Matrices: inversa y ecuación matricial

, por dani

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– a) Para que una matriz tenga inversa, su determinante tiene que ser distinto de cero.

|C| = \left| \begin{array}{ccc} 2&0&-1\\1&1&-1\\1&3&2 \end{array}\right|=4-3+0+1+6-0=8 \neq 0 \longrightarrow \exists C^{-1}
Por lo tanto C tiene inversa.

– b) Calculamos la inversa usando la fórmula:

C^{-1}=\frac{1}{|C|} \cdot \left( Adj(C)\right)^t

\alpha_{11}=\left|\begin{array}{cc}1&-1\\3&2\end{array}\right|=5 \qquad \alpha_{12}=\left|\begin{array}{cc}1&-1\\1&2\end{array}\right|=3 \qquad \alpha_{13}=\left|\begin{array}{cc}1&1\\1&3\end{array}\right|=2

\alpha_{21}=\left|\begin{array}{cc}0&-1\\3&2\end{array}\right|=3 \qquad \alpha_{22}=\left|\begin{array}{cc}2&-1\\1&2\end{array}\right|=5 \qquad \alpha_{23}=\left|\begin{array}{cc}2&0\\1&3\end{array}\right|=2

\alpha_{31}=\left|\begin{array}{cc}0&-1\\1&-1\end{array}\right|=1 \qquad \alpha_{32}=\left|\begin{array}{cc}2&-1\\1&-1\end{array}\right|=-1 \qquad \alpha_{33}=\left|\begin{array}{cc}2&0\\1&1\end{array}\right|=2

Adj(C)=\left(\begin{array}{ccc}5&-3&2\\-3&5&-6\\1&1&2 \end{array}\right) \qquad \left(Adj(C)\right)^t=\left(\begin{array}{ccc}5&-3&1\\-3&5&1\\2&-6&2 \end{array}\right)

C^{-1}=\frac{1}{|C|} \cdot \left( Adj(C)\right)^t=\frac{1}{8} \cdot \left(\begin{array}{ccc}5&-3&1\\-3&5&1\\2&-6&2 \end{array}\right) =\left(\begin{array}{ccc}\frac{5}{8}&\frac{-3}{8}&\frac{1}{8}\\\frac{-3}{8}&\frac{5}{8}&\frac{1}{8}\\\frac{1}{4}&\frac{-3}{4}&\frac{1}{4} \end{array}\right)

– c) BA + 2X = C
2X = C-BA
X = \frac{1}{2} \cdot \left(C-BA\right)

BA=\left( \begin{array}{ccc} 2&-4&6\\-1&2&-3\\1&-2&3 \end{array}\right)

C-BA=\left( \begin{array}{ccc} 0&4&-7\\2&-1&2\\0&5&-1 \end{array}\right)

X=\left( \begin{array}{ccc} 0&2&\frac{-7}{2}\\1&\frac{-1}{2}&1\\0&\frac{5}{2}&\frac{-1}{2} \end{array}\right)

Dadas las matrices:

A=\left( \begin{array}{ccc} 1&-2&3 \end{array}\right) \qquad B=\left( \begin{array}{c} 2\\-1\\1 \end{array}\right) \qquad C=\left( \begin{array}{ccc} 2&0&-1\\1&1&-1\\1&3&2 \end{array}\right)

– a) Justifica si la matriz C tiene inversa
– b) Halla la inversa de C
– c) Resuelve la ecuación matricial BA + 2X = C