_1$

$\left[\begin{array}{ccc|ccc}1 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\ 2 & 0 & 1 & 0 & 0 & 1\end{array}\right]$

$F_3 = F_3 - 2\cdot F_1$

$\left[\begin{array}{ccc|ccc}1 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & \frac{3}{2} & -1 & \frac{1}{2} & 1 & 0 \\ 0 & -1 & 1 & -1 & 0 & 1\end{array}\right]$

$F_2 = \dfrac{2}{3}\cdot F_2$

$\left[\begin{array}{ccc|ccc}1 & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & -1 & 0 & 1\end{array}\right]$

$F_1 = F_1 - \frac{1}{2}\cdot F_2$

$\left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & -1 & 1 & -1 & 0 & 1\end{array}\right]$

$F_3 = F_3 + F_2$

$\left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & \frac{1}{3} & -\frac{2}{3} & \frac{2}{3} & 1\end{array}\right]$

$F_3 = 3\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{3} & \frac{1}{3} & -\frac{1}{3} & 0 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & -2 & 2 & 3\end{array}\right]$

$F_1 = F_1 - \frac{1}{3}\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & -\frac{2}{3} & \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 & -2 & 2 & 3\end{array}\right]$

$F_2 = F_2 + \frac{2}{3}\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & -1 & -1 \\ 0 & 1 & 0 & -1 & 2 & 2 \\ 0 & 0 & 1 & -2 & 2 & 3\end{array}\right]$

$A^{-1} = \begin{pmatrix}1 & -1 & -1 \\ -1 & 2 & 2 \\ -2 & 2 & 3\end{pmatrix}$

Usando determinantes: A⁻¹ = (1/|A|) · Adj(A)

$A^{-1}=\dfrac{1}{|A|}\cdot\left(\text{Adj}\,A\right)^t$

Paso 1 · Determinante de A:

$|A| = 1$

Paso 2 · Cofactores:

$C_{11}=(-1)^{1+1}\cdot\left|\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right|=(+1)\cdot1=1$

$C_{12}=(-1)^{1+2}\cdot\left|\begin{array}{cc}-1 & -1 \\ 2 & 1\end{array}\right|=(-1)\cdot1=-1$

$C_{13}=(-1)^{1+3}\cdot\left|\begin{array}{cc}-1 & 1 \\ 2 & 0\end{array}\right|=(+1)\cdot-2=-2$

$C_{21}=(-1)^{2+1}\cdot\left|\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right|=(-1)\cdot1=-1$

$C_{22}=(-1)^{2+2}\cdot\left|\begin{array}{cc}2 & 0 \\ 2 & 1\end{array}\right|=(+1)\cdot2=2$

$C_{23}=(-1)^{2+3}\cdot\left|\begin{array}{cc}2 & 1 \\ 2 & 0\end{array}\right|=(-1)\cdot-2=2$

$C_{31}=(-1)^{3+1}\cdot\left|\begin{array}{cc}1 & 0 \\ 1 & -1\end{array}\right|=(+1)\cdot-1=-1$

$C_{32}=(-1)^{3+2}\cdot\left|\begin{array}{cc}2 & 0 \\ -1 & -1\end{array}\right|=(-1)\cdot-2=2$

$C_{33}=(-1)^{3+3}\cdot\left|\begin{array}{cc}2 & 1 \\ -1 & 1\end{array}\right|=(+1)\cdot3=3$

Paso 3 · Matriz de cofactores:

$C = \begin{pmatrix}1 & -1 & -2 \\ -1 & 2 & 2 \\ -1 & 2 & 3\end{pmatrix}$

Paso 4 · Adjunta (traspuesta de C):

$\text{Adj}\,A = C^t = \begin{pmatrix}1 & -1 & -1 \\ -1 & 2 & 2 \\ -2 & 2 & 3\end{pmatrix}$

Paso 5 · Inversa:

$A^{-1}=\dfrac{1}{1}\begin{pmatrix}1 & -1 & -1 \\ -1 & 2 & 2 \\ -2 & 2 & 3\end{pmatrix}=\begin{pmatrix}1 & -1 & -1 \\ -1 & 2 & 2 \\ -2 & 2 & 3\end{pmatrix}$

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