Problema 23

Considera las matrices A=\begin{pmatrix}1&2\\2&1\\0&1\end{pmatrix}, B=\begin{pmatrix}3&1&1\\2&-1&1\end{pmatrix} y C=\begin{pmatrix}1&1&0\\-1&2&1\\1&-1&1\end{pmatrix}.

Determina, si existe, la matriz X que verifique que ABX-2C=CX.


Solución:

Primero despejaremos la matriz X

\begin{aligned}ABX-2C&=CX\\ABX&=CX+2C\\ABX-CX&=2C\\(AB-C)X&=2C\\X&=(AB-C)^{-1}2C\end{aligned}

Ahora, poco a poco, calculamos lo que se necesita para obtener X.

AB=\begin{pmatrix}1&2\\2&1\\0&1\end{pmatrix}\cdot \begin{pmatrix}3&1&1\\2&-1&1\end{pmatrix}=\begin{pmatrix}7&-1&3\\8&1&3\\2&-1&1\end{pmatrix}

AB-C=\begin{pmatrix}7&-1&3\\8&1&3\\2&-1&1\end{pmatrix}-\begin{pmatrix}1&1&0\\-1&2&1\\1&-1&1\end{pmatrix}=\begin{pmatrix}6&-2&3\\9&-1&2\\1&0&0\end{pmatrix}

Calculamos la matriz inversa de ABC

|AB-C|=\begin{vmatrix}6&-2&3\\9&-1&2\\1&0&0\end{vmatrix}=-4+3=-1

\mbox{Adj }(AB-C)=\begin{pmatrix}\begin{vmatrix}-1&2\\0&0\end{vmatrix}&-\begin{vmatrix}9&2\\1&0\end{vmatrix}&\begin{vmatrix}9&-1\\1&0\end{vmatrix}\\-\begin{vmatrix}-2&3\\0&0\end{vmatrix}&\begin{vmatrix}6&3\\1&0\end{vmatrix}&-\begin{vmatrix}6&-2\\1&0\end{vmatrix}\\\begin{vmatrix}-2&3\\-1&2\end{vmatrix}&-\begin{vmatrix}6&3\\9&2\end{vmatrix}&\begin{vmatrix}6&-2\\9&-1\end{vmatrix}\end{pmatrix}=\begin{pmatrix}0&2&1\\0&-3&-2\\-1&15&12\end{pmatrix}

(\mbox{Adj }(AB-C))^t=\begin{pmatrix}0&0&-1\\2&-3&15\\1&-2&12\end{pmatrix}

\displaystyle (AB-C)^{-1}=\frac 1{-1}\begin{pmatrix}0&0&-1\\2&-3&15\\1&-2&12\end{pmatrix}=\begin{pmatrix}0&0&1\\-2&3&-15\\-1&2&-12\end{pmatrix}

Por último calculamos X

X=(AB-C)^{-1}2C=\begin{pmatrix}0&0&1\\-2&3&-15\\-1&2&-12\end{pmatrix}\cdot \begin{pmatrix}2&2&0\\-2&4&2\\2&-2&2\end{pmatrix}=\begin{pmatrix}2&-2&2\\-40&38&-24\\-30&30&-20\end{pmatrix}

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