高専の数学解説

第15章 15.6

(1)

f:(3112)g:(2−33−7)f:\begin{pmatrix}3&1\\1&2\end{pmatrix} \qquad g:\begin{pmatrix}2&-3\\3&-7\end{pmatrix}

(2)

f∘g=(3112)(2−33−7)=(6+3−9−72+6−3−14)=(9−168−17)f\circ g=\begin{pmatrix}3&1\\1&2\end{pmatrix}\begin{pmatrix}2&-3\\3&-7\end{pmatrix}=\begin{pmatrix}6+3&-9-7\\2+6&-3-14\end{pmatrix}=\begin{pmatrix}9&-16\\8&-17\end{pmatrix}

g∘f=(2−33−7)(3112)=(6−32−69−73−14)=(3−42−11)g\circ f=\begin{pmatrix}2&-3\\3&-7\end{pmatrix}\begin{pmatrix}3&1\\1&2\end{pmatrix}=\begin{pmatrix}6-3&2-6\\9-7&3-14\end{pmatrix}=\begin{pmatrix}3&-4\\2&-11\end{pmatrix}

(3)

P′=(3112)(2−1)=(6−12−2)=(50)Q′=(2−33−7)(30)=(69)P'=\begin{pmatrix}3&1\\1&2\end{pmatrix}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}6-1\\2-2\end{pmatrix}=\begin{pmatrix}5\\0\end{pmatrix} \qquad Q'=\begin{pmatrix}2&-3\\3&-7\end{pmatrix}\begin{pmatrix}3\\0\end{pmatrix}=\begin{pmatrix}6\\9\end{pmatrix}

(4)

逆変換 f−1=16−1(2−1−13)=15(2−1−13)g−1=1−14+9(−73−32)=15(7−33−2)f^{-1}=\dfrac{1}{6-1}\begin{pmatrix}2&-1\\-1&3\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}2&-1\\-1&3\end{pmatrix} \qquad g^{-1}=\dfrac{1}{-14+9}\begin{pmatrix}-7&3\\-3&2\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}7&-3\\3&-2\end{pmatrix}

これから

15(2−1−13)(74)=15(14−4−7+12)=(21)15(7−33−2)(54)=15(35−1215−8)=15(237)\dfrac{1}{5}\begin{pmatrix}2&-1\\-1&3\end{pmatrix}\begin{pmatrix}7\\4\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}14-4\\-7+12\end{pmatrix}=\begin{pmatrix}2\\1\end{pmatrix} \qquad \dfrac{1}{5}\begin{pmatrix}7&-3\\3&-2\end{pmatrix}\begin{pmatrix}5\\4\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}35-12\\15-8\end{pmatrix}=\dfrac{1}{5}\begin{pmatrix}23\\7\end{pmatrix}

解説: r31bn1z