Ecuaciones Matriciales

ecuacion_matricial Ejercicios_Resueltos matrices

Dadas las matrices:

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

Resuelve las siguientes ecuaciones matriciales:

– $AX + B = C$
– $XA+B=C$

SOLUCIÓN

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

Tenemos que calcular la inversa de A , la resta C-B y por último hacer el producto
$|A| = \left| \begin{array}{cc} 1 & 1 \\ 3 & 4 \end{array} \right| = 4-3 = 1$

$A^ {-1} = \frac{1}{|A|} \cdot (Adj(A))^t$

$\alpha_{11} = 4$ ; $\alpha_{12} = 3$ ; $\alpha_{21} = 1$ ; $\alpha_{22} = 1$

$Adj(A) = \left( \begin{array}{cc} 4 & -3 \\ -1 & 1 \end{array} \right)$

$(Adj(A))^t = \left( \begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array} \right)$

$A^ {-1} = \frac{1}{1} \cdot \left( \begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array} \right) = \left( \begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array} \right)$

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

$X =A^{-1} \cdot (C-B) = \left( \begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array} \right) \cdot \left( \begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array} \right) = \left( \begin{array}{cc} -4 & 2 \\ 3 & -1 \end{array} \right)$

$XA+B=C$
$XA=C-B$
$XA \cdot A^{-1}=(C-B) \cdot A^{-1}$
$X=(C-B) \cdot A^{-1} = \left( \begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array} \right) \cdot \left( \begin{array}{cc} 4 & -1 \\ -3 & 1 \end{array} \right)=\left( \begin{array}{cc} -7 & 2 \\ -6 & 2 \end{array} \right)$

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