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Ecuaciones Matriciales

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)