Matrices Selectividad

castilla_mancha ecuacion_matricial Ejercicios_Resueltos matrices selectividad

– Despeja la matriz $X$ en la ecuación $A \cdot X - X = B \cdot X + C$

– Halla la matriz $X$ sabiendo que
$A = \left( \begin{array}{ccc} 1 & 1 & 0\\ 1 & 0 & 1\\ 1 & 1 & 1 \end{array} \right)$

$B = \left( \begin{array}{ccc} 2 & 0 & 0\\ -1 & 1 & 2 \\ 0 & 0 & 1 \end{array} \right)$

$C = \left( \begin{array}{ccc} -2 & 2 & 0\\ 2 & -4 & -3 \\ 1 & 2 & -3 \end{array} \right)$

SOLUCIÓN

Matrices:

$A = \begin{pmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{pmatrix}\qquad B = \begin{pmatrix}2 & 0 & 0 \\ -1 & 1 & 2 \\ 0 & 0 & 1\end{pmatrix}\qquad C = \begin{pmatrix}-2 & 2 & 0 \\ 2 & -4 & -3 \\ 1 & 2 & -3\end{pmatrix}$

Ecuación:

$AX-X = BX+C$

Despejamos $X$ :

$(A-I-B)\cdot X = C$

Multiplicamos a la izquierda por $(A-I-B)^{-1}$ en ambos miembros:

$(A-I-B)^{-1}\cdot (A-I-B)\cdot X = (A-I-B)^{-1}\cdot C$

$I\cdot X = (A-I-B)^{-1}\cdot C$

$\boxed{X = (A-I-B)^{-1}\cdot C}$

Calculamos numéricamente:

Suma de coeficientes de $X$ :

Calculamos $(A-I-B)$ :

$(A-I-B) = \begin{pmatrix}1 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1\end{pmatrix}-\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}-\begin{pmatrix}2 & 0 & 0 \\ -1 & 1 & 2 \\ 0 & 0 & 1\end{pmatrix} = \begin{pmatrix}-2 & 1 & 0 \\ 2 & -2 & -1 \\ 1 & 1 & -1\end{pmatrix}$

Calculamos $(A-I-B)^{-1}$ (Gauss-Jordan):

Matriz ampliada $[M|I]$ :

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

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

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

$F_2=F_2-2\cdot F_1$

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

$F_3=F_3-F_1$

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

$F_2=-1\cdot F_2$

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

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

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

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

$\left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{2} & -1 & -\frac{1}{2} & 0 \\ 0 & 1 & 1 & -1 & -1 & 0 \\ 0 & 0 & -\frac{5}{2} & 2 & \frac{3}{2} & 1\end{array}\right]$

$F_3=-\dfrac{2}{5}\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & \frac{1}{2} & -1 & -\frac{1}{2} & 0 \\ 0 & 1 & 1 & -1 & -1 & 0 \\ 0 & 0 & 1 & -\frac{4}{5} & -\frac{3}{5} & -\frac{2}{5}\end{array}\right]$

$F_1=F_1-\frac{1}{2}\cdot F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & -\frac{3}{5} & -\frac{1}{5} & \frac{1}{5} \\ 0 & 1 & 1 & -1 & -1 & 0 \\ 0 & 0 & 1 & -\frac{4}{5} & -\frac{3}{5} & -\frac{2}{5}\end{array}\right]$

$F_2=F_2-F_3$

$\left[\begin{array}{ccc|ccc}1 & 0 & 0 & -\frac{3}{5} & -\frac{1}{5} & \frac{1}{5} \\ 0 & 1 & 0 & -\frac{1}{5} & -\frac{2}{5} & \frac{2}{5} \\ 0 & 0 & 1 & -\frac{4}{5} & -\frac{3}{5} & -\frac{2}{5}\end{array}\right]$

$(A-I-B)^{-1} = \begin{pmatrix}-\frac{3}{5} & -\frac{1}{5} & \frac{1}{5} \\ -\frac{1}{5} & -\frac{2}{5} & \frac{2}{5} \\ -\frac{4}{5} & -\frac{3}{5} & -\frac{2}{5}\end{pmatrix}$

Calculamos $X = (A-I-B)^{-1}\cdot C$ :

$X = \begin{pmatrix}-\frac{3}{5} & -\frac{1}{5} & \frac{1}{5} \\ -\frac{1}{5} & -\frac{2}{5} & \frac{2}{5} \\ -\frac{4}{5} & -\frac{3}{5} & -\frac{2}{5}\end{pmatrix}\begin{pmatrix}-2 & 2 & 0 \\ 2 & -4 & -3 \\ 1 & 2 & -3\end{pmatrix}$

$X_{11} = (-\frac{3}{5})\cdot (-2) + (-\frac{1}{5})\cdot 2 + \frac{1}{5}\cdot 1 = {\color{blue}1}$

$X_{12} = (-\frac{3}{5})\cdot 2 + (-\frac{1}{5})\cdot (-4) + \frac{1}{5}\cdot 2 = {\color{blue}0}$

$X_{13} = (-\frac{3}{5})\cdot 0 + (-\frac{1}{5})\cdot (-3) + \frac{1}{5}\cdot (-3) = {\color{blue}0}$

$X_{21} = (-\frac{1}{5})\cdot (-2) + (-\frac{2}{5})\cdot 2 + \frac{2}{5}\cdot 1 = {\color{blue}0}$

$X_{22} = (-\frac{1}{5})\cdot 2 + (-\frac{2}{5})\cdot (-4) + \frac{2}{5}\cdot 2 = {\color{blue}2}$

$X_{23} = (-\frac{1}{5})\cdot 0 + (-\frac{2}{5})\cdot (-3) + \frac{2}{5}\cdot (-3) = {\color{blue}0}$

$X_{31} = (-\frac{4}{5})\cdot (-2) + (-\frac{3}{5})\cdot 2 + (-\frac{2}{5})\cdot 1 = {\color{blue}0}$

$X_{32} = (-\frac{4}{5})\cdot 2 + (-\frac{3}{5})\cdot (-4) + (-\frac{2}{5})\cdot 2 = {\color{blue}0}$

$X_{33} = (-\frac{4}{5})\cdot 0 + (-\frac{3}{5})\cdot (-3) + (-\frac{2}{5})\cdot (-3) = {\color{blue}3}$

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

Resultado:

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

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