画像に示された行列計算の問題を解きます。問題は全部で9問あります。それぞれの計算結果を求めます。代数学行列行列計算線形代数2025/5/251. 問題の内容画像に示された行列計算の問題を解きます。問題は全部で9問あります。それぞれの計算結果を求めます。2. 解き方の手順(9) (3−2)(41)(3-2)\left(\begin{array}{l}4 \\ 1\end{array}\right)(3−2)(41)=(1)(41)=(41)=(1)\left(\begin{array}{l}4 \\ 1\end{array}\right)=\left(\begin{array}{l}4 \\ 1\end{array}\right)=(1)(41)=(41)(10) (24)(−13)\left(\begin{array}{l}2 \\ 4\end{array}\right)(-1 \quad 3)(24)(−13)=(−26−412)=\left(\begin{array}{cc}-2 & 6 \\ -4 & 12\end{array}\right)=(−2−4612)(11) (24)(210)\left(\begin{array}{l}2 \\ 4\end{array}\right)\left(\begin{array}{lll}2 & 1 & 0\end{array}\right)(24)(210)=(420840)=\left(\begin{array}{ccc}4 & 2 & 0 \\ 8 & 4 & 0\end{array}\right)=(482400)(12) (2−13)(32−3)\left(\begin{array}{lll}2 & -1 & 3\end{array}\right)\left(\begin{array}{c}3 \\ 2 \\ -3\end{array}\right)(2−13)32−3=2(3)+(−1)(2)+3(−3)=6−2−9=−5=2(3)+(-1)(2)+3(-3)=6-2-9=-5=2(3)+(−1)(2)+3(−3)=6−2−9=−5(13) (30−1)(123)\left(\begin{array}{c}3 \\ 0 \\ -1\end{array}\right)\left(\begin{array}{lll}1 & 2 & 3\end{array}\right)30−1(123)=(369000−1−2−3)=\left(\begin{array}{ccc}3 & 6 & 9 \\ 0 & 0 & 0 \\ -1 & -2 & -3\end{array}\right)=30−160−290−3(14) (30−1)(1−3)\left(\begin{array}{c}3 \\ 0 \\ -1\end{array}\right)\left(\begin{array}{ll}1 & -3\end{array}\right)30−1(1−3)=(3−900−13)=\left(\begin{array}{cc}3 & -9 \\ 0 & 0 \\ -1 & 3\end{array}\right)=30−1−903(15) (2−11301)(30−1)\left(\begin{array}{ccc}2 & -1 & 1 \\ 3 & 0 & 1\end{array}\right)\left(\begin{array}{c}3 \\ 0 \\ -1\end{array}\right)(23−1011)30−1=(2(3)−1(0)+1(−1)3(3)+0(0)+1(−1))=(6−0−19+0−1)=(58)=\left(\begin{array}{l}2(3)-1(0)+1(-1) \\ 3(3)+0(0)+1(-1)\end{array}\right)=\left(\begin{array}{l}6-0-1 \\ 9+0-1\end{array}\right)=\left(\begin{array}{l}5 \\ 8\end{array}\right)=(2(3)−1(0)+1(−1)3(3)+0(0)+1(−1))=(6−0−19+0−1)=(58)(16) (2−11301)(1002−1−1)\left(\begin{array}{ccc}2 & -1 & 1 \\ 3 & 0 & 1\end{array}\right)\left(\begin{array}{cc}1 & 0 \\ 0 & 2 \\ -1 & -1\end{array}\right)(23−1011)10−102−1=(2(1)−1(0)+1(−1)2(0)−1(2)+1(−1)3(1)+0(0)+1(−1)3(0)+0(2)+1(−1))=(2−0−10−2−13+0−10+0−1)=(1−32−1)=\left(\begin{array}{ll}2(1)-1(0)+1(-1) & 2(0)-1(2)+1(-1) \\ 3(1)+0(0)+1(-1) & 3(0)+0(2)+1(-1)\end{array}\right)=\left(\begin{array}{cc}2-0-1 & 0-2-1 \\ 3+0-1 & 0+0-1\end{array}\right)=\left(\begin{array}{cc}1 & -3 \\ 2 & -1\end{array}\right)=(2(1)−1(0)+1(−1)3(1)+0(0)+1(−1)2(0)−1(2)+1(−1)3(0)+0(2)+1(−1))=(2−0−13+0−10−2−10+0−1)=(12−3−1)(17) (1002−11)(2−11301)\left(\begin{array}{cc}1 & 0 \\ 0 & 2 \\ -1 & 1\end{array}\right)\left(\begin{array}{lll}2 & -1 & 1 \\ 3 & 0 & 1\end{array}\right)10−1021(23−1011)=(1(2)+0(3)1(−1)+0(0)1(1)+0(1)0(2)+2(3)0(−1)+2(0)0(1)+2(1)−1(2)+1(3)−1(−1)+1(0)−1(1)+1(1))=(2−11602110)=\left(\begin{array}{ccc}1(2)+0(3) & 1(-1)+0(0) & 1(1)+0(1) \\ 0(2)+2(3) & 0(-1)+2(0) & 0(1)+2(1) \\ -1(2)+1(3) & -1(-1)+1(0) & -1(1)+1(1)\end{array}\right)=\left(\begin{array}{ccc}2 & -1 & 1 \\ 6 & 0 & 2 \\ 1 & 1 & 0\end{array}\right)=1(2)+0(3)0(2)+2(3)−1(2)+1(3)1(−1)+0(0)0(−1)+2(0)−1(−1)+1(0)1(1)+0(1)0(1)+2(1)−1(1)+1(1)=261−1011203. 最終的な答え(9) (41)\left(\begin{array}{l}4 \\ 1\end{array}\right)(41)(10) (−26−412)\left(\begin{array}{cc}-2 & 6 \\ -4 & 12\end{array}\right)(−2−4612)(11) (420840)\left(\begin{array}{lll}4 & 2 & 0 \\ 8 & 4 & 0\end{array}\right)(482400)(12) −5-5−5(13) (369000−1−2−3)\left(\begin{array}{ccc}3 & 6 & 9 \\ 0 & 0 & 0 \\ -1 & -2 & -3\end{array}\right)30−160−290−3(14) (3−900−13)\left(\begin{array}{cc}3 & -9 \\ 0 & 0 \\ -1 & 3\end{array}\right)30−1−903(15) (58)\left(\begin{array}{l}5 \\ 8\end{array}\right)(58)(16) (1−32−1)\left(\begin{array}{cc}1 & -3 \\ 2 & -1\end{array}\right)(12−3−1)(17) (2−11602110)\left(\begin{array}{ccc}2 & -1 & 1 \\ 6 & 0 & 2 \\ 1 & 1 & 0\end{array}\right)261−101120