第14章 14.12

(1)

AB=(120110)(001101100)AB = \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} これは計算できない

(2)

BA=(001101100)(120110)=(101+1212)=(102212)BA = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 1+1 & 2 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 2 & 2 \\ 1 & 2 \end{pmatrix}

(3)

DC=(010)(130)=3DC = \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} = 3

(4)

CD+B=(130)(010)+(001101100)=(010030000)+(001101100)=(011131100)CD+B = \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 0 & 0 \end{pmatrix}

(5)

2A+B=2(120110)+(001101100)2A+B = 2\begin{pmatrix} 1 & 2 \\ 0 & 1 \\ 1 & 0 \end{pmatrix} + \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} これは計算できない

(6)

BC−4C=(001101100)(130)−4(130)=(011)−(4120)=(−4−111)BC-4C = \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} - 4\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 4 \\ 12 \\ 0 \end{pmatrix} = \begin{pmatrix} -4 \\ -11 \\ 1 \end{pmatrix}

解説: r31bn1z