Ejemplo resuelto: Operaciones con matrices
Desarrollo paso a paso (estilo Word/MathType): igualdades encadenadas, producto fila × columna y cálculo de la adjunta.
\(A=\begin{bmatrix}2 & 1 & -1\\ 1 & 0 & -3\\ 2 & 1 & -3\end{bmatrix}\)
\(B=\begin{bmatrix}1 & -2 & -2\\ 3 & 1 & -1\\ 0 & -1 & 0\end{bmatrix}\)
\(C=\begin{bmatrix}-1 & 0 & 0\\ 2 & 2 & 0\\ 3 & 3 & 0\end{bmatrix}\)
Cuando la igualdad cabe, se muestra en una sola línea; si no, se presenta en varias líneas alineadas.
a) Suma: \(A+B\)
\[A+B\;=\;\begin{bmatrix}2 & 1 & -1\\ 1 & 0 & -3\\ 2 & 1 & -3\end{bmatrix}+\begin{bmatrix}1 & -2 & -2\\ 3 & 1 & -1\\ 0 & -1 & 0\end{bmatrix}\;=\;\begin{bmatrix}3 & -1 & -3\\ 4 & 1 & -4\\ 2 & 0 & -3\end{bmatrix}\]
b) Producto por escalar: \(-\frac{2}{3}B\)
\[-\frac{2}{3}B\;=\;-\frac{2}{3}\begin{bmatrix}1 & -2 & -2\\ 3 & 1 & -1\\ 0 & -1 & 0\end{bmatrix}\;=\;\begin{bmatrix}- \frac{2}{3} & \frac{4}{3} & \frac{4}{3}\\ -2 & - \frac{2}{3} & \frac{2}{3}\\ 0 & \frac{2}{3} & 0\end{bmatrix}\]
c) Potencia: \(A^3\)
\[A^3\;=\;\begin{bmatrix}2 & 1 & -1\\ 1 & 0 & -3\\ 2 & 1 & -3\end{bmatrix}^{3}\;=\;\begin{bmatrix}3 & 1 & 0\\ 6 & 4 & -14\\ 5 & 3 & -8\end{bmatrix}\]
d) Diferencia: \(B-C\)
\[B-C\;=\;\begin{bmatrix}1 & -2 & -2\\ 3 & 1 & -1\\ 0 & -1 & 0\end{bmatrix}-\begin{bmatrix}-1 & 0 & 0\\ 2 & 2 & 0\\ 3 & 3 & 0\end{bmatrix}\;=\;\begin{bmatrix}2 & -2 & -2\\ 1 & -1 & -1\\ -3 & -4 & 0\end{bmatrix}\]
e) Combinación lineal: \(2A+3B\)
\[2A+3B\;=\;2\begin{bmatrix}2 & 1 & -1\\ 1 & 0 & -3\\ 2 & 1 & -3\end{bmatrix}+3\begin{bmatrix}1 & -2 & -2\\ 3 & 1 & -1\\ 0 & -1 & 0\end{bmatrix}\;=\;\begin{bmatrix}7 & -4 & -8\\ 11 & 3 & -9\\ 4 & -1 & -6\end{bmatrix}\]
f) Traspuesta: \(C^t\)
\[C^t\;=\;\begin{bmatrix}-1 & 0 & 0\\ 2 & 2 & 0\\ 3 & 3 & 0\end{bmatrix}^{t}\;=\;\begin{bmatrix}-1 & 2 & 3\\ 0 & 2 & 3\\ 0 & 0 & 0\end{bmatrix}\]
g) Producto: \(CB\)
\[\begin{aligned}\text{Producto fila}\times\text{columna: }(CB)_{ij}=\sum_{k=1}^3 c_{ik}b_{kj}.\\ (CB)_{11} &= (-1)\cdot(1) + (0)\cdot(3) + (0)\cdot(0) = -1\\ (CB)_{12} &= (-1)\cdot(-2) + (0)\cdot(1) + (0)\cdot(-1) = 2\\ (CB)_{13} &= (-1)\cdot(-2) + (0)\cdot(-1) + (0)\cdot(0) = 2\\ \;\\ (CB)_{21} &= (2)\cdot(1) + (2)\cdot(3) + (0)\cdot(0) = 8\\ (CB)_{22} &= (2)\cdot(-2) + (2)\cdot(1) + (0)\cdot(-1) = -2\\ (CB)_{23} &= (2)\cdot(-2) + (2)\cdot(-1) + (0)\cdot(0) = -6\\ \;\\ (CB)_{31} &= (3)\cdot(1) + (3)\cdot(3) + (0)\cdot(0) = 12\\ (CB)_{32} &= (3)\cdot(-2) + (3)\cdot(1) + (0)\cdot(-1) = -3\\ (CB)_{33} &= (3)\cdot(-2) + (3)\cdot(-1) + (0)\cdot(0) = -9\\ CB &= \begin{bmatrix}-1 & 2 & 2\\ 8 & -2 & -6\\ 12 & -3 & -9\end{bmatrix}\end{aligned}\]
h) Inversa: \(B^{-1}\)
\[\begin{aligned}\det(B) &= 5\\ \text{Ejemplo: }M_{11}=\det\!\begin{bmatrix}1&-1\\-1&0\end{bmatrix}=-1,\ C_{11}=(-1)^{1+1}M_{11}=-1\\ \operatorname{Cof}(B) &= \begin{bmatrix}-1 & 0 & -3\\ 2 & 0 & 1\\ 4 & -5 & 7\end{bmatrix}\\ \operatorname{adj}(B) &= \operatorname{Cof}(B)^t = \begin{bmatrix}-1 & 2 & 4\\ 0 & 0 & -5\\ -3 & 1 & 7\end{bmatrix}\\ B^{-1} &= \frac{1}{\det(B)}\operatorname{adj}(B)\\ &= \frac{1}{5}\begin{bmatrix}-1 & 2 & 4\\ 0 & 0 & -5\\ -3 & 1 & 7\end{bmatrix}\\ &= \begin{bmatrix}- \frac{1}{5} & \frac{2}{5} & \frac{4}{5}\\ 0 & 0 & -1\\ - \frac{3}{5} & \frac{1}{5} & \frac{7}{5}\end{bmatrix}\end{aligned}\]
i) Expresión: \(A(2B+B^{-1})-C^t\)
\[\begin{aligned}2B &= \begin{bmatrix}2 & -4 & -4\\ 6 & 2 & -2\\ 0 & -2 & 0\end{bmatrix}\\ B^{-1} &= \begin{bmatrix}- \frac{1}{5} & \frac{2}{5} & \frac{4}{5}\\ 0 & 0 & -1\\ - \frac{3}{5} & \frac{1}{5} & \frac{7}{5}\end{bmatrix}\\ 2B+B^{-1} &= \begin{bmatrix}2 & -4 & -4\\ 6 & 2 & -2\\ 0 & -2 & 0\end{bmatrix}+\begin{bmatrix}- \frac{1}{5} & \frac{2}{5} & \frac{4}{5}\\ 0 & 0 & -1\\ - \frac{3}{5} & \frac{1}{5} & \frac{7}{5}\end{bmatrix} = \begin{bmatrix}\frac{9}{5} & - \frac{18}{5} & - \frac{16}{5}\\ 6 & 2 & -3\\ - \frac{3}{5} & - \frac{9}{5} & \frac{7}{5}\end{bmatrix}\\ A(2B+B^{-1}) &= \begin{bmatrix}2 & 1 & -1\\ 1 & 0 & -3\\ 2 & 1 & -3\end{bmatrix}\,\begin{bmatrix}\frac{9}{5} & - \frac{18}{5} & - \frac{16}{5}\\ 6 & 2 & -3\\ - \frac{3}{5} & - \frac{9}{5} & \frac{7}{5}\end{bmatrix} = \begin{bmatrix}\frac{51}{5} & - \frac{17}{5} & - \frac{54}{5}\\ \frac{18}{5} & \frac{9}{5} & - \frac{37}{5}\\ \frac{57}{5} & \frac{1}{5} & - \frac{68}{5}\end{bmatrix}\\ C^t &= \begin{bmatrix}-1 & 2 & 3\\ 0 & 2 & 3\\ 0 & 0 & 0\end{bmatrix}\\ A(2B+B^{-1})-C^t &= \begin{bmatrix}\frac{51}{5} & - \frac{17}{5} & - \frac{54}{5}\\ \frac{18}{5} & \frac{9}{5} & - \frac{37}{5}\\ \frac{57}{5} & \frac{1}{5} & - \frac{68}{5}\end{bmatrix}-\begin{bmatrix}-1 & 2 & 3\\ 0 & 2 & 3\\ 0 & 0 & 0\end{bmatrix} = \begin{bmatrix}\frac{56}{5} & - \frac{27}{5} & - \frac{69}{5}\\ \frac{18}{5} & - \frac{1}{5} & - \frac{52}{5}\\ \frac{57}{5} & \frac{1}{5} & - \frac{68}{5}\end{bmatrix}\end{aligned}\]