高専の数学解説

第16章 16.21

(1)

X=A−1BX = A^{-1}B

∣212301−243∣=24−2−8−9=5\begin{vmatrix} 2 & 1 & 2 \\ 3 & 0 & 1 \\ -2 & 4 & 3 \end{vmatrix} = 24 - 2 - 8 - 9 = 5

∣0143∣=−4∣31−23∣=9+2=11∣30−24∣=12\begin{vmatrix} 0 & 1 \\ 4 & 3 \end{vmatrix} = -4 \quad \begin{vmatrix} 3 & 1 \\ -2 & 3 \end{vmatrix} = 9 + 2 = 11 \quad \begin{vmatrix} 3 & 0 \\ -2 & 4 \end{vmatrix} = 12

∣1243∣=3−8=−5∣22−23∣=6+4=10∣21−24∣=8+2=10\begin{vmatrix} 1 & 2 \\ 4 & 3 \end{vmatrix} = 3 - 8 = -5 \quad \begin{vmatrix} 2 & 2 \\ -2 & 3 \end{vmatrix} = 6 + 4 = 10 \quad \begin{vmatrix} 2 & 1 \\ -2 & 4 \end{vmatrix} = 8 + 2 = 10

∣1201∣=1∣2231∣=2−6=−4∣2130∣=−3\begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} = 1 \quad \begin{vmatrix} 2 & 2 \\ 3 & 1 \end{vmatrix} = 2 - 6 = -4 \quad \begin{vmatrix} 2 & 1 \\ 3 & 0 \end{vmatrix} = -3

A−1=15(−451−1110412−10−3)A^{-1} = \frac{1}{5}\begin{pmatrix} -4 & 5 & 1 \\ -11 & 10 & 4 \\ 12 & -10 & -3 \end{pmatrix}

A−1B=15(−451−1110412−10−3)(101110001)=15(15−3−110−72−109)A^{-1}B = \frac{1}{5}\begin{pmatrix} -4 & 5 & 1 \\ -11 & 10 & 4 \\ 12 & -10 & -3 \end{pmatrix}\begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 1 & 5 & -3 \\ -1 & 10 & -7 \\ 2 & -10 & 9 \end{pmatrix}

(2)

Y=BA−1Y = BA^{-1}

BA−1=(101110001)15(−451−1110412−10−3)=15(8−5−2−1515512−10−3)BA^{-1} = \begin{pmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}\frac{1}{5}\begin{pmatrix} -4 & 5 & 1 \\ -11 & 10 & 4 \\ 12 & -10 & -3 \end{pmatrix} = \frac{1}{5}\begin{pmatrix} 8 & -5 & -2 \\ -15 & 15 & 5 \\ 12 & -10 & -3 \end{pmatrix}

解説: r31bn1z