Ecuaciones Matriciales

ecuacion_matricial Ejercicios_Resueltos matrices

Dadas las matrices:

$A = \left( \begin{array}{cc} 1 & 1 \\ 3 & 4 \end{array} \right)$ $\qquad$
$B = \left( \begin{array}{cc} 2 & 1 \\ 1 & 1 \end{array} \right)$ $\qquad$
$C = \left( \begin{array}{cc} 1 & 2 \\ 3 & 3 \end{array} \right)$

Resuelve la siguiente ecuación matricial:

– $XAB - XC = 2C$

SOLUCIÓN

Matrices:

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

Ecuación:

$XAB-XC = 2C$

Despejamos $X$ :

$X\cdot (AB-C) = 2C$

Multiplicamos a la derecha por $(AB-C)^{-1}$ en ambos miembros:

$X\cdot (AB-C)\cdot (AB-C)^{-1} = 2C\cdot (AB-C)^{-1}$

$X\cdot I = 2C\cdot (AB-C)^{-1}$

$\boxed{X = 2C\cdot (AB-C)^{-1}}$

Calculamos numéricamente:

Calculamos $2C$ :

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

Suma de coeficientes de $X$ :

Calculamos $AB$ :

$AB = \begin{pmatrix}1 & 1 \\ 3 & 4\end{pmatrix}\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix} = \begin{pmatrix}3 & 2 \\ 10 & 7\end{pmatrix}$

Calculamos $(AB-C)$ :

$(AB-C) = \begin{pmatrix}3 & 2 \\ 10 & 7\end{pmatrix}-\begin{pmatrix}1 & 2 \\ 3 & 3\end{pmatrix} = \begin{pmatrix}2 & 0 \\ 7 & 4\end{pmatrix}$

Calculamos $(AB-C)^{-1}$ (Gauss-Jordan):

Matriz ampliada $[M|I]$ :

$\left[\begin{array}{cc|cc}2 & 0 & 1 & 0 \\ 7 & 4 & 0 & 1\end{array}\right]$

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

$\left[\begin{array}{cc|cc}1 & 0 & \frac{1}{2} & 0 \\ 7 & 4 & 0 & 1\end{array}\right]$

$F_2=F_2-7\cdot F_1$

$\left[\begin{array}{cc|cc}1 & 0 & \frac{1}{2} & 0 \\ 0 & 4 & -\frac{7}{2} & 1\end{array}\right]$

$F_2=\dfrac{1}{4}\cdot F_2$

$\left[\begin{array}{cc|cc}1 & 0 & \frac{1}{2} & 0 \\ 0 & 1 & -\frac{7}{8} & \frac{1}{4}\end{array}\right]$

$(AB-C)^{-1} = \begin{pmatrix}\frac{1}{2} & 0 \\ -\frac{7}{8} & \frac{1}{4}\end{pmatrix}$

Calculamos $X = 2C\cdot (AB-C)^{-1}$ :

$X = \begin{pmatrix}2 & 4 \\ 6 & 6\end{pmatrix}\begin{pmatrix}\frac{1}{2} & 0 \\ -\frac{7}{8} & \frac{1}{4}\end{pmatrix}$

$X_{11} = 2\cdot \frac{1}{2} + 4\cdot (-\frac{7}{8}) = {\color{blue}-\frac{5}{2}}$

$X_{12} = 2\cdot 0 + 4\cdot \frac{1}{4} = {\color{blue}1}$

$X_{21} = 6\cdot \frac{1}{2} + 6\cdot (-\frac{7}{8}) = {\color{blue}-\frac{9}{4}}$

$X_{22} = 6\cdot 0 + 6\cdot \frac{1}{4} = {\color{blue}\frac{3}{2}}$

$X = \begin{pmatrix}-\frac{5}{2} & 1 \\ -\frac{9}{4} & \frac{3}{2}\end{pmatrix}$

Resultado:

$\boxed{X = \begin{pmatrix}-\frac{5}{2} & 1 \\ -\frac{9}{4} & \frac{3}{2}\end{pmatrix}}$

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