Ecuaciones Matriciales

, por dani

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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)

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